S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 1 Strength of Materials
SKP Engineering College
Tiruvannamalai – 606611
A Course Material
on
Strength Of Materials
By
S.Sankar
Assistant Professor
Civil Engineering Department
Quality Certificate
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 2 Strength of Materials
This is to Certify that the Electronic Study Material
Subject Code:CE6402
Subject Name: Strength of Materials
Year/Sem: II / IV
Being prepared by me and it meets the knowledge requirement of the University
curriculum.
Signature of the Author
Name: S.Sankar
Designation: Assistant Professor
This is to certify that the course material being prepared by Mr.S.Sankari is of the
adequate quality. He has referred more than five books and one among them is from
abroad author.
Signature of HD Signature of the Principal
Name: Mr. A.Saravanan Name: Dr.V.Subramania Bharathi
Seal: Seal:
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 3 Strength of Materials
CE6402 STRENGTH OF MATERIALS L T P C 3 1 0 4
OBJECTIVES: The student should be made to:
To know the method of finding slope and deflection of beams and trusses using energy theorems and to know the concept of analysing indeterminate beam
To estimate the load carrying capacity of columns, stresses due to unsymmetrical bending and various theories for failure of material.
UNIT I ENERGY PRINCIPLES 9
Strain energy and strain energy density –strain energy due to axial load, shear, flexure and torsion –
Castigliano‟stheorems–Maxwell‟s reciprocal - Principleof virtual theorems work – application
application of energy theorems for computing deflections in beams and trusses Williot Mohr's Diagram.
UNIT II INDETERMINATE BEAMS 9
Concept of Analysis - Propped cantilever and fixed beams-fixed end moments and reactions – Theorem of three moments –analysis of continuous beams –shear force and bending moment diagrams.
UNIT III COLUMNS AND CYLINDER 9
Euler‟s theory–criticalof loadslongfor prismaticcolumns with different end conditions;
Rankine-Gordon formula for eccentrically loaded columns –Eccentrically loaded short
columns – middle third rule –core section –Thick cylinders –Compound cylinders
UNIT IV STATE OF STRESS IN THREE DIMENSIONS 9
Determination of principal stresses and principal planes –Volumetric strain –Theories of failure – Principal stress - Principal strain –shear stress –Strain energy and distortion energy theories – application in analysis of stress, load carrying capacity.
UNIT V ADVANCED TOPICS IN BENDING OF BEAMS 9
Unsymmetrical bending of beams of symmetrical and unsymmetrical sections –Shear Centre - curved beams –Winkler Bach formula.
TOTAL (L:45+T:15): 60 PERIODS
TEXT BOOKS:
1. Rajput R.K. "Strength of Materials (Mechanics of Solids)", S.Chand & company Ltd., New Delhi, 2010.
2. Egor PPopov, “Engineering 2nd
edition, Mechanic sPHI Learning Pvt.Ltd of.,New Solid Delhi, 2012
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 4 Strength of Materials
CONTENTS
S.No Particulars Page
1 Unit – I 5
2 Unit – II 25
3 Unit – III 52
4 Unit – IV 91
5 Unit – V 119
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 5 Strength of Materials
Unit – I
Energy Principles
Part A
1. Define strain energy and Proof stress. Strain energy
Whenever a body is strained, the energy is absorbed in the body. The energy which is absorbed in the body due to straining effect is known as strain energy. The strain energy
stored in the body is equal to the work done by the applied load in stretching the body Proof stress
The stress induced in an elastic body when it possesses maximum strain energy is termed
as its proof stress.
2. Define: Resilience
The resilience is defined as the capacity of a strained body for doing work on the removal of the straining force. The total strain energy stored in a body is commonly known as
resilience 3. Define Resilience, Proof Resilience and Modulus of Resilience.
Resilience
The resilience is defined as the capacity of a strained body for doing work on the removal
of the straining force. The total strain energy stored in a body is commonly known as resilience. Proof Resilience
The proof resilience is defined as the quantity of strain energy stored in a body when strained up to elastic limit. The maximum strain energy stored in a body is known as proof
resilience. Modulus of Resilience
It is defined as the proof resilience of a material per unit volume.
Proof resilience
Modulus of resilience = -------------------
Volume of the body
4. State the two methods for analyzing the statically indeterminate structures.
Displacement method (equilibrium method (or) stiffness coefficient method Force method (compatibility method (or) flexibility coefficient method)
5. Define Castigliano’s first theorem second Theorem.
First Theorem.
It states that the deflection caused by any external force is equal to the partial derivative of the strain energy with respect to that force. Second Theorem
It states that “If U is the total strain energy stored up in a frame work in equilibrium under an
external force; its magnitude is always a minimum.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 6 Strength of Materials
6. State the Principle of Virtual work.
It states that the workdone on a structure by external loads is equal to the internal energy
stored in a structure (Ue = Ui)
Work of external loads = work of internal loads
7. What is the strain energy stored in a rod of length l and axial rigidity AE to an
axial force P?
8. Strain energy stored
P2 L U= -------- 2AE
9. State the various methods for computing the joint deflection of a perfect frame.
The Unit Load method
Deflection by Castigliano’s First Theorem
Graphical method : Willot – Mohr Diagram
10. State the deflection of the joint due to linear deformation.
n δv = Σ U x ∆
1 n δH = Σ U’ x ∆
1 PL
∆ = --------- Ae
U= vertical deflection U’= horizontal deflection
11. State the deflection of joint due to temperature variation.
n
δ = Σ U X A 1
= U1∆1 + U2 ∆2 + …………+ Un ∆n
If the change in length (∆) of certain member is zero, the product U.∆ for those members will be substituted as zero in the above equation.
12. State the deflection of a joint due to lack of fit.
n
δ = Σ U ∆ 1
= U1∆1 + U2 ∆2 + …………+ Un ∆n
If there is only one member having lack of fit ∆1, the deflection of a particular joint will be equal to U1∆1.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 7 Strength of Materials
13. What is the effect of change in temperature in a particular member of a redundant frame?
When any member of the redundant frame is subjected to a change in temperature, it
will cause a change in length of that particular member, which in turn will cause lack of fit stresses in all other members of the redundant frame.
14. State the difference between unit load and strain energy method in the
determination of structures.
In strain energy method, an imaginary load P is applied at the point where the deflection is desired to be determined. P is equated to zero in the final step and the
deflection is obtained. In the Unit Load method, a unit load (instead of P) is applied at the point where the deflection is desired.
15. State the assumptions made in the Unit Load method.
1. The external and internal forces are in equilibrium 2. Supports are rigid and no movement is possible
3. The material is strained well within the elastic limit.
16. State the comparison of Castigliano’s first theorem and unit load method .
The deflection by the unit load method is given by
n PUL δ = Σ ------- 1 AE
n PL
δ = Σ ------- x U 1 AE
n = Σ ∆ x U ----- (i)
1 The deflection by castigliano’s theorem is given by
n
W
P
AE
PL
1
--------- (ii)
By comparing (i) & (ii)
UW
P
17. State Maxwell’s Reciprocal Theorem.
The Maxwell’s Reciprocal theorem states as “ The work done by the first
system of loads due to displacements caused by a second system of loads equals the
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 8 Strength of Materials
work done by the second system of loads due to displacements caused by the first system of loads.
18. Define degree of redundancy.
A frame is said to be statically indeterminate when the no of unknown reactions or stress components exceed the total number of condition equations of equilibrium.
20. Define Perfect Frame.
If the number of unknowns is equal to the number of conditions equations available, the frame is said to be a perfect frame.
21. State the two types of strain energies.
a. strain energy of distortion (shear strain energy)
b. strain energy of uniform compression (or) tension (volumetric strain energy)
22. State in which cases, Castigliano’s theorem can be used.
1. To determine the displacements of complicated structures. 2. To find the deflection of beams due to shearing (or) bending forces (or) bending moments are unknown.
3. To find the deflections of curved beams springs etc.
23. Define Proof stress.
The stress induced in an elastic body when it possesses maximum strain energy is termed as its proof stress.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 9 Strength of Materials
Part B
1. Derive the expression for strain energy in Linear Elastic Systems for the following cases. (i) Axial loading (ii) Flexural Loading (moment (or) couple)
(i)Axial Loading
Let us consider a straight bar of Length L, having uniform cross- sectional area
A. If an axial load P is applied gradually, and if the bar undergoes a deformation ∆, the work done, stored as strain energy (U) in the body, will be equal to
average force (1/2 P) multiplied by the deformation ∆. Thus U = ½ P. ∆ But ∆ = PL / AE
U = ½ P. PL/AE = P2 L / 2AE ---------- (i)
If, however the bar has variable area of cross section, consider a small of length dx and area of cross section Ax. The strain energy dU stored in this small element of length dx will be, from equation (i)
P2 dx dU = ---------
2Ax E The total strain energy U can be obtained by integrating the above expression
over the length of the bar.
U = EA
dxP
x
L
2
2
0
(ii) Flexural Loading (Moment or couple )
Let us now consider a member of length L subjected to uniform bending moment M. Consider an element of length dx and let d i be the change in the slope of
the element due to applied moment M. If M is applied gradually, the strain energy stored in the small element will be dU = ½ Mdi
But di d
------ = ----- (dy/dx) = d2y/d2x = M/EI dx dx M
di = ------- dx EI
Hence dU = ½ M (M/EI) dx
= (M2/2EI) dx
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 10 Strength of Materials
Integrating
U = L
EI
dxM
0
2
2
2. State and prove the expression for castigliano’s first theorem.
Castigliano’s first theorem:
It states that the deflection caused by any external force is equal to the partial
derivative of the strain energy with respect to that force. A generalized statement of the theorem is as follows:
“ If there is any elastic system in equilibrium under the action of a set of a forces W1 , W2, W3 ………….Wn and corresponding displacements δ1 , δ2,
δ3…………. δn and a set of moments M1 , M2, M3………Mn and corresponding rotations Φ1 , Φ2, Φ3,…….. Φn , then the partial derivative of the total strain energy U with respect to any one of the forces or moments taken individually
would yield its corresponding displacements in its direction of actions.”
Expressed mathematically,
1
1
W
U ------------- (i)
1
1
M
U ------------- (ii)
Proof:
Consider an elastic body as show in fig subjected to loads W1, W2, W3
………etc. each applied independently. Let the body be supported at A, B etc.
The reactions RA ,RB etc do not work while the body deforms because the hinge reaction is fixed and cannot move (and therefore the work done is zero) and the roller reaction is perpendicular to the displacements of the roller. Assuming that
the material follows the Hooke’s law, the displacements of the points of loading will be linear functions of the loads and the principles of superposition will hold.
Let δ1, δ2, δ3……… etc be the deflections of points 1, 2, 3, etc in the direction of the loads at these points. The total strain energy U is then given by
U = ½ (W1δ1 + W2 δ2 + ……….) --------- (iii)
Let the load W1 be increased by an amount dW1, after the loads have been applied. Due to this, there will be small changes in the deformation of the body, and the strain energy will be increased slightly by an amount dU. expressing this
small increase as the rate of change of U with respect to W1 times dW1, the new strain energy will be
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 11 Strength of Materials
U + 1
1
xdWW
U
--------- (iv)
On the assumption that the principle of superposition applies, the final strain
energy does not depend upon the order in which the forces are applied. Hence assuming that dW1 is acting on the body, prior to the application of W1 , W2, W3
………etc, the deflections will be infinitely small and the corresponding strain
energy of the second order can be neglected. Now when W1 , W2, W3 ………etc, are applied (with dW1 still acting initially), the points 1, 2, 3 etc will move through
δ1, δ2, δ3……… etc. in the direction of these forces and the strain energy will be given as above. Due to the application of W1, rides through a distance δ1 and produces the external work increment dU = dW1 . δ1. Hence the strain energy,
when the loads are applied is
U+dW1.δ1 ----------- (v) Since the final strain energy is by equating (iv) & (v).
U+dW1.δ1= U + 1
1
xdWW
U
δ1=1W
U
Which proves the proportion. Similarly it can be proved that Φ1=1M
U
.
Deflection of beams by castigliano’s first theorem:
If a member carries an axial force the energies stored is given by
U = EA
dxP
x
L
2
2
0
In the above expression, P is the axial force in the member and is the function of
external load W1, W2,W3 etc. To compute the deflection δ1 in the direction of W1
δ1=1W
U
= dx
W
p
AE
PL
10
If the strain energy is due to bending and not due to axial load
U = EI
dxML
2
2
0
δ1=1W
U
=
EI
dx
W
MM
L
10
If no load is acting at the point where deflection is desired, fictitious load W is
applied at the point in the direction where the deflection is required. Then after differentiating but before integrating the fictitious load is set to zero. This method is sometimes known as the fictitious load method. If the rotation Φ1 is required in
the direction of M1.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 12 Strength of Materials
Φ1=1M
U
=
EI
dx
M
MM
L
10
3. Calculate the central deflection and the slope at ends of a simply supported
beam carrying a UDL w/ unit length over the whole span.
Solution: a) Central deflection:
Since no point load is acting at the center where the deflection is required, apply the fictitious load W, then the reaction at A and B will (WL/2 + W/2)↑ each.
δc=W
U
=
EI
dx
W
ML
0
Consider a section at a distance x from A. Bending moment at x,
M=222
2wxx
WwL
2
x
x
M
dxxwx
xWwL
EI
l
c2222
2 22
0
Putting W=0,
dxxwx
xwL
EI
l
c222
2 22
0
= 2
0
43
1612
2
l
wxwLx
EI
EI
wlc
4
384
5
b) Slope at ends
To obtain the slope at the end A, say apply a frictions moment A as shown in
fig. The reactions at A and B will be
l
mwl
2 and
l
mwl
2
Measuring x from b, we get
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 13 Strength of Materials
A =
l
MxEIm
u
0
1 Dx
M
Mx.
-------------------------------- 2
Where Mx is the moment at a point distant x from the origin (ie, B) is a function of M.
Mx =
l
mwl
2 x -
2
2Wx
inl
x
m
Mx
2
A = l
EI0
1
l
mwl
2 x -
2
2Wx X/2 Dx
Putting M=0
dxl
xWXx
wl
Eia
l
2
2
2
1
0
L
AL
wxwx
EI0
43
86
1
EI
wLA
24
3
4. State and prove the Castigliano’s second Theorem.
Castigliano’s second theorem:
It states that the strain energy of a linearly elastic system that is initially unstrained will have less strain energy stored in it when subjected to a total load
system than it would have if it were self-strained.
t
u
= 0
For example, if is small strain (or) displacement, within the elastic limit in the
direction of the redundant force T,
t
u
=
=0 when the redundant supports do not yield (or) when there is no initial lack of fit
in the redundant members.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 14 Strength of Materials
Proof:
Consider a redundant frame as shown in fig.in which Fc is a redundant
member of geometrical length L.Let the actual length of the member Fc be (L- ),
being the initial lack of fit.F2 C represents thus the actual length (L- ) of the
member. When it is fitted to the truss, the member will have to be pulled such that F2 and F coincide.
According to Hooke’s law
F2 F1 = Deformation = )()(
approxAE
TL
AE
lT
Where T is the force (tensile) induced in the member.
Hence FF1=FF2-F1 F2
=AE
TL ------------------------------------ ( i )
Let the member Fc be removed and consider a tensile force T applied at the corners F and C as shown in fig.
FF1 = relative deflection of F and C
= T
u
1 ------------------------------------------ ( ii )
According to castigliano’s first theorem where U1 is the strain energy of the whole
frame except that of the member Fc. Equating (i) and (ii) we get
T
u
1 = --
AE
TL
(or)
T
u
1 +
AE
TL= ----------------------- ( iii )
To strain energy stored in the member Fc due to a force T is
UFC = ½ T. AE
TL =
AE
LT
2
2
T
U FC
AE
TL
Substitute the value of AE
TL in (iii) we get
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 15 Strength of Materials
T
U
T
u FC' (or)
T
U
When U= U1 + U Fc.If there is no initial lack of fit, =0 and hence 0
T
U
Note:
i) Castigliano’s theorem of minimum strain energy is used for the for analysis of
statically indeterminate beam ands portal tranes,if the degree of redundancy is not more than two.
ii) If the degree of redundancy is more than two, the slope deflection method or the moment distribution method is more convenient. 5. A beam AB of span 3mis fixed at both the ends and carries a point load of 9KN at C distant 1m from A. The M.O.I. of the portion AC of the beam is 2I and
that of portion CB is I. calculate the fixed end moments and reactions.
Solution:
There are four unknowns Ma, Ra, Mb and Rb.Only two equations of static are
available (ie) 0v and 0M
This problem is of second degree indeterminacy.
First choose MA and MB as redundant.
δA=
dxR
M
EI
Mx
R
UA
x
A
AB 0 -----------(1)
θA= dxM
M
EI
M
M
U
A
xx
B
AA
AB
0 -------------(2)
1) For portion AC:
Taking A as the origin
Mx = -MA + RA x
1;
A
x
A
x
M
Mx
R
M
IIOM 2.. Limits of x: 0 to 1m
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 16 Strength of Materials
Hence
dxEI
dxR
M
EI
M
A
x
C
A
x
1
0
AA
2
x xR M-
232
1
3
1
2
1
2
132
AA
AA
MR
EI
RM
EI
And
dx
EIdx
R
M
EI
M
A
x
C
A
x
1
0
AA
2
1 xR M-
22
1
2
11
2
12
AA
AA
RM
EI
RM
EI
For portion CB, Taking A as the origin we have
xM = )1(9 XXRM AA
1;
A
x
A
x
M
Mx
R
M
M.O.I = I Limits of x : 1 to 3 m
Hence
dxEI
dxR
M
EI
M
A
x
B
C
x
3
1
AA x1)-9(x- xR M-
=
42
3
264
1AA RM
EI
And
dx
EIdx
M
M
EI
M
A
x
B
C
x
3
1
AA 1-1)-9(x- xR M-
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 17 Strength of Materials
= 18421
AA RMEI
Subs these values in (1) & (2) we get
0
A
AB
R
U
042
3
264
1
23
1
AA
AA RMEI
MR
EI
2.08 – MA = 9.88 __________ (3)
0
A
AB
M
U
01842
1
212
1
AA
AA RMEI
RM
EI
MA – 1.7RA = -7.2 -------------- (4)
Solving (3) & (4)
MA = 4.8 KN – M (assumed direction is correct) RA = 7.05 KN
To find MB, take moments at B, and apply the condition 0M there. Taking
clockwise moment as positive and anticlockwise moment as negative. Taking MB
clockwise, we have MB – MA =RA (3) – 9x2 = 0
MB – 4.8 + (7.05x 3) -18 = 0 MB = 1.65 KN – m (assumed direction is correct)
To find RB Apply 0V for the whole frame.
RB = 9 – RA = 9-7.05 = 1.95 KN
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 18 Strength of Materials
6. Determine the vertical deflection of the joint C of the frame shown in fig. due to temperature rise of 60º F in the upper chords only. The coefficient of expansion = 6.0 x 10-6 per 1º F and E = 2 x 10 6 kg /cm2.
Sol:
Increase in length of each member of the upper chord = L α t
= 400 x 6x 10-6 x 60 = 0.144 cm
The vertical deflection of C is given by
u
To find u, apply unit vertical load at C. Since the change in length (∆) occurs only in the three top chord members, stresses in these members only need be found
out.
Reaction at A = 4/12 = 1/3 Reaction at B = 8/12 = 2/3
Passing a section cutting members 1 and 4, and taking moments at D, we get
U1 = (1/3 x 4) 1/3 = 4/9 (comp) Similarly, passing a section cutting members 3 and 9 and taking moments at
C, we get
Also
332211
12
3
)(9
4
)(9
8
3
14
3
2
uuu
compuu
compxu
C
cm
x
C
C
256.0
)144.0(9
8
9
4
9
4
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 19 Strength of Materials
7. A simply supported beam of span 6m is subjected to a concentrated load of 45 KN at 2m from the left support. Calculate the deflection under the load
point. Take E = 200 x 106 KN/m2 and I = 14 x 10-6 m4.
Solution: Taking moments about B.
VA x 6 – 45 x 4=0 VA x 6 -180 = 0
VA = 30 KN VB = Total Load – VA = 15 KN
Virtual work equation:
EI
mMdxL
c 0
V
Apply unit vertical load at c instead of 45 KN
RA x 6-1 x 4 =0 RA = 2/3 KN
RB = Total load –RA = 1/3 KN Virtual Moment:
Consider section between AC
M1 = 2/3 X1 [limit 0 to 2]
Section between CB M2 = 2/3 X2-1 (X2-2 ) [limit 2 to 6 ]
Real Moment:
The internal moment due to given loading
M1= 30 x X1
M2 = 30 x X2 -45 (X2 -2)
6
2
222111
2
0
VEI
dxMm
EI
dxMmc
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 20 Strength of Materials
2
0
6
2
22222
2
1
2
0
6
2
2
222
1
11
90453023
220
1
2453023
230
3
2
dxxxxxxEI
dxEI
xxxx
dxEI
xx
2
0
222
6
2
2
1 901523
201
dxxx
xEI
2
0
222
2
2
6
2
2
1 18030305201
dxxxxxEI
6
2
2
3
2
3
2
3
0
1 1802
60
3
5
3
201
x
xxx
EI
=
216180263026
3
51
3
820 2233
EIEI
mmormxxxEI
EI
1.57)(0571.0101410200
160160
72096067.34633.531
66
The deflection under the load = 57.1 mm
8. Define and prove the Maxwell’s reciprocal theorem.
The Maxwell’s reciprocal theorem stated as “ The work done by the first
system loads due to displacements caused by a second system of loads equals the work done by the second system of loads due to displacements caused by the first
system of loads”. Maxwell’s theorem of reciprocal deflections has the following three versions:
1. The deflection at A due to unit force at B is equal to deflection at B due to
unit force at A. δAB = δBA
2. The slope at A due to unit couple at B is equal to the slope at B due to
unit couple A ΦAB = ΦBA
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 21 Strength of Materials
3. The slope at A due to unit load at B is equal to deflection at B due to unit couple.
'
' ABAB
Proof:
By unit load method,
EI
Mmdx
Where,
M= bending moment at any point x due to external load. m= bending moment at any point x due to unit load applied at the point
where deflection is required. Let mXA=bending moment at any point x due to unit load at A
Let mXB = bending moment at any point x due to unit load at B. When unit load (external load) is applied at A,
M=mXA To find deflection at B due to unit load at A, apply unit load at B.Then m= mXB
Hence,
dxEI
mm
EI
Mmdx XBXABA
. ____________ (i)
Similarly,
When unit load (external load) is applied at B, M=mXB
To find the deflection at A due to unit load at B, apply unit load at A.then m= mXA
dxEI
mmB
EI
Mmdx XAAB
. ____________ (ii)
Comparing (i) & (ii) we get
δAB = δBA
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 22 Strength of Materials
9. Fig shows a cantilever, 8m long, carrying a point loads 5 KN at the center and an udl of 2 KN/m for a length 4m from the end B. If EI is the flexural rigidity of the cantilever find the reaction at the prop. (NOV/DEC – 2004)
Solution:
To find Reaction at the prop, R (in KN)
Portion AC: ( origin at A )
EI
R
EI
R
EI
xR
EI
dxRxU
3
32
6
64
62
224
0
3224
0
1
Portion CB: ( origin at C ) Bending moment Mx = R (x+4) – 5x – 2x2/2
= R (x+4) – 5x –x2
EI
dxMU x
2
24
0
2
Total strain energy = U1 +U2
At the propped end 0
R
U
dxdR
dMx
EI
M
EI
R
R
U xx
4
03
64
= dxxxxxREIEI
R)4(54
1
3
64 22
4
0
dxxxxxxREIEI
R)4(454
1
3
64 224
0
dxxxxxxxREIEI
R)4()4(5168
1
3
64 2322
4
0
0
4
0
342
32
3
)3
4
4()2
3(5164
3
1
3
64
xxx
xxx
xR
EIEI
R
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 23 Strength of Materials
)
3
256
4
256()32
3
64(56464
3
64
3
64R
R
= 21.33 R + (149.33R – 266.67 – 149.33) = 21.33 R + (149.33 R – 416)
21.33 R +149.33 R – 416 =0 R = 2.347 KN
10. A simply supported beam of span L is carrying a concentrated load W at the centre and a uniformly distributed load of intensity of w per unit length. Show that Maxwell’s reciprocal theorem holds good at the centre of the beam.
Solution:
Let the load W is applied first and then the uniformly distributed load w.
Deflection due to load W at the centre of the beam is given by
EI
WlW
384
5 4
Hence work done by W due to w is given by:
EI
wlWxU BA
384
5 4
,
Deflection at a distance x from the left end due to W is given by
22 4348
xxlEI
WxW
Work done by w per unit length due to W,
dxxxlEI
WwxU
l
AB )43(48
2 22
2/
0
,
422
,222
3
24
lll
EI
WwU AB
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 24 Strength of Materials
168
3
24
44
,
ll
EI
WwU AB
EI
WwlU BA
4
,384
5
Hence proved.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 25 Strength of Materials
Unit– II
Indeterminate Beams
Part A
1. Define statically indeterminate beams.
If the numbers of reaction components are more than the conditions equations, the structure is
defined as statically indeterminate beams.
E = R – r
E = Degree of external redundancy
R = Total number of reaction components
r = Total number of condition equations available.
A continuous beam is a typical example of externally indeterminate structure.
2. State the degree of indeterminacy in propped cantilever.
For a general loading, the total reaction components (R) are equal to (3+2) =5, While the total
number of condition equations (r) are equal to 3. The beam is statically indeterminate, externally
to second degree. For vertical loading, the beam is statically determinate to single degree.
E = R – r
= 5 – 3 = 2
3. State the degree of indeterminacy in a fixed beam. For a general system of loading, a fixed beam is statically indeterminate to third degree. For
vertical loading, a fixed beam is statically indeterminate to second degree.
E = R – r
For general system of
loading: R = 3 + 3
and r = 3
E = 6-3 = 3
For vertical loading:
R = 2+2 and r = 2
E = 4 – 2 = 2
4. State the degree of indeterminacy in the given beam.
The beam is statically indeterminate to third degree of general system of loading.
R = 3+1+1+1 = 6
E = R-r
= 6-3 = 3
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 26 Strength of Materials
5. State the degree of indeterminacy in the given beam.
The beam is statically determinate. The total numbers of condition equations are equal to 3+2 =
5. Since, there is a link at B. The two additional condition equations are at link.
E = R-r
= 2+1+2-5
= 5-5
E = 0
6. State the methods available for analyzing statically indeterminate structures.
i. Compatibility method
ii. Equilibrium method
7. Explain the effect of settlement of supports in a continuous beam. (Nov/Dec 2003)
Due to the settlement of supports in a continuous beam, the bending stresses will alters
appreciably. The maximum bending moment in case of continuous beam is less when
compare to the simply supported beam.
8. What are the advantages of Continuous beams over Simply Supported beams?
(i)The maximum bending moment in case of a continuous beam is much less than in case
of a simply supported beam of same span carrying same loads.
(ii) In case of a continuous beam, the average B.M is lesser and hence lighter
materials of construction can be used it resist the bending moment.
9. Give the procedure for analyzing the continuous beams with fixed ends using
three moment equations?
The three moment equations, for the fixed end of the beam, can be modified by
imagining a span of length l 0 and moment of inertia, beyond the support the and applying
the theorem of three moments as usual.
10. Define Flexural Rigidity of Beams.
The product of young’s modulus (E) and moment of inertia (I) is called Flexural
Rigidity (EI) of Beams. The unit is N mm2.
11. What is a fixed beam? (AUC Apr/May 2011)
A beam whose both ends are fixed is known as a fixed beam. Fixed beam is also called as
built-in or encaster beam. Incase of fixed beam both its ends are rigidly fixed and the slope
and deflection at the fixed ends are zero.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 27 Strength of Materials
12. What are the advantages of fixed beams?
(i) For the same loading, the maximum deflection of a fixed beam is less than that
of a simply supported beam.
(ii) For the same loading, the fixed beam is subjected to lesser maximum
bending moment.
(iii) The slope at both ends of a fixed beam is zero.
(iv) The beam is more stable and stronger.
13. What are the disadvantages of a fixed beam?
(i) Large stresses are set up by temperature changes.
(ii) Special care has to be taken in aligning supports accurately at the same
leveL
(iii) Large stresses are set if a little sinking of one support takes place.
(iv) Frequent fluctuations in loading render the degree of fixity at the ends very uncertain.
14. . Define: Continuous beam.
A Continuous beam is one, which is supported on more than two supports. For usual
loading on the beam hogging ( - ive ) moments causing convexity upwards at the supports
and sagging ( + ve ) moments causing concavity upwards occur at mid span.
15. What is mean by prop? . (AUC Nov/Dec 2012)
When a beam or cantilever carries some load , maximum deflection occurs at the free end . the
deflection can be reduced by providing vertical support at these points or at any suitable points.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 28 Strength of Materials
L = 6m
Load at C, W C
Load at D, W C
= 120 kN
= 160 kN
Distance AC = 2m
Distance AD =4m
Part B 1. A fixed beam AB of length 6m carries point load of 160 kN and 120 kN at a distance of 2m
and 4m from the left end A. Find the fixed end moments and the reactions at the supports.
Draw B.M and S.F diagrams. (AUC Apr/May
2008)(AUC Nov/Dec2006)
Solution
:
Given:
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 29 Strength of Materials
First calculate the fixed end moments due to loads at C and D separately
and then add up the moments.
Fixed End Moments:
For the load at C, a=2m and b=4m 2
M A1
M A1
WC ab
L2
60x2x(4) 2
(6) 2
42.22
kNm
M WC a b
B1
L2
M 60x2
x(4)
1.11 kNm
B1 (6)
2
For the load at D, a = 4m and b = 2m 2
M A2 WD a
b
L2
M 20x2 x(4)
3.33 kNmA2
(6) 2
2
M B 2 WD a
b
L2
2M
60 x2 x(4) 06.66
kNmB 2
Total fixing moment at A,
(6) 2
MA = MA1 + MA2
= 142.22 + 53.33
MA = 195.55 kNm
Total fixing moment at B,
MB =MB1 + MB2
= 71.11 + 106.66
= 177.77 kN m
B.M diagram due to vertical loads: * *
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 30 Strength of Materials
Consider the beam AB as simply supported. Let RA
and RB
are the reactions at A
and B due to simply supported beam. Taking moments about A, we get *
RB x6
*
60x2
00
20x4
RB 6
33.33 kN
* * RA = Total load - RB =(160 +120) – 133.33 = 146.67 kN
B.M at A = 0 *B.M at C =
RA x 2 = 146.67 x 2 = 293.34 kN m
S.F
Diagram:
B.M at D = 133.33 x 2 = 266.66 kN m
B.M at B= 0 Let RA = Resultant reaction at A due to fixed end moments and vertical loads
RB = Resultant reaction at B
Equating the clockwise moments and anti-clockwise moments about A, RB x 6 + MA = 160 x 2 + 120 x 4 + MB
RB= 130.37 kN
RA = total load – RB =
149.63 kN S.F at A = RA = 149.63
kN S.F at C = 149.63- 160 = -10.37 kN
S.F at D = -10.37 – 120 = -130.37
kN S.F at B= 130.37 KN
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 31 Strength of Materials
2. A fixed beam AB of length 6m carries two point loads of 30 kN each at a
distance of 2m from the both ends. Determine the fixed end moments.
Sloution:
Given:
Length L = 6m
Point load at C = W1 = 30
kN Point load at D = W2=
30 Kn
Fixed end moments:
MA = Fixing moment due to load at C + Fixing moment due to load at D
W a b 2 W a b
21 1 1
L2
2
2 2 2
L2
20x2x4
62
0x4x2
62
0kN m
Since the
beam is symmetric
al, MA = MB
= 40 kNm
B.M Diagram:
To draw the B.M diagram due to vertical loads, consider the beam
AB as simply supported. The reactions at A and B is equal to 30kN.
B.M at A and B = 0
B.M at C =30 x 2 = 60 kNm
B.M at D = 30 x 2 = 60 kNm
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 32 Strength of Materials
3. A cantilever AB of span 6m is fixed at the end and proposal at the end B . it carries a
point load of 50KnN at mid span . level of the prop is the same as that of the fixed end .
(i) Determine The Reaction At The
Prop. (ii) Draw SFD AND BMD.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 33 Strength of Materials
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 34 Strength of Materials
4. Analysis the propped cantilever beam of the length 10m is subjected to point load of 10KN acting at a 6m from fixed and draw SFD and BMD. (AUC Nov/Dec 2010 )
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 35 Strength of Materials
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 36 Strength of Materials
5. A propped cantilever of span 6m having the prop at the is subjected to two concentrated
loads of 24 KN and 48KN at one third points respective from left end (fixed support )
draw SFD and BMD .
(AUC Nov/Dec 2010 )
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 37 Strength of Materials
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 38 Strength of Materials
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 39 Strength of Materials
6. A propped cantilever of span 6m is subjected to a UDL of 2KN/m over a length of fixed the end.
Determine the prop reaction and draw the SFD and BMD. (AUC May/June 2012)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 40 Strength of Materials
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 41 Strength of Materials
7. A fixed beam of a 6m span supports two point loads 300KN each at a two meters from each end
. find the fixing moments at the ends and draw the BMD and SFD . Find also the deflection. Take
I = 9 x 108 mm2 and E = 200KN /m2. (AUC Apr / M AY 2010)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 42 Strength of Materials
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 43 Strength of Materials
8. For The fixed beam shown in fig. Draw the BMD and SFD.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 44 Strength of Materials
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 45 Strength of Materials
9. A cantilever beam ABC of span 6m fixed at a and propped at c is loaded with an UDL of
10KN/m for the length of 4m from the fixed end . the prop reaction .the find the
maximum sagging and point of concentrations .
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 46 Strength of Materials
10. A continuous beam consists of three successive span of 6m and 12m and 4m and
carries load of 2KN/m , 1KN/m and 3KN/M respectively on the spans . Draw BMD and
SFD for the beam .
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 47 Strength of Materials
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 48 Strength of Materials
10. Find the support moments and reactions for the continuous beam shown in fig . Draw the BMD and SFD. ( AUC APR/May 2010)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 49 Strength of Materials
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 50 Strength of Materials
11. A continuous beam ABC, is loaded as shown in fig . Find the support moments
three moment equation. Draw SFD and BMD.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 51 Strength of Materials
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 52 Strength of Materials
Unit– III
Columns
Part A
1. Define columns
If the member of the structure is vertical and both of its ends are fixed rigidly while subjected to axial compressive load, the member is known as column.
Example: A vertical pillar between the roof and floor.
2. Define struts.
If the member of the structure is not vertical and one (or) both of its ends is Linged (or) pin
jointed, the bar is known as strut. Example: Connecting rods, piston rods etc,
3. Mention the stresses which are responsible for column failure.
i. Direct compressive stresses ii. Buckling stresses
iii. Combined of direct compressive and buckling stresses.
4. State the assumptions made in the Euler’s column theory.
i. The column is initially perfectly straight and the load is applied axially. ii. The cross-section of the column is uniform throughout its length.
iii. The column material is perfectly elastic, homogeneous and isotropic and obeys Hooke’s law.
iv. The self weight of column is negligible.
5. What are the important end conditions of columns?
i. Both the ends of the column are linged (or pinned)
ii. One end is fixed and the other end is free. iii. Both the ends of the column are fixed.
iv. One end is fixed and the other is pinned.
6. Write the expression for crippling load when the both ends of the column are hinged.
2
2
l
EIP
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 53 Strength of Materials
P = Crippling load
E = Young’s Modulus I = Moment of inertia l = Length of column
7. Write the expression for buckling load (or) Crippling load when both ends of the
column are fixed?
2
24
L
EIP
P = Crippling load E = Young’s Modulus
I = Moment of inertia l = Length of column
8. Write the expression for crippling load when column with one end fixed and other
end linged.
2
22
l
EIP
P = Crippling load
E = Young’s Modulus I = Moment of inertia
l = Length of column
9. Write the expression for buckling load for the column with one fixed and
other end free.
2
2
4l
EIP
P = Crippling load E = Young’s Modulus I = Moment of inertia
l = Length of column
10. Explain equivalent length (or) Effective length.
If l is actual length of a column, then its equivalent length (or) effective length L may be
obtained by multiplying it with some constant factor C, which depends on the end fixation of the column (ie) L = C x l.
11. Write the Equivalent length (L) of the column in which both ends hinged and write
the crippling load.
Crippling Load 2
2
L
EIP
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 54 Strength of Materials
Equivalent length (L) = Actual length (l)
P = Crippling load
E = Young’s Modulus
I = Moment of inertia L= Length of column
12. Write the relation between Equivalent length and actual length for all end conditions
of column.
Both ends linged L = l Constant = 1
Both ends fixed 2
lL Constant =
2
1
One end fixed and other end hinged 2
lL Constant =
2
1
One end fixed and other end free
lL 2 Constant = 2
13. Define core (or) Kernel of a section. (April/May 2003)
When a load acts in such a way on a region around the CG of the section So that in that region
stress everywhere is compressive and no tension is developed anywhere, then that area is called the core (or) Kernal of a section. The kernel of the section is the area within which the line of action of the eccentric load P must cut the cross-section if the stress is not to become
tensile.
14. Derive the expression for core of a rectangular section.(Nov/Dec 2003)
The limit of eccentricity of a rectangular section b x d on either side of XX axis (or) YY axis is d/6 to avoid tension at the base core of the rectangular section.
Core of the rectangular section = Area of the shaded portion
632
12
db
18
bd
15. Derive the expression for core of a solid circular section of diameter D.
The limit of eccentricity on either side of both XX (or) YY axis = D/8 to avoid tension of the base.
Core of the circular section = Area of the shaded portion
28/D
64
2D
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 55 Strength of Materials
16. A steel column is of length 8m and diameter 600 mm with both ends hinged.
Determine the crippling load by Euler’s formula. Take 5101.2 E N/mm2.
49441036.6600
6464mmdI
Since the column is hinged at the both ends,
Equivalent length L = l
2
2
L
EIPcr
2
952
8000
1036.6101.2
N81006.2
17. Define Slenderness ratio.
It is defined as the ratio of the effective length of the column (L) to the least radius of gyration
of its cross –section (K) (i.e) the ratio of K
L is known as slenderness ratio.
Slenderness ratio = K
L
18. State the Limitations of Euler’s formula.(April /May 2005)
a. Euler’s formula is applicable when the slenderness ratio is greater than or equal to 80
b. Euler’s formula is applicable only for long column c. Euler’s formula is thus unsuitable when the slenderness ratio is less than a certain
value.
19. Write the Rankine’s formula for columns.
2
1
K
L
AfP c
K = Least radius of gyration A
I
P = Crippling load A = Area of the column fc = Constant value depends upon the material.
= Rankine’s constant E
fc
2
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 56 Strength of Materials
20. Write the Rankine’s formula for eccentric column.
2
211
k
L
k
ey
AfP
c
c
K = Least radius of gyration A
I
P = Crippling load
A = Area of the column fc = Constant value depends upon the material.
= Rankine’s constant E
fc
2
21. Define thick cylinder.
If the ratio of thickness of the internal diameter of a cylindrical or spherical shell exceeds
1/20, it is termed as a thick shell. The hoop stress developed in a thick shell varies from a maximum value at the inner
circumference to a minimum value at the outer circumference. Thickness > 1/20
22. State the assumptions involved in Lame’s Theory
i. The material of the shell is Homogeneous and isotropic. ii. Plane section normal to the longitudinal axis of the cylinder remains plane after the
application of internal pressure.
iii. All the fibers of the material expand (or) contact independently without being constrained by there adjacent fibers.
23. What is the middle third rule? (Nov/Dec 2003)
In rectangular sections, the eccentricity ‘e’ must be less than or equal to b/6. Hence the
greatest eccentricity of the load is b/6 form the axis Y-Y and with respect to axis X –X1 the eccentricity does not exceed d/6. Hence the load may be applied with in the middle third of
the base (or) Middle d/3.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 57 Strength of Materials
16 MARKS QUESTIONS AND ANSWERS
1. Explain the failure of long column.
Solution:
A long column of uniform cross-sectional area A and of length l, subjected to an axial compressive load P, as shown in fig. A column is known as long column if the length of the column in comparison to its lateral dimensions is very large. Such columns do not fail y crushing
alone, but also by bending (also known buckling)
The load, at which the column just buckles, is known as buckling load and it is less than the crushing load is less than the crushing load for a long column.
Buckling load is also known as critical just (or) crippling load. The value of buckling load for long columns are long columns is low whereas for short columns the value of buckling load is
high. Let
l = length of the long column
p = Load (compressive) at which the column has jus buckled. A = Cross-sectional area of he column
e = Maximum bending of the column at the centre.
0 = Stress due to direct load A
P
b = Stress due to bending at the centre of the column
= Z
eP
Where Z = Section modulus about the axis of bending.
The extreme stresses on the mid-section are given by
Maximum stress = 0 + b
Minimum stress = 0 - b
The column will fail when maximum stress (i.e) 0 + b is more the crushing stress fc.
In case of long column, the direct compressive stresses are negligible as compared to buckling
stresses. Hence very long columns are subjected to buckling stresses.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 58 Strength of Materials
2. State the assumptions made in the Euler’s column Theory. And explain the sign
conventions considered in columns. (April/May2003)
The following are the assumptions made in the Euler’s column theory:
1. The column is initially perfectly straight and the load is applied axially 2. The cross-section of the column is uniform throughout its length. 3. The column material is perfectly elastic, homogeneous and isotropic and
obeys Hooke’s law. 4. The length of the column is very large as compared to its lateral dimensions
5. The direct stress is very small as compared to the bending stress 6. The column will fail by buckling alone. 7. The self-weight of column is negligible.
The following are the sign conventions considered in columns:
1. A moment which will tend to bend the column with its convexity towards its initial
centre line is taken as positive.
2. A moment which will tend to bend the column with its concavity towards its initial center line is taken as negative.
3. Derive the expression for crippling load when the both ends of the column are
hinged.
Solution:
Consider a column AB of length L hinged at both its ends A and B carries an axial crippling load at A.
Consider any section X-X at a distance of x from B.
Let the deflection at X-X is y.
The bending moment at X-X due to the load P, M = yP.
ykEI
Py
dx
yd 2
2
2
Where EI
pk 2
` 02
2
2
ykdx
yd
Solution of this differential equation is kxBkxAy sincos
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 59 Strength of Materials
EI
pxB
EI
pxAy sincos
By using Boundary conditions,
At B, x = 0, y = 0 A = 0
At A, x = l, y = 0
EI
plB sin0
0EI
pSinl
......3,2,,0 EI
pl
Now taking the lest significant value (i.e)
EI
pl ;
22
EI
pl
2
2
l
EIp
`The Euler’s crippling load for long column with both ends hinged.
2
2
l
EIp
4. Derive the expression for buckling load (or) crippling load when both ends of the column
are fixed.
Solution:
Consider a column AB of length l fixed at both the ends A and B and caries an axial
crippling load P at A due to which buckling occurs. Under the action of the load P the column will deflect as shown in fig.
Consider any section X-X at a distance x from B.Let the deflection at X-X is y.
Due to fixity at the ends, let the moment at A or B is M.
Total moment at XX = M – P.y Differential equation of the elastic curve is
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 60 Strength of Materials
PyMdx
ydEI
2
2
IE
M
EI
py
dx
yd
2
2
p
p
IE
M
EI
py
dx
yd
2
2
P
M
EI
P
EI
py
dx
yd
2
2
The general solution of the above differential equation is
P
MEIPxBEIPxAy /sin/cos (i)
Where A and B are the integration constant
At, N. x = 0 and y = 0
From (i)
p
MBA 010
p
MA
Differentiating the equation (i) with respect to x,
0./.
EI
PxCos
EI
PBEIPxSin
EI
PA
dx
dy
At the fixed end B, x = 0 and 0dx
dy
0EI
PB
Either B = 0 (or) 0EI
P
Since 0EI
P as p 0
B = 0
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 61 Strength of Materials
Subs p
MA and B = 0 in equation (i)
P
M
EI
Px
P
My
.cos
EI
Px
P
My ..cos1
Again at the fixed end A, x = l, y = 0
EIPlCosP
M/.10
........6,4,2,0/. EIPl
Now take the least significant value 2
2. EI
Pl
22 4.
EI
Pl
2
24
l
EIP
The crippling load for long column when both the ends of the column are fixed
5. Derive the expression for crippling load when column with one end fixed and other
end hinged. (April/May 2003)
Solution:
Consider a column AB of length l fixed at B and hinged at A. It carries an axial crippling load P at A for which the column just buckles.
As here the column AB is fixed at B, there will be some fixed end moment at B. Let it be M. To balance this fixing moment M, a horizontal push H will be exerted at A. Consider any section X-X at a distance x from the fixed end B. Let the deflection at xx is
y.
Bending moment at xx = H (l-x) - Py
2
24
L
EIP
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 62 Strength of Materials
Differential equation of the elastic curve is,
PyxlHdx
ydEI
2
2
EI
xly
EI
P
dx
yd
142
2
P
p
EI
xlHy
EI
P
dx
yd
2
2
EI
p
EI
xlHy
EI
P
dx
yd
2
2
The general solution of the above different equation is
P
xlH
EI
pxB
EI
pxAy
.sin.cos
Where A and B are the constants of integration. (i)
At B, x = 0, y = 0
From (i) P
HlA
P
H
EI
PB
p
EI
P
HB
Again at the end A, x = l, y=0. substitute these values of x, y, A and B in equation (i)
EIPlSinP
EI
P
HEIPlCos
P
Hl/./.0
EIPlCosP
HlEIPlSin
p
EI
P
H/./..
lEIPlEIPl ././.tan
The value of lEIP ./tan in radians has to be such that its tangent is equal to itself. The
only angle whose tangent is equal to itself, is about 4.49 radians.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 63 Strength of Materials
49.4./ lEIP
22 49.4lEI
P
22 2l
EI
P(approx)
2
22
l
EIP
The crippling load (or) buckling load for the column with one end fixed and one end hinged.
6. Derive the expression for buckling load for the column with one end fixed and other
end free. (April/May 2003)
Solution:
Consider a column AB of length l, fixed at B and free at A, carrying an axial rippling load
P at D de to which it just buckles. The deflected form of the column AB is shown in fig. Let the new position of A is A1.
Let a be the deflection at the free end. Consider any section X-X at a distance x from B.
Let the deflection at xx is y.
Bending moment due to critical load P at xx,
yaPdx
ydEIM
2
2
pyPadx
ydEI
2
2
EI
pq
EI
py
dx
yd
2
2
The solution of the above differential equation is,
aEI
PxB
EI
PxAy
.sin.cos Where A and B are constants of integration.
At B, x = 0, y = 0
2
22
l
EIP
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 64 Strength of Materials
From (i), A = 0
Differentiating the equation (I w.r. to x
EI
PxCos
EI
PB
EI
PxSin
EI
PA
dx
dy..
At the fixed end B, x = 0 and 0dx
dy
EI
PB0
0EI
PAs 0 p
Substitute A = -a and B = 0 in equation (i) we get,
aEI
Pxay
.cos
EI
Pxay ..cos1 (ii)
At the free end A, x = l, y = a, substitute these values in equation (ii)
EI
Paa ..1cos1
0..1cos
EI
P
2
5,
2
3,
21
EI
P
Now taking the least significant value,
2
1
EI
P
4
12
2
EI
P
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 65 Strength of Materials
2
2
4l
EIP
The crippling load for the columns with one end fixed and other end free.
7. A steel column is of length 8 m and diameter 600 mm with both ends hinged. Determine
the crippling load by Euler’s formula. Take E =2.1 x 105 N/mm2
Solution:
Given, Actual length of the column, l = 8m = 8000 mm
Diameter of the column d= 600 mm
E = 2.1 x 105 N/mm2
464
dI
460064
491036.6 mmI
Since the column is hinged at the both ends,
Equivalent length L =l
Euler’s crippling load,
2
2
L
EIPcr
2952
8000
1036.6101.22
= 2.06 x 108 N
8. A mild steel tube 4m long, 3cm internal diameter and 4mm thick is used as a strut
with both ends hinged. Find the collapsing load, what will be the crippling load if
i. Both ends are built in?
ii. One end is built –in and one end is free?
2
2
4l
EIP
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 66 Strength of Materials
Solution:
Given:
Actual length of the mild steel tube, l = 4m = 400 cm
Internal diameter of the tube, d = 3 cm Thickness of the tube, t = 4mm = 0.4cm.
External diameter of the tube, D = d + 2t
= 3+2(0.4) = 3.8 cm.
Assuming E for steel = 2 x 106 Kg/cm2
M.O.I of the column section,
44
64dDI
2438.3
64
I = 6.26 cm 4
i. Since the both ends of the tube are hinged, the effective length of the column when both ends are hinged.
L = l = 400 cm
Euler’s crippling load 2
2
L
EIPcr
2
62
400
26.6102
.30.772 KgPcr
The required collapsed load = 772.30 Kg.
ii. When both ends of the column are built –in , then effective length of the column,
cml
L 2002
400
2
Euler’s crippling load,
2
2
L
EIPcr
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 67 Strength of Materials
2
62
200
26.6102
Pcr = 3089.19 Kg.
iii. When one end of the column is built in and the other end is free, effective length of the column, L = 2l
= 2 x 400 = 800 cm
Euler’s crippling load,
2
2
L
EIPcr
2
62
800
26.6102
Pcr = 193.07 Kg.
9. A column having a T section with a flange 120 mm x 16 mm and web 150 mm x 16
mm is 3m long. Assuming the column to be hinged at both ends, find the crippling
load by using Euler’s formula. E = 2 x 106 Kg/cm2.
Solution:
Given:
Flange width = 120 mm = 12 cm
Flange thickness = 16 mm = 1.6 cm Length of the web = 150 mm = 15cm
Width of the web = 16mm = 1.6cm E = 2 106 Kg/cm2
Length of the column, l = 3m = 300 cm.
Since the column is hinged at both ends, effective length of the column.
L = l = 300 cm.
From the fig. Y-Y is the axis of symmetry. The C.G of the whole section lies on Y-Y axis.
Let the distance of the C.G from the 16 mm topmost fiber of the section = Y
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 68 Strength of Materials
6.1156.112
2
156.16.115
2
6.16.112
Y
cmY 41.5
Distance of C.G from bottom fibre = (15+1.6) - 5.41 = 11.19cm
Now M.O.I of the whole section about X-X axis.
2323
2
1519.11156.1
12
156.1
2
6.141.56.112
12
6.112XXI
492.1188 cmIXX
M.I of the whole section about Y-Y axis
433
52.23512
10615
12
126.1cmI yy
4
min 52.235 cmI
Euler’s Crippling load,
2
2
L
EIPcr
2
62
300
52.235102
; .32.51655 KgPcr
10. A steel bar of solid circular cross-section is 50 mm in diameter. The bar is pinned at
both ends and subjected to axial compression. If the limit of proportionality of the material
is 210 MPa and E = 200 GPa, determine the m minimum length to which Euler’s
formula is valid. Also determine the value of Euler’s buckling load if the column has this
minimum length.
Solution:
Given,
Dia of solid circular cross-section, d = 50 mm Stress at proportional limit, f = 210 Mpa
= 210 N/mm2
Young’s Modulus, E = 200 GPa = 200 x 10 3 N/mm2
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 69 Strength of Materials
Area of cross –section, 2249.196350
4mmA
Least moment of inertia of the column section,
4341079.6.350
64mmI
Least radius of gyration,
243
2 25.1565049.1963
1079.306mm
A
Ik
The bar is pinned at both ends,
Effective length, L = Actual length, l
Euler’s buckling load,
2
2
L
EIPcr
2
2
/ KL
E
A
Pcr
For Euler’s formula to be valid, value of its minimum effective length L may be found out by equating the buckling stress to f
210
2
2
K
L
E
210
222 kE
L
210
25.156102 522
L
L = 1211.89 mm = 1212 mm = 1.212 m
The required minimum actual length l =L = 1.212 m
For this value of minimum length,
Euler’s buckling load 2
2
L
EI
2352
1212
1075.306102
= 412254 N = 412.254 KN
Result:
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 70 Strength of Materials
Minimum actual length l = L = 1.212 m Euler’s buckling Load =412.254 KN
11. Explain Rankine’s Formula and Derive the Rankine’s formula for both short and
long column.
Solution:
Rankine’s Formula:
Euler’s formula gives correct results only for long columns, which fail mainly due to buckling. Whereas Rankine’s devised an empirical formula base don practical experiments for
determining the crippling or critical load which is applicable to all columns irrespective of whether they a short or long.
If P is the crippling load by Rankine’s formula.
Pc is the crushing load of the column material PE is the crippling load by Euler’s formula.
Then the Empirical formula devised by Rankine known as Rankine’s formula stand as:
Ee PPP
111
For a short column, if the effective length is small, the value of PE will be very high and
the value of EP
1 will be very small as compared to
CP
1and is negligible.
For the short column, (i.e) P = PC
Thus for the short column, value of crippling load by Rankine is more or less equal to the
value of crushing load:
For long column having higher effective length, the value of PE is small and EP
1will
be large enough in comparison to CP
1. So
CP
1 is ignored.
For the long column, CP
1
EP
1 (i.e) p PE
Thus for the long column the value of crippling load by Rankine is more or less equal to
the value of crippling load by Euler.
cPP
11
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 71 Strength of Materials
Ec PPP
111
Ec
cE
PP
PP
P
1
cE
Ec
PP
PPp
;
E
c
c
P
P
Pp
1
Substitute the value of Pc = fc A and 2
2
L
EIPE
in the above equation,
22 /1
LEI
Af
Afp
c
c
Where,
fc = Ultimate crushing stress of the column material.
A = Cross-sectional are of the column L = Effective length of the column
I = Ak2
Where k = Least radius of gyration.
22
2
22 1/
1EAk
LAf
Af
LEI
Af
Afp
c
c
c
c
2
1
K
L
Afp c
where = Rankine’s constant E
f c
2
P = 2
/1 kL
LoadCrushing
When Rankine’s constant is not given then find
E
f c
2
The following table shows the value of fc and for different materials.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 72 Strength of Materials
Material fc N/mm2 E
f c
2
Wrought iron 250 9000
1
Cast iron 550 1600
1
Mild steel 320 7500
1
Timber 50 750
1
12. A rolled steel joist ISMB 300 is to be used a column of 3 meters length with both ends
fixed. Find the safe axial load on the column. Take factor of safety 3, fc = 320 N/mm2
and7500
1 . Properties of the column section.
Area = 5626 mm2, IXX = 8.603 x 107 mm4
Iyy =4.539 x 107 mm4
Solution:
Given: Length of the column, l = 3m = 3000 mm
Factor of safety = 3
fc = 320 N/mm2, 7500
1
Area, A = 5626 mm2
IXX = 8.603 x 107 mm4
Iyy =4.539 x 107 mm4
The column is fixed at both the ends,
Effective length, mml
L 15002
3000
2
Since Iyy is less then Ixx, The column section,
47
min 10539.4 mmIII yy
Least radius of gyration of the column section,
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 73 Strength of Materials
mmA
IK 82.89
5626
10539.4 7
Crippling load as given by Rakine’s formula,
22
82.89
1500
7500
11
5626320
1
K
L
Afp c
cr
Pcr = 1343522.38 N
Allowing factor of safety 3,
Safe load = safetyofFactor
LoadCrippling
N79.4478403
38.1343522
Result:
i. Crippling Load (Pcr) = 1343522.38 N
ii. Safe load =447840.79N
13. A built up column consisting of rolled steel beam ISWB 300 with two plates 200 mm
x 10 mm connected at the top and bottom flanges. Calculate the safe load the column
carry, if the length is 3m and both ends are fixed. Take factor of safety 3 fc = 320
N/mm2 and 7500
1
Take properties of joist: A = 6133 mm2
IXX = 9821.6 x 104 mm4 ; Iyy = 990.1 x 104 mm4
Solution:
Given:
Length of the built up column, l = 3m = 3000 mm
Factor of safety = 3 fc =320 N/mm2
7500
1
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 74 Strength of Materials
Sectional area of the built up column,
2101331020026133 mmA
Moment of inertia of the built up column section abut xx axis,
23
4 1551020012
102002106.9821XXI
= 1.94 x 108 mm4 Moment of inertia of the built up column section abut YY axis,
12
200102101.990
34
YYI
= 0.23 x 108 mm4
Since Iyy is less than Ixx , The column will tend to buckle about Y-Y axis.
Least moment of inertia of the column section,
48
min 1023.0 mmIII YY
The column is fixed at both ends.
Effective length,
mml
L 15002
3000
2
Least radius of gyration o the column section,
mmA
JK 64.47
10133
1023.0 8
Crippling load as given by Rankine’s formula,
22
64.47
1500
7500
11
10133320
1
K
L
Afp c
cr
= 2864023.3 N
Safe load = N43.9546743
3.2864023
Result:
Crippling load
Factor of safety
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 75 Strength of Materials
i. Crippling load = 2864023.3 N
ii. Safe load = 954674.43 N
14. Derive Rankine’s and Euler formula for long columns under long columns under
Eccentric Loading?
i. Rankine’s formula:
Consider a short column subjected to an eccentric load P with an eccentricity e form the
axis. Maximum stress = Direct Stress + Bending stress
Z
M
A
Pf c
y
IZ
2
..
Ak
yep
A
P c 2AkI
A
Ik
where
A = Sectional are of the column
Z = Sectional modulus of the column yc = Distance of extreme fibre from N.A k = Least radius of gyration.
21
k
ey
A
Pf cc
Where
21
k
eyc
is the reduction factor for eccentricity of loading.
For long column, loaded with axial loading, the crippling load,
Eccentric load,
21
k
ey
AfP
c
c
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 76 Strength of Materials
2
1
K
L
AfP c
Where
2
1K
L is the reduction factor for buckling of long column.
Hence for a long column loaded with eccentric loading, the safe load,
ii. Euler’s formula
Maximum stress n the column = Direct stress + Bending stress
Z
lEIPeP
A
P 2/sec
Hence, the maximum stress induced in the column having both ends hinged and an
eccentricity of e is
2/sec
lEIP
Z
Pe
A
P
The maximum stress induced in the column with other end conditions are determined by changing the length in terms of effective length.
15. A column of circular section has 150 mm dia and 3m length. Both ends of the
column are fixed. The column carries a load of 100 KN at an eccentricity of 15 mm
from the geometrical axis of the column. Find the maximum compressive stress in
the column section. Find also the maximum permissible eccentricity to avoid tension
in the column section. E = 1 x 105 N/mm2
Solution:
Given,
Diameter of the column, D = 150 mm
Actual length of the column, l = 3m = 3000 mm Load on the column, P = 100 KN = 1000 x 103 N E = 1 x 105 N/mm2
Eccentricity, e = 15 mm
2
211
K
L
K
ey
AfP
c
c
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 77 Strength of Materials
Area of the column section 4
2DA
2150
4
= 17671 mm2 Moment of inertia of the column section N.A.,
44 1506464
DI
= 24.85 x 106 mm4 Section modulus,
2/D
I
y
IZ
= 3
6
331339
2
150
1085.24mm
Both the ends of the column 2 are fixed.
Effective length of the column, mml
L 15002
3000
2
Now, the angle
2
1500
1085.24101
10100
2/
65
3
LEIP
= 0.1504 rad = 8.61 o
Maximum compressive stress,
2/sec
LEIP
Z
eP
A
P
331339
61.8sec1510100
17671
10100 33 o
= 10.22 N/mm2 To avoid tension we know,
Z
M
A
P
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 78 Strength of Materials
Z
ep
A
P o61.8.sec
331339
61.8.sec10100
17671
10100 33 oe
e = 18.50 mm
Result: i. Maximum compressive stress = 10.22 N/mm2
ii. Maximum eccentricity = 18.50 mm 16. State the assumptions and derive Lame’s Theory?
1. The assumptions involved in Lame’s Theory.
i. The material of the shell is homogenous and isotropic ii. Plane sections normal to the longitudinal axis of the cylinder remain plane after
the application of internal pressure. iii. All the fibres of the material expand (or) contract independently without being
constrained by their adjacent fibres.
2 Derivation of Lame’s Theory
Consider a thick cylinder
Let
rc = Inner radius of the cylinder r0 = Outer radius of the cylinder
Pi = Internal radial pressure Po = External radial pressure
L = Length of the cylinder f2 = Longitudinal stress.
Lame’s Equation:
axx pf 2
axx
bP
2
aa
x
bf x 2
2
a
x
bf x
2
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 79 Strength of Materials
where
fx = hoop stress induced in the ring. px = Internal radial pressure in the fig. Px + dPx = External radial pressure in the ring.
The values of the two constants a and to b are found out using the following boundary
conditions: i. Since the internal radial pressure is Pi,
At x = ri, Px = Pi
ii. Since the external radial pressure is P0,
At x = r0, Px = P0
17. A thick steel cylinder having an internal diameter of 100 mm an external diameter of
200 mm is subjected to an internal pressure of 55 M pa and an external pressure of 7
Mpa. Find the maximum hoop stress.
Solution:
Given,
Inner radius of the cylinder, mmri 502
100
Outer radius of the cylinder, mmro 1002
200
Internal pressure, Pi = 55 Mpa
External pressure, P0 = 7 Mpa
In the hoop stress and radial stress in the cylinder at a distance of x from the centre is fx
and px respectively, using Lame’s equations,
ax
bf x
2 (i)
ax
bPx
2 (ii)
where a and b are constants,
Now by equation, at x = 50 mm, Px = 55 MPa (Boundary condition)
Using these boundary condition in equation (ii)
ax
bPx
2
ab
2
5055 (iii)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 80 Strength of Materials
Then x = 100 mm, px = 7 Mpa Using these boundary condition is equation (ii)
ab
2100
7 (iv)
Solving (iii) & (iv)
7100/2
ab
5550/2
ab
(- ) (+)
10000
3b = - 48
Substitute a & b in equation (i)
9160000
2
xf x
The value of fx is maximum when x is minimum
Thus fx is maximum for x = ri = 50 mm
Maximum hoop stress
950
1600002
= 73 Mpa (tensile) Result:
Maximum hoop stress = 73 MPa (tensile)
18. A cast iron pipe has 200 mm internal diameter and 50 mm metal thickness. It carries
water under a pressure of 5 N/mm2. Find the maximum and minimum intensities of
circumferential stress. Also sketch the distribution of circumferential stress and
radial stress across the section.
Solution:
Given:
Internal diameter, di = 200 mm
Wall thickness, t = 50 mm Internal pressure, Pi = 5 N/mm2
b = 160000
a = 9
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 81 Strength of Materials
External pressure, P0 = 0.
Internal radius mmdi
ri 1002
200
2
External radius mmtrr i 150501000
Let fx and Px be the circumferential stress and radial stress at a distance of x from the centre of the
pipe respectively.
Using Lame’s equations,
ax
bf x
2 (i)
ax
bpx
2 (ii)
where, a & b are arbitrary constants.
Now at x = 100 mm, Px = 5 N/mm2 At x = 150 mm, Px = 0
Using boundary condition is (ii)
ab
2
1005 (ii)
ab
2
1500 (iv)
By solving (iii) & (iv) a = 4 ; b = 90000
,490000
2
xf x ,4
900002
x
Px
Putting x = 100 mm, maxi circumferential stress.
tensilemmNf x
2
2/134
100
90000
Putting x = 150 mm, mini circumferential stress.
tensilemmNf x
2
2/84
150
90000
19. Explain the stresses in compound thick cylinders.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 82 Strength of Materials
Solution:
Consider a compound thick cylinder as shown in fig. Let, r1 = Inner radius of the compound cylinder
r2 = Radius at the junction of the two cylinders
r3 = Outer radius of the compound cylinder When one cylinder is shrunk over the other, thinner cylinder is under compression and the
outer cylinder is under tension. Due to fluid pressure inside the cylinder, hoop stress will develop. The resultant hoop stress in the compound stress is that algebraic sum of the hoop stress
due to initial shrinkage and that due to fluid pressure. a. Stresses due to initial shrinkage:
Applying Lame’s Equations for the outer cylinder,
121 a
x
bPx
12
1 ax
bf x
At x = r3, Px = 0 and at x = r2, px = p
Applying Lame’s Equations for the inner cylinder
22
2 ax
bPx
22
2 ax
bf x
At x = r2, Px = p and at x = r3, px = 0
b. Stresses due to Internal fluid pressure.
To find the stress in the compound cylinder due to internal fluid pressure alone, the inner and outer cylinders will be considered together as one thick shell. Now applying Lame’s
Equation,
A
x
BPx
2
A
x
Bf x
2
At x = r1, Px = pf ( Pf being the internal fluid pressure)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 83 Strength of Materials
At x = r3, px = 0 The resultant hoop stress is the algebraic sum of the hoop stress due to shrinking and due
internal fluid pressure.
20. A compound cylinder is composed of a tube of 250 mm internal diameter at 25 mm
wall thickness. It is shrunk on to a tube of 200 mm internal diameter. The radial
pressure at the junction is 8 N/mm2. Find the variation of hoop stress across the wall
of the compound cylinder, if it is under an internal fluid pressure of 60 N/mm2
Solution:
Given:
Internal diameter of the outer tube, d1 = 250 mm Wall thickness of the outer tuber , t = 25 mm Internal diameter of the inner tube , d2 = 200 mm
Radial pressure at the junction P = 8 N/mm2 Internal fluid pressure within the cylinder Pf = 60 N/mm2
External radius of the compound cylinder,
2
212
tdr
mm1502522502
1
Internal radius of the compound cylinder,
mmd
r 1002
200
2
21
Radius at the junction, mmd
r 1252
250
2
11
Let the radial stress and hoop stress at a distance of x from the centre of the cylinder be px
and fx respectively.
i. Hoop stresses due to shrinking of the outer and inner cylinders before fluid pressure is admitted.
a. Four outer cylinder:
Applying Lame’s Equation
12
1 ax
bPx (i)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 84 Strength of Materials
12
1 ax
bf x (ii)
Where a1 and b1 are arbitrary constants for the outer cylinder.
Now at x = 150 mm, Px = 0 X = 125 mm, Px = 8 N/mm2
12
1
150a
bo (iii)
12
1
1258 a
b (iv)
Solving equation (iii) & (iv) a1 = 18 ; b1 = 409091
18409091
2
xf x (v)
Putting x = 150 mm in the above equation stress at the outer surface,
2
2/3618
150
409091mmNf x (tensile)
Again putting x = 125 mm in equation (v), stress at junction,
2
2/4418
125
409091mmNf x (tensile)
b). For inner cylinder: Applying Lame’s Equation with usual Notations.
22
2 ax
bPx (iv)
22
2 ax
bf x (v)
Now at x = 125 mm, Px = 8 N/mm2
x =100 mm, Px = 0
22
2
1258 a
b (vi)
22
2
100a
bo (vii)
By solving (vi) & (vii) a2 = -22
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 85 Strength of Materials
b2 = -222222
2
2/2.4422
100
222222mmNf x
(comp)
2
2/2.3622
125
222222mmNf x
(comp)
iii. Hoop stresses due to internal fluid pressure alone for the compound cylinder:
In this case, the two tubes will be taken as a single thick cylinder. Applying Lame’s equations with usual notations.
Ax
BPx
2 (viii)
Ax
Bf x
2 (ix)
At x = 150 mm, Px = 0 x = 100 mm, Px = pf = 60 N/mm2
From Equation (viii)
AB
O 2
150 (x)
AB
2
10060 (xi)
By solving (x) & (xi)
A = 133, B = 3 x 106
1331032
6
x
f x
Putting x = 150 mm, hoop stress at the outer surface
2
2
6
/266133150
103mmNf x
(Tensile)
Again putting x = 125 mm, hoop stress at the junction
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 86 Strength of Materials
TensilemmNf x2
2
6
/325133125
103
Putting x = 100 mm, hoop stress at the inner surface
TensilemmNf x2
2
6
/433133100
103
iii. Resultant hoop stress (shrinkage +Fluid pressure):
a. Outer cylinder Resultant hoop stress at the outer surface = 36 + 266
= 302 N/ mm2 (Tensile) Resultant hoop stress at the junction = 44 + 325 = 369 N/mm2 (tensile)
b. Inner cylinder;
Resultant hoop stress at the inner face = - 44.2 + 433 = 388.8 N/mm2 (Tensile)
Resultant hoop stress at the junction = - 36.2 + 325
= 288.8 N/mm2 (Tensile)
21. A column with alone end hinged and the other end fixed has a length of 5m and a
hollow circular cross section of outer diameter 100 mm and wall thickness 10 mm. If E
= 1.60 x 105 N/mm2 and crushing strength 2
0 /350 mmN , Find the load that the
column may carry with a factor of safety of 2.5 according to Euler theory and Rankine –
Gordon theory. If the column is hinged on both ends, find the safe load according to
the two theories. (April/May 2003) Solution:
Given: L = 5 m = 5000 mm
Outer diameter D = 100 mm Inner diameter d = D-2t = 100 – 2 (10) = 80 mm Thickness = 10 mm
I = 1.60 x 105 N/mm2
2
0 /350 mmN
f = 2.5
i. Calculation of load by Euler’s Theory:
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 87 Strength of Materials
Column with one end fixed and other end hinged.
2
22
L
EIP
mm
lL 60.3536
2
5000
2
2
52
06.3536
1060.114.32 IP
44
64dDI
44 8010064
4096000010000000064
I = 28.96 x 105 mm4
14.12503716
1096.281060.114.32 552
P
p = 73.074 x 103 N ii. Calculation of load by Rankine-Gordon Theory:
Rankine’s Constant 7500
1a (assume the column material is mild steel.)
2
1
K
La
Afp c
K = lest radius of Gyration
01.322826
1096.28 5
A
I
22 801004
A
6400100004
fc = c
= 2826 mm2
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 88 Strength of Materials
2
01.32
06.3536
7500
11
26.28350
P
036.122031033.1
9891004
P
NP 41094.60 iii. Both ends are hinged
Euler’s theory
2
2
L
EIP
L = l
2
552
5000
1096.281060.114.3
P = 18.274 x 104 N ; Safe Load =5.2
10274.18 4
= 73096 N Rankine’s Theory
2
1
K
La
Afp c
2
01.32
5000
7500
11
2826350
81.243981033.1
9891004
Safe load
5.2
10480.30 4 = 121920 N
P = 30.480 x 104
Result: i. Euler’s Theory
One end fixed & one end hinged P = 73.074 x 103 N Both ends hinged P = 18.274 x 104 N
ii. Rankine’s Theory One end fixed & one end hinged P = 60.94 x 104 N
Both ends hinged P = 30.480 x 104 N
iii. Safe Load
Euler’s Theory = 73096 N
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 89 Strength of Materials
Rankine’s theory = 121920 N
22. A column is made up of two channel ISJC 200 mm and two 25 cm x 1 cm flange plate
as shown in fig. Determine by Rankine’s formula the safe load, the column of 6m
length, with both ends fixed, can carry with a factor of safety 4. The properties of
one channel are A = 17.77 cm2, Ixx = 1,161.2 cm4 and Iyy = 84.2 cm4. Distance of
centroid from back of web = 1.97 cm. Take fc = 0.32 KN/mm2 and Rankine’s
Constant 7500
1 (April /May 2003)
Solution:
Given: Length of the column l = 6 m = 600 mm
Factor of safety = 4 Yield stress, fc = 0.32 KN/mm2
Rankine’s constant, 7500
1a
Area of column, A = 2 (17.77+25 x 1) A = 85.54 cm2
A = 8554 mm2
Moment of inertia of the column about X-X axis
2
3
5.1012512
1252.161,12XXI = 7839.0 cm4
2
3
97.1577.1742.812
2512YYI = 4,499.0 cm4
Iyy < IXX The column will tend to buckle in yy-direction
I = Iyy =4499.0 cm4 Column is fixed at both the ends
mml
L 30002
6000
2
mmA
IK 5.72
855
1044994
4
2
1
.
L
Ka
AfP c
2
5.72
3000
75000
11
.855432.0
A = 2228 KN
Safe load of column SOF
P
..
4
2228 =557 KN
Result:
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 90 Strength of Materials
Safe load = 557 KN
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 91 Strength of Materials
Unit– IV
State Of Stress In Three Dimensions
Part A
1. Define stress When a certain system of external forces act on a body then the body offers resistance to these forces. This internal resistance offered by the body per unit area is called the stress
induced in the body.
2. Define principal planes. The plane in which the shear stress is zero is called principal planes. The plane which is independent of shear stress is known as principal plane.
3. Define spherical tensor.
0
0
m
iiij
0
0
m
m
0
0
It is also known as hydrostatic stress tensor
zyxm 3
1
m is the mean stress.
4. Define Deviator stress tensor
xz
xy
mx
ij
1
yz
my
xy
l
l
mz
yz
xz
5. Define volumetric strain It is defined as the ratio between change in volume and original volume of the body and is denoted by e v
Change in volume v
e v =
Original volume v
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 92 Strength of Materials
6. State the principal theories of failure.
i. Maximum principal stress theory
ii. Maximum shear stress (or) stress difference theory iii. Strain energy theory
iv. Shear strain energy theory v. Maximum principal strain theory
vi. Mohr’s Theory
7. State the Limitations of Maximum principal stress theory
1. On a mild steel specimen when spiel tension test is carried out sliding occurs
approximately 45o to the axis of the specimen; this shows that the failure in this case
is due to maximum shear stress rather than the direct tensile stress.
2. It has been found that a material which is even though weak in simple compression yet can sustain hydrostatic pressure for in excess of the elastic limit in simple compression.
8. Explain maximum principal stress theory. According to this theory failure will occur when the maximum principle tensile stress (1)
in the complex system reaches the value of the maximum stress at the elastic limit (et) in
the simple tension.
9. Define maximum shear stress theory This theory implies that failure will occur when the maximum shear stress maximum in
the complex system reaches the value of the maximum shear stress in simple tension at elastic limit (i.e)
22
31
max
etl
(or) et 31
10. State the limitations of maximum shear stress theory.
i. The theory does not give accurate results for the state of stress of pure shear in which the maximum amount of shear is developed (i.e) Torsion test.
ii. The theory does not give us close results as found by experiments on ductile
materials. However, it gives safe results.
11. Explain shear strain Energy theory.
This theory is also called “ Distortion energy Theory” or “Von Mises - Henky Theory.”
According to this theory the elastic failure occurs where the shear strain energy per unit
volume in the stressed material reaches a value equal to the shear strain energy per unit volume at the elastic limit point in the simple tension test.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 93 Strength of Materials
12. State the limitations of Distortion energy theory.
1. The theory does to agree the experiment results for the material for which at is
quite different etc. 2. This theory is regarded as one to which conform most of the ductile material under
the action of various types of loading.
13. Explain Maximum principal strain theory
The theory states that the failure of a material occurs when the principal tensile strain in the material reaches the strain at the elastic limit in simple tension (or) when the min minimum
principal strain (ie ) maximum principal compressive strain reaches the elastic limit in simple compression.
14. State the Limitations in maximum principal strain theory
i. The theory overestimates the behaviour of ductile materials. ii. The theory does no fit well with the experimental results except for brittle
materials for biaxial tension.
15. State the stress tensor in Cartesian components
xz
xy
x
ij
.'
yz
y
xy
z
yz
xz
16. Explain the three stress invariants.
The principal stresses are the roots of the cubic equation,
032
2
1
3 III
where
zyxI 1
xzyxyI zzxzyy222
2
xzyzxyxyzxzyxyxZyxI 22223
17. State the two types of strain energy
i. Strain energy of distortion (shear strain energy) ii. Strain energy of dilatation.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 94 Strength of Materials
18. Explain Mohr’s Theory
Let f
The enveloping curve f must represent in this abscissa and ordinates e, the
normal and shearing stresses in the plane of slip.
2
312
2
31
22
Let 312
1 P
312
1 m
222lmp
19. State the total strain energy theory.
The total strain energy of deformation is given by
133221
2
3
2
2
2
1 22
1 v
EU
and strain energy in simple tension is
E
U2
20
20. State the shear strain energy per unit volume
2
132
322
2112
1
Cs
where
m
EC
112
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 95 Strength of Materials
21. Explain the concept of stress?
When certain system of external forces act on a body then the body offers resistance to these forces. This internal resistance offered by the body per unit area is called the stress
induced in the body.
The stress may be resolved into two components. The first one is the normal stress
n, which is the perpendicular to the section under examination and the second one is the shear
stress , which is operating in the plane of the section.
22. State the Theories of failure.
The principal theories are:
1. Maximum principal stress theory
2. Maximum shear stress (or) stress difference theory 3. Strain energy theory 4. Shear strain energy theory
5. Maximum principal strain theory 6. Mohr’s Theory
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 96 Strength of Materials
Part B
SIXTEEN MARKS QUESTIONS AND ANSWERS:
1. The stress components at a point are given by the following array.
6
5
10
10
8
5
6
10
6
Mpa
Calculate the principal stress and principal planes.
Solution:
The principal stresses are the roots of the cubic equation
0322
13 III (1)
where,
zyxI 1
222
2 zxzyyxxzzyyxI
xzyzxyxyzxzyyzxzyxI 22223
are three stress invariants
The stress tensor
zx
yx
x
ij
.
zy
y
xy
z
yz
xz
By comparing stress tensor and the given away,
zyxI 1
= 10 + 8 +6 =24
xzyzxyxzzyyxI 2222
= (10 x 8) + (8 x 6) + (6 x 10) - (5)2 – (10)2 – (6)2 =80 + 48 + 60 - 25 – 100 -36
=27
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 97 Strength of Materials
xzyzxyxyzxzyyzxzyxI 22223
= 10 x 8 x 6 -10 (10)2 -8 (6 )2 - 6 (5)2 + 2(5) (10) (6) =480 -1000-288-150+600
=-358 Substitute these values in (1) equation
03582724 23 (2) We know that
From this
CosCosCos 3334
04
33
4
13 CosCosCos (3)
put,
3
1IrCos
3
24 rCos
8 rCos
Equation (2) becomes
82642419224512 222233 rCosCosrrCosCosrCosr
27 (r cos + 8) + 358 =0
r3 Cos3 + 512 - 24 r2 Cos2+ + 192 r Cos - 24 r2 Cos2 - 1536 -
384 r Cos + 27 r Cos + 216 + 358 =0
r3 Cos3 - 165 r Cos - 450 = 0
Divided by r3
0450165
32
3 r
Cosr
Cos (4)
Comparing equation (3) and (4) ,w e get,
4
31652
r
r = 14.8324 and
CosCosCos 3343
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 98 Strength of Materials
4
34503
Cos
r
3
8324.14
44503
Cos
Cos 3 = 0.551618
1 = 18.84o
2 = 1 + 120
2 = 138.84o
3 = 2 +120
3 = 258.84o
1 = r Cos 1 + 8
= 14.8324 Cos (18.84o) + 8
1 = 22.04 MPa = 14.8324 Cos 138. 84o + 8
= - 3.17 MPa
3 = r cos 3 + 8 = 14.8324 Cos 258. 84o + 8
= 5.13 MPa Result:
1 = 18.84 o 1 = 22.04 MPa
2 = 138.84 o 2 = -3.17 MPa
3 = 258.84 o 3 = 5.13 MPa
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 99 Strength of Materials
2. Obtain the principal stresses and the related direction cosines for the following
state of stress.(April / May 2003)
6
4
.3
5
2
4
MPa
1
5
6
Solution:
The principal stresses are the roots of the cubic equation.
0322
13 III (1)
zyxI 1
= 3 + 2 + 1 = 6
xzyzxzyyx yxI 222
2
= (3 x 2 ) + (2 x 1) + (1 x 3) - (4)2 - (5)2 - (6)2 = 11 – 16 - 25 - 36
I2 = -66
xzyzxyxyzxzyyzxzyxI 22223
=(3 x 2 x 1) - 3(5)2 - 2(6)2 - 1 (4)3 + 2 (4 x 6 x 5)
= 6 - 75 - 72 - 16 + 240 I3 = 83
Substitute these values in equation (1)
083666 23 (2) We know that
CosCosCos 3334
CosCosCos4
33
4
13
CosCosCos4
33
4
13 (3)
Put 3
1IrCos
2 rCos
CosCosCos 3343
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 100 Strength of Materials
Equation (2) becomes
083666 23
083266262333
rCosrCosrCos
2232233 643238 CosrrCosCosrCosr
083132cos6624cos24 rr
0179662733 rCosrCosCosr
01793933 rCosCosr Divided by r3
017939
32
3 r
Cosr
Cos (4)
By comparing (3) and (4)
4
1392
r
r2 = 156
r = 12.48
and 4
31793
Cos
r
716 = Cos 3 x (12.48 )3
765.1943
7163 Cos
Cos 3 = 0.3683573
3 = 68.38565
1 = 22.79o
2 = 1 + 120
2 = 142.79
3 = 2 +120
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 101 Strength of Materials
3 = 262.79
2cos 11 r
= 12.48 Cos (22.790) + 2
MPa506.131
222 rCos
= 12.48 Cos (142.79) + 2
MPa939.72
23 3 rCos
= 12.48 Cos (262.79) + 2
= 0.433680 MPa
Result:
1 = 22. 79o 1 = 13.506 MPa
2 = 142. 79o 2 = -7.939 MPa
3 = 262. 79o 3 = 0.433680 MPa
3. The state of stress at a point is given by
10
6
.20
8
10
6
MPa
7
8
10
Determine the principal stresses and principal direction. Solution:
The cubic equation
032
2
1
3 III (1)
zyxI 1
= 20 + 10 + 7 = 37
222
2 zxyzxyxzzyyxI
=(20 x 10) + (10 x 7) + (7) x 20 + (36) + (64) + (100)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 102 Strength of Materials
=200 + 70 + 140 + 26 + 64 + 100 I2=610
zxyzxyxyzxzyyzxzyxI 2222
3
=(20 x 10 x 7) - 20 (64) - 10 (100) - 7 (36) + 2 (6) (8) (10)
=1400 - 1280 - 1000 – 252 + 960
=1308 Substitute these values in equation (1)
0130861037 23 (2) We know that
CosCosCos 334 3
CosCosCos4
33
4
13
CosCosCos4
33
4
13 (3)
Put 3
1IrCos
33.12 rCos Equation (2) becomes
0130861037 23
0130833.1261033.123733.1223
rCosrCosrCos
0289.15266.2437087.45699.36516.1874 222233 rCosCosrrCosCosrCosr
160 r Cos + 1972.80 - 1308 = 0
0693.562512.937087.45699.36516.1874 222222233 CosrCosrrCosCosrCosr
160 r Cos + 1972.80 - 1308 = 0
CosCosCos 343 3
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 103 Strength of Materials
02693.496029533 Cosr 3r
02693.496029532
3 r
Cosr
Cos (4)
By comparing (3) & (4)
2
295
4
1
r
r2 = 1180
r = 34.35 and
3
2693.4960
4
3
r
Cos
331.40534
2693.4960
4
3
Cos
3 = 60.6930
1 = 20.231o
2 = 1 + 120
2 = 140 .23 o
3 = 26.231 o
33.1211 rCos
= 34.35 Cos (140.23o) + 12.33
MPa530.441
33.1222 rCos
= 34.35 Cos (140.231o) + 12.33
MPa217.142
33.1233 rCos
= 34.35 Cos (260.231o) + 12.33
5016.63
Result:
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 104 Strength of Materials
1 = 20.231o 3 = 260.231o 2 = - 14.217 MPa
2 = 140.23o 1 = 44.530 MPa 3 = 6.5016 MPa 4. Explain the Energy of Distortion ( shear strain energy ) and Dilatation
The strain energy can be split up on the following two strain energies. i. Strain energy of distortion (shear strain energy)
ii. Strain energy of Dilatation (Strain energy of uniform compression (or)) tension (or) volumetric strain energy )
Let e1 e2 an d e3 be the principal strain in the directions of principal stresses 1, 2 and
3.
Then
3211
1
Ee
1322
1
Ee
2133
1
Ee
Adding the above equation we get,
321321321 21
E
eee
21321
E
But e1 + e2 + e3 = e v (Volumetric strain)
321
21
Eev
If 0,0321 ve . This means that if sum of the three principal stress is
zero there is no volumetric change, but only the distortion occurs.
From the above discussion,
1. When the sum of three principal stresses is zero, there is no volumetric change but only the distortion occurs.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 105 Strength of Materials
2. When the three principal stresses are equal to one another there is no distortion but only volumetric change occurs.
Note:
In the above six theories,
et , ec = Tensile stress at the elastic limit in simple tension and
compression;
1, 2, 3 = Principal stresses in any complex system (such that e1 > e2 > e3 )
It may be assumed that the loading is gradual (or) static (and there is no cyclic (or) impact load.)
5. Explain the Maximum Principal stress Theory: ( Rankine’s Theory)
This is the simplest and the oldest theory of failure
According to this theory failure will occur when the maximum principle tensile
stress (1) in the complex system reaches the value of the maximum stress at the
elastic limit (et) in the simple tension (or) the minimum principal stress (that is,
the maximum principal compressive stress), reaches the elastic limit stress () in simple compression.
(ie.) 1 = et (in simple tension)
ac 3 (In simple compression)
3 Means numerical value of 3
If the maximum principal stress is the design criterion, the maximum principal stress
must not exceed the working for the material. Hence,
1
This theory disregards the effect of other principal stresses and of the shearing stresses on other plane through the element. For brittle materials which do not fail by yielding but fail by brittle fracture, the maximum principal stress theory is considered
to be reasonably satisfactory.
This theory appears to be approximately correct for ordinary cast – irons and brittle metals.
The maximum principal stress theory is contradicted in the following cases:
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 106 Strength of Materials
1. On a mild steel specimen when simple tension test is carried out sliding occurs approximately 45o to the axis of the specimen; this shows that the failure in the
case is due to maximum shear stress rather than the direct tensile stress. 2. It has been found that a material which is even though weak in simple
compression yet can sustain hydrostatic pressure for in excess of the elastic limit in simple compression.
6. Explain the Maximum shear stress (or) Stress Difference theory (April / May 2003)
This theory is also called Guesti’s (or) Tresca’s theory.
This theory implies that failure will occur when the maximum shear stress maximum in the complex system reaches the value of the maximum shear
stress in simple tension at the elastic limit i.e.
22
31
max
et
in simple tension.
(or) et 31
In actual design et in the above equation is replaced by the safe stress.
This theory gives good correlation with results of experiments on ductile materials. In the case of two dimensional tensile stress and then the maximum
stress difference calculated to equate it to et. Limitations of this theory:
i. The theory does not give accurate results for the state of stress of pure shear in which the maximum amount of shear is developed (ie) Torsion test.
ii. The theory is not applicable in the case where the state of stress consists of triaxial tensile stresses of nearly equal magnitude reducing, the shearing stress
to a small magnitude, so that failure would be by brittle facture rather than by yielding.
iii. The theory does not give as close results as found by experiments on ductile
materials. However, it gives safe results.
7. Explain the Shear strain Energy Theory (April / May 2003)
This theory is also called “Distortion Energy Theory”: (or) “Von Mises – Henky Theory”
According to this theory the elastic failure occurs where the shear strain energy per unit volume in the stressed material reaches a value equal to the shear strain
energy per unit volume at the elastic limit point in the simple tension test.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 107 Strength of Materials
Shear strain energy due to the principal stresses 1, 2, and 3 per unit volume of the stress material.
2
13
2
32
2
2112
1
CU S
But for the simple tension test at the elastic limit point where there is only one principal
stress (ie) et we have the shear strain energy per unit volume which is given by
2221 0000
12
1tates
CU
Equating the two energies, we get
0
0
3
2
1
et
22
13
2
32
2
21 2 et
The above theory has been found to give best results for ductile material for which ecet
approximately. Limitations of Distortion energy theory:
1. Te theory does to agree with the experimental results for the material for which et
is quite different from ec.
2. The theory gives 0et for hydrostatic pressure (or) tension, which means that the
material will never fail under any hydrostatic pressure (or) tension. When three equal tensions are applied in three principal directions, brittle facture occurs and as
such maximum principal stress will give reliable results in this case. 3. This theory is regarded as one to which conform most of the ductile material under
the action of various types of loading.
8. Explain the Maximum principal strain Theory?
This theory associated with St Venent
The theory states that the failure of a material occurs when the principal tensile strain in the material reaches the strain at the elastic limit in simple tension (or)
when the minimum principal strain (ie) maximum principal compressive strain reaches the elastic limit in simple compression.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 108 Strength of Materials
Principal strain in the direction of principal stress 1,
3211
11
mEe
Principal strain in the direction of the principal stress 3,
2133
11
mEe
The conditions to cause failure according to eh maximum principal strain theory are:
E
e et1 (e1 must be +Ve)
and
E
e ec3 (e3 must be -Ve)
EmE
et
321
11
EmE
et
213
11
etm
311
1
ecm
313
1
To prevent failure:
etm
321
1
cem
213
1
At the point of elastic failure:
etm
321
1
and cem
213
1
For design purposes,
tm
213
1
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 109 Strength of Materials
cm
213
1
(where, t and c are the safe stresses)
Limitations:
i. The theory overestimates the behavior of ductile materials. ii. Te theory does not fit well with the experimental results except for brittle materials for
biaxial tension.
9. Explain the Strain energy theory?
The total stain energy of deformation is given by
133221
2
3
2
2
2
1 22
1 v
EU
and the strain energy under simple tension is
E
U e
2
2
Hence for the material to yield,
133221
2
3
2
2
2
1 2 v
The total elastic energy stored in a material before it reaches the plastic state can have
no significance as a limiting condition, since under high hydrostatic pressure, large amount of strain energy ma be stored without causing either fracture (or) permanent deformation.
10. Explain Mohr’s Theory?
A material may fail either through plastic slip (or) by fracture when either the shearing
stress in the planes of slip has increased.
Let f
The enveloping curve f must represent in their abscissa and ordinates , the
normal and shearing stresses in the plane of slip. Now
2
312
2
31
22
Let 312
1 P
312
1 m
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 110 Strength of Materials
then
222mp
This equation represents the family of major principal stress circles in parameter form. The equation of this envelope is obtained by partially differentiating with respect to P
222mP
2222 2 mPp
dp
dp m
m
.
dp
d mm
2
1.
This is to equation of Mohr’s envelope of the major
principal stress in parameter form.
11. In a steel member, at a point the major principal stress is 180 MN/m2 and the minor
principal stresses is compressive. If the tensile yield point of the steel is 225
MN/m2, find the value of the minor principal stress at which yielding will
commence, according to each of the following criteria of failure.
i. Maximum shearing stress
ii. Maximum total strain energy
iii. Maximum shear strain energy
Take Poisson’s ratio = 0.26
Solution:
Major principal stress, 2
1 /180 mMN
Yield point stress 2
2 /225 mMN
26.01
m
To calculate minor principal stress (2)
(i) Maximum shearing stress criterion
e 12
= 180 - 225
2 = - 45 MN/m2
2 = 45 MN/m2 (comp)
e 21
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 111 Strength of Materials
ii. Maximum total strain energy criterion:
2
133221
2
3
2
2
2
1
2e
m
3 = 0
(180)2 + 22 - 2 x 0.26 x 180 2 = (225)2
32400 + 22 -93.6 2 = 50625
22 - 93.6 2 - 18225 = 0
2
1822546.9336.92
2
2/08.962
76.28536.9mMN
(Only –Ve sign is taken as 2 is compressive)
2 = 96.08 MN/ m2 (compressive)
iii. Maximum shear strain energy criterion:
putting 3 = 0
221
22
221 2 e
22
121
2
2
2
1 22 e
(180)2 + (2)2+ - 180 2 = (225)2
(2)2 - 180 2 - 18225 = 0
2
21
2
2
2
1
2e
m
2213
232
221 2 e
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 112 Strength of Materials
2
1822541801802
2
2
2 /25.722
5.324180mMN
2 = 72.25 MN/m2 (Compressive)
12. In a material the principal stresses are 60 MN/m2, 48 MN/m2 and - 36 MN/m2.
Calculate
i. Total strain energy
ii. Volumetric strain energy
iii. Shear strain energy
iv. Factor of safety on the total strain energy criteria if the material
yields at 120 MN/m2.
Take E = 200 GN/m2+ and 1/m = 0.3
Solution: Given Data: Principal stresses:
1 = + 60 MN/m2
2 = + 48 MN/m2
3 = - 36 MN/m2
Yield stress, e = 120 MN /m2
E = 200 GN/m2, 1/m = 0.3
i. Total strain energy per unit volume:
133221
2
3
2
2
2
1
2
2
1
mEU
60364848603.02364860102002
10 222
9
12
U
2160172828806.01296230436005.2U
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 113 Strength of Materials
U = 19.51 KNm/m3 ii. Volumetric strain energy per unit volume:
3
9
12210
102002
3.02110364860
3
1
ve
e v = 1.728 KN/m3
iii. shear strain energy per unit volume
Where,
2/923.763.012
200
112
mGN
m
EC
222
9
12
60363648486010923.7612
101
se
31092167056144083.1 se
3/78.17 mKNmes
iv. Factor of safety (F.O.S)
Strain energy per unit volume under uniaxial loading is
33
9
262
/3610102002
10120
2mKNm
E
e
F.O.S 845.151.19
36
13. In a material the principal stresses are 50 N/mm2, 40 N/mm2 and - 30 N/mm2,
calculate:
E
mev
2
/21
3
1 2
321
213
232
221
12
1
ces
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 114 Strength of Materials
i. Total strain energy
ii. Volumetric strain energy
iii. Shear strain energy and
iv. Factor of safety on the total strain energy criterion if the material
yield at 100 N/mm2.
Take E = 200 x 103 N/mm2 and poission ratio = 0 .28
Solution:
Given, Principal stresses:
2
1 /50 mmN
2
2 /40 mmN
2
3 /30 mmN
Yield stress, 2/100 mmNe
i. Total strain energy per unit volume:
133221
2
3
2
2
2
1
2
2
1
mEU
5030304040503.02304050102002
1 222
3
1500120020006.09001600250010400
13
7006.0500010400
13
542010400
13
U = 13.55 KNm/m3
ii)Volumetric strain energy per unit volume:
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 115 Strength of Materials
E
mev
2
/21
3
1 2221
3
2
102002
3.021304050
3
1
ve
3
2
10400
4.060
3
1
310
001.03600
3
1ve
ev = 1.2 K N m / m3
iii. Shear strain energy
2
13
2
32
2
2112
1
Ces
where
233
/10923.763.012
10200
/112mmN
m
EC
222
3503030404050
10923.7612
1
se
6400490010010076.923
13
se
3/35.12 mKNnes
iv. Factor of safety (F.O.S)
Strain energy per unit volume under uniaxial loading is
3
3
22
/254102002
100
2mKNm
E
e
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 116 Strength of Materials
845.155.13
25.. SOF
14. In a material the principal stresses are 50 N/mm2, 40 N/mm2 and - 30 N/mm2,
calculate:
v. Total strain energy
vi. Volumetric strain energy
vii. Shear strain energy and
viii. Factor of safety on the total strain energy criterion if the material
yield at 100 N/mm2.
Take E = 200 x 103 N/mm2 and poission ratio = 0 .28
Solution:
Given,
Principal stresses:
2
1 /50 mmN
2
2 /40 mmN
2
3 /30 mmN
Yield stress, 2/100 mmNe
i. Total strain energy per unit volume:
133221
2
3
2
2
2
1
2
2
1
mEU
5030304040503.02304050102002
1 222
3
1500120020006.09001600250010400
13
7006.0500010400
13
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 117 Strength of Materials
542010400
13
U = 13.55 KNm/m3
ii)Volumetric strain energy per unit volume:
E
mev
2
/21
3
1 2221
3
2
102002
3.021304050
3
1
ve
3
2
10400
4.060
3
1
310
001.03600
3
1ve
ev = 1.2 K N m / m3
iii. Shear strain energy
2
13
2
32
2
2112
1
Ces
where
233
/10923.763.012
10200
/112mmN
m
EC
222
3503030404050
10923.7612
1
se
6400490010010076.923
13
se
3/35.12 mKNnes
iv. Factor of safety (F.O.S)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 118 Strength of Materials
Strain energy per unit volume under uniaxial loading is
3
3
22
/254102002
100
2mKNm
E
e
845.155.13
25.. SOF
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 119 Strength of Materials
Unit– V
Advanced Topics In Bending Of Beams
Part A
1. Define Unsymmetrical bending
The plane of loading (or) that of bending does not lie in (or) a plane that contains the principle centroidal axis of the cross- section; the bending is called Unsymmetrical bending.
2. State the two reasons for unsymmetrical bending.
(i) The section is symmetrical (viz. Rectangular, circular, I section) but the load line is
inclined to both the principal axes. (ii) The section is unsymmetrical (viz. Angle section (or) channel section vertical web) and
the load line is along any centroidal axes.
3. Define shear centre.
The shear centre (for any transverse section of the beam) is the point of intersection of the bending axis and the plane of the transverse section. Shear centre is also known as “centre of
twist”
4. Write the shear centre equation for channel section.
f
w
A
A
be
6
3
e = Distance of the shear centre (SC ) from the web along the symmetric axis XX Aw = Area of the web
Af = Area of the flange
5. A channel Section has flanges 12 cm x 2 cm and web 16 cm x 1 cm. Determine the
shear centre of the channel.
Solution:
b= 12-0.5 = 11.5 cm t1 = 2cm, t2 = 1cm, h= 18 cm Af = bt1 = 11.5 x 2 = 23 cm2
Aw = ht2 = 18 x 1= 18 cm2
f
w
A
A
be
6
3
cme 086.5
23
186
)5.11(3
6. Write the shear centre equation for unsymmetrical I section.
xxI
bbhte
4
)( 212
21
e = Distance of the shear centre (SC) from the web along the symmetric axis XX
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 120 Strength of Materials
t1 = thickness of the flange h = height of the web b1 = width of the flange in right portion.
b2 = width of the flange in left portion. Ixx = M.O.I of the section about XX axis.
7. State the assumptions made in Winkler’s Bach Theory.
i. Plane sections (transverse) remain plane during bending.
ii. The material obeys Hooke’s law (limit state of proportionality is not exceeded)
iii. Radial strain is negligible. iv. The fibres are free to expand (or) contract without any constraining effect
from the adjacent fibres.
8. State the parallel Axes and Principal Moment of inertia.
If the two axes about which the product of inertia is found, are such , that the product of inertia becomes zero, the two axes are then called the principle axes. The moment of inertia about a principal axes is called the principal moment of inertia.
9. Define stress concentration.
The term stress gradient is used to indicate the rate of increase of stress as a stress raiser is approached. These localized stresses are called stress concentration.
10. Define stress – concentration factor.
It is defined as the ratio of the maximum stress to the nominal stress.
nom
tK
max
max = maximum stress nom = nominal stress
11. Define fatigue stress concentration factor.
The fatigue stress – concentration factor (Kf ) is defined as the ratio of flange limit of
unnotched specimen to the fatigue limit of notched specimen under axial (or) bending loads. )1(1 tf KqK
Value of q ranges from zero to one.
12. Define shear flow.
Shear flow is defined as the ratio of horizontal shear force H over length of the beam x.
Shear flow is acting along the longitudinal surface located at discharge y1.Shear flow is defined by q.
z
zy
I
QV
x
Hq
H = horizontal shear force
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 121 Strength of Materials
13. Explain the position of shear centre in various sections.
(i) In case of a beam having two axes of symmetry, the shear centre coincides
with the centroid. (ii) In case of sections having one axis of symmetry, the shear centre does not
coincide with the centroid but lies on the axis of symmetry. 14. State the principles involved in locating the shear centre.
The principle involved in locating the shear centre for a cross – section of a beam is that the loads acting on the beam must lie in a plane which contains the resultant shear force
on each cross-section of the beam as computed from the shearing stresses.
15. State the stresses due to unsymmetrical bending.
VVUUb
I
u
I
vM
sincos
σb = bending stress in the curved bar M = moment due to the load applied
IUU = Principal moment of inertia in the principal axes UU IVV = Principal moment of inertia in the principal axes VV
16. Define the term Fatigue.
Fatigue is defined as the failure of a material under varying loads, well below the ultimate
static load, after a finite number of cycles of loading and unloading. 17. State the types of fatigue stress.
(i) Direct stress (ii) Plane bending
(iii) Rotating bending (iv) Torsion (v) Combined stresses
(a) Fluctuating or alternating stress (b) Reversed stress.
18. State the reasons for stress- concentration.
When a large stress gradient occurs in a small, localized area of a structure, the high stress is referred to as a stress concentration. The reasons for stress concentration are (i)
discontinuities in continuum (ii) contact forces. 19. Define creep.
Creep can be defined as the slow and progressive deformation of a material with time under a constant stress.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 122 Strength of Materials
SIXTEEN MARKS QUESTIONS AND ANSWERS
1. Explain the stresses induced due to unsymmetrical bending.
Fig. shows the cross-section of a beam under the action of a bending moment M acting in plane YY.
Also G = centroid of the section, XX, YY = Co-ordinate axes passing through G,
UU, VV = Principal axes inclined at an angle θ to XX and YY axes respectively
The moment M in the plane YY can be resolved into its components in the planes UU and VV as follows:
Moment in the plane UU, M’ = M sinθ Moment in the plane VV, M’ = M cosθ
The components M’ and M” have their axes along VV and UU respectively. The resultant bending stress at the point (u,v) is given by,
UUVVUUVV
bI
M
I
M
I
vM
I
uM
cossin"'
vvUUb
I
uSin
I
VCosM
At any point the nature of σb will depend upon the quadrant in which it lies. The equation
of the neutral axis (N.A) can be found by finding the locus of the points on which the resultant stress is zero. Thus the points lying on neutral axis satisfy the condition that σb = 0
0
vvUU I
uSin
I
VCosM
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 123 Strength of Materials
0vvUU I
uSin
I
VCos
uCos
Sin
I
Iv
vv
UU
(or) u
I
Iv
vv
UU
tan
This is an equation of a straight line passing through the centroid G of the section and
inclined at an angle with UU where
tantan
vv
UU
I
I
Following points are worth noting:
i. The maximum stress will occur at a point which is at the greatest distance form the neutral
ii. All the points of the section on one side of neutral axis will carry stresses of the same nature and on the other side of its axis, of opposite nature.
iii. In the case where there is direct stress in addition to the bending stress, the neutral
axis will still be a straight line but will not pass through G (centroid of section.)
2. Derive the equation of Shear centre for channel section. April/May 2005
Fig shows a channel section (flanges: b x t1 ; Web h x t2) with XX as the horizontal symmetric axis.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 124 Strength of Materials
Let S = Applied shear force. (Vertical downward X) (Then S is the shear force in the web in the upward direction)
S1 = Shear force in the top flange (there will be equal and opposite shear force in the bottom flange as shown.)
Now, shear stress () in the flange at a distance of x from the right hand edge (of the top flange)
tI
ySA
xa
2
.1
hxtyA (where t = t1 , thickness of flange)
xx
xh
xx I
Sh
tI
xSt
22.
.
.
1
1
Shear force is elementary area
dztddxtd AA 11 ..
Total shear force in top flange
dxt
b
..
0
1 (where b = breadth of the flange)
b
xx
b
xx
xdxI
shtdxt
I
hSS
0
11
0
12
.;2
(or) 4
.2
11
b
I
ShtS
xx
Let e = Distance of the shear centre (sc) from taking moments of shear forces about the centre O of the web,We get
hSeS .. 1
xxxx I
bhtSh
b
I
Sht
4
..
4.
221
21
xxI
thbe
4
122
(1)
Now, 122.
122
32
2
1
31 hth
tbtb
Ixx
122
.
6
32
21
31 hthtbbt
122
32
21 hthbt
(neglecting the term 3
3
1bt, being negligible in comparison to other
terms)(or) 12
2
12bbtht
hI xx
Substitute the value of Ixx in equation (1) we get,
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 125 Strength of Materials
12
12
122
122
6
3
6
12
4 htht
tb
bthth
thbe
Let bt1 = Af (area of the flange)
ht2 = A (area of the web)
Then
f
wfw
f
A
A
b
AA
bAe
6
3
6
3
i.e
3. Derive the equation of Shear center for unequal I-section
Solution:
Fig. shows an unequal I – section which is symmetrical about XX axis.
f
w
A
A
be
6
3
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 126 Strength of Materials
Shear stress in any layer,
It
ySA
where I = IXX =
12122
3
121
31
21
hxtbb
tbb
Shear force S1 :
2
... 11
hxtyAdxtdA
S1 =
1
0
11
1
2
..b
XX
dxxth
tI
txSdA
= 1
0
12
..b
XX
dxth
I
xS =
XX
b
XX I
bShtx
I
Sht
422
211
0
21
1
Similarly the shear force (S2) in the other part of the flange,
S2 =XXI
bSht
4
221
Taking moments of the shear forces about the centre of the web O, we get
S2. h = S1. h + S .e (S3 = S for equilibrium) (where, e = distance of shear centre from the centre of the web) or, (S2 – S1) h = S.e
eSI
bbtSh
XX
.4
)( 21
221
2
4. Derive the stresses in curved bars using Winkler – Bach Theory.
The simple bending formula, however, is not applicable for deeply curved beams where the neutral and centroidal axes do not coincide. To deal with such cases Winkler – Bach Theory
is used. Fig shows a bar ABCD initially; in its unstrained state. Let AB’CD’ be the strained
position of the bar.
xxI
bbhte
4
21
22
21
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 127 Strength of Materials
Let R = Radius of curvature of the centroidal axis HG.
Y = Distance of the fiber EF from the centroidal layer HG. R’ = Radius of curvature of HG’ M = Uniform bending moment applied to the beam (assumed
positive when tending to increase the curvature)
= Original angle subtended by the centroidal axis HG at its
centre of curvature O and
’ = Angle subtended by HG’ (after bending) a t the center of curvature
’ For finding the strain and stress normal to the section, consider the fibre EF at a distance
y from the centroidal axis.
Let σ be the stress in the strained layer EF’ under the bending moment M and e is strain in the same layer.
Strain,
)(
)(')''('
yR
yRyR
EF
EFEFe
or 1
'.
''
yR
yRe
e0 = strain in the centroidal layer i.e. when y = 0
1'
.'
R
R or
'.
''1
yR
yRe
--------- (1)
and 1+e =
'.
'
R
R --------- (2)
Dividing equation (1) and (2) , we get
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 128 Strength of Materials
01
1
e
e
'.
''
R
R
yR
yR
or
R
y
R
ye
R
y
R
ye
e
1
'
'
'
'. 00
According to assumption (3) , radial strain is zero i.e. y = y’
Strain,
R
y
R
ye
R
y
R
ye
e
1
''. 00
Adding and subtracting the term e0. y/R, we get
R
y
R
ye
R
ye
R
ye
R
y
R
ye
e
1
.''
. 0000
R
y
yRR
e
ee
1
)1
'
1)(1( 0
0 ------------- (3)
From the fig. the layers above the centroidal layer is in tension and the layers below the
centroidal layer is in compression.
Stress , σ = Ee = )
1
)1
'
1)(1(
(0
0
R
y
yRR
e
eE
___________ (4)
Total force on the section, F = dA.
Considering a small strip of elementary area dA, at a distance of y from the centroidal layer HG, we have
dA
R
y
yRR
e
EdAeEF
1
)1
'
1)(1(
.0
0
dA
R
y
y
RReEdAeEF
1
)1
,
1(1. 00
dA
R
y
y
RReEAeEF
1
)1
,
1(1. 00 ____________ (5)
where A = cross section of the bar
The total resisting moment is given given by
dA
R
y
yRR
e
EydAeEdAyM
1
)1
'
1)(1(
...
20
0
dA
R
y
y
RReEeEM
1
)1
,
1(10.
2
00 (since )0ydA
M = E (1+e0)
dA
R
y
y
RR1
1
'
12
Let
22
1
AhdA
R
y
y
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 129 Strength of Materials
Where h2 = a constant for the cross section of the bar
M = E (1+e0) 21
'
1Ah
RR
----------- (6)
Now,
dAyR
yydA
yR
RydA
R
y
y 2
..
1
= dA
yR
yydA .
2
dA
R
y
y
1
dA
R
y
y
R.
1
10
2
= 21Ah
R ---------- (7)
Hence equation (5) becomes
F = Ee0 .A – E (1+e0 )R
Ah
RR
21
'
1
Since transverse plane sections remain plane during bending F = 0
0 = Ee0 .A – E (1+e0 )R
Ah
RR
21
'
1
E e0 .A = E (1+e0 )R
Ah
RR
21
'
1
e0 = (1+e0 )R
Ah
RR
21
'
1
(or)
2
0
h
Re(1+e0 )
RR
1
'
1
Substituting the value of 2
0
h
Re(1+e0 )
RR
1
'
1 in the equation (6)
M = E 2
2
0 Ahh
Re = e0 EAR
Or EAR
Me 0 substituting the value of e0 in equation (4)
2
0*
1
*h
Re
R
y
yE
AR
M
(or) EAR
M
h
R
R
y
yE
AR
M**
1
*2
2
1*
1
*h
R
y
Ry
AR
M
AR
M
yR
y
h
R
AR
M2
2
1 (Tensile)
yR
y
h
R
AR
M2
2
1 (Compressive)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 130 Strength of Materials
5. The curved member shown in fig. has a solid circular cross –section 0.10 m in
diameter. If the maximum tensile and compressive stresses in the member are not to
exceed 150 MPa and 200 MPa. Determine the value of load P that can safely be
carried by the member.
Solution:
Given,
d = 0.10 m; R = 0.10 m; G = 150 MPa = 150 MN / m2 (tensile )
2 = 200 MPa = 200 MN / m2 (Compressive)
Load P:
Refer to the fig . Area of cross section,
2322
10854.710.044
md
A
Bending moment, m = P (0.15 + 0.10) =0.25 P
2
422
10.0
10.0.
128
1
16
dh = 7.031 x 10-4 m2
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 131 Strength of Materials
Direct stress, compA
pd
Bending stress at point 1 due to M:
yR
y
h
R
AR
Mb 2
2
1 1 (tensile)
Total stress at point 1,
11 bd
yR
y
h
R
AR
M
A
P2
2
1150 (tensile)
05.010.0
05.0
10031.7
10.01
10.010854.7
25.0
10854.7150
4
2
33
PP
= -127.32 P + 318.31 P x 5. 74
= 1699.78 P
KNP 25.8878.1699
10150 3
(i)
Bending stress at point 2 due to M:
1
2
2
2yR
y
h
R
AR
Mb (comp)
Total stress at point 2,
22 bd
1200
2
2
yR
y
h
R
AR
M
A
P
1
05.010.0
05.0
10031.7
10.0
10.010854.7
25.0
10854.7 4
2
33
PP
=127.32 P + 318. 31 P x 13.22
= 4335.38 P
MNP38.4335
200
KNP 13.4638.4335
10200 3
(ii)
By comparing (i) & (ii) the safe load P will be lesser of two values
Safe load = 46.13 KN.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 132 Strength of Materials
6. Fig. shows a ring carrying a load of 30 kN. Calculate the stresses at 1 and 2.
Solution:
Area of cross-section = 2222 01131.01.113124
mcmcmx
Bending moment M = 30*103 * (13.5*10-2)Nm = 4050 Nm
h2 = ......*128
1
16 2
42
R
dd
Here d = 12 cm, R = 7.5 +6 = 13.5 cm
h2 =2
42
5.13
12*
128
1
16
12 = 9.89 cm2 = 9.89*10-4 m2
Direct Stress σd = 263
/65.210*01131.0
10*30mMN
A
P
Bending stress at point 1 due to M,
yR
y
h
R
AR
Mb 2
2
1 1
6
4
2
1 10*06.0135.0
06.0
10*89.9
135.01
135.0*01131.0
4050
b 2.65*6.6
7 = 17.675 MN/m2 (tensile)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 133 Strength of Materials
Bending stress at point 2 due to M,
yR
y
h
R
AR
Mb 2
2
2 1
6
4
2
1 10*06.0135.0
06.0
10*89.9
135.01
135.0*01131.0
4050
b 2.65*13.
74 = 36.41 MN/m2 (comp) Hence σ1 = σd + σb1 = -2.65 + 17.675
= 15.05 MN /m2 (tensile) and σ2 = σd + σb2 = -2.65 – 36.41 = 39.06 MN/m2 (comp)
8. A curved bar is formed of a tube of 120 mm outside diameter and 7.5 mm thickness. The
centre line of this is a circular arc of radius 225 mm. The bending moment of 3 kNm
tending to increase curvature of the bar is applied. Calculate the maximum tensile and
compressive stresses set up in the bar.
Solution:
Outside diameter of the tube, d2 = 120 mm = 0.12 m
Thickness of the tube = 7.5 mm Inside diameter of the tube, d1 = 120-2*7.5 = 105 mm = 0.105m Area of cross-section,
222 00265.015.012.04
mA
Bending moment M = 3 kNm Area of inner circle,
221 00866.0105.0
4mA
Area of outer circle,
222 01131.012.0
4mA
For circular section,
h2 = ......*128
1
16 2
42
R
dd
For inner circle,
h2 = ......*128
1
16 2
41
21
R
dd
h2 = 4
2
42
10*08.7225.0
105.0*
128
1
16
105.0
For outer circle,
h2 = ......*128
1
16 2
42
22
R
dd; h2= 4
2
42
10*32.9225.0
12.0*
128
1
16
12.0
211
222
2 hAhAAh
0.00265 h2 = 0.01131*9.32*10-4 – 0.00866*7.078*10-4
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 134 Strength of Materials
h2 = 0.00166 m2, and R2/h2 = 0.2252/0.00166 = 30.49 Maximum stress at A,
yR
y
h
R
AR
MA 2
2
1 (where, y = 60 mm = 0.06 m)
263
/10*06.0225.0
06.049.301
225.0*00265.0
10*3mMNA
σA = 37.32 MN/m2 (tensile)
Maximum stress at B,
yR
y
h
R
AR
MB 2
2
1
263
/10*06.0225.0
06.049.301
225.0*00265.0
10*3mMNB
σB = 50.75 MN/m2 (comp)
9. A curved beam has a T-section (shown in fig.). The inner radius is 300 mm. what is the
eccentricity of the section?
Solution:
Area of T-section, = b1t1 + b2t2
= 60*20 + 80*20 = 2800 mm2 To find c.g of T- section, taking moments about the edge LL, we get
21
2211
AA
xAxAx
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 135 Strength of Materials
)20*80()20*60(
)10*20*80)(20*80()202
60)(20*60(
x =27.14 mm
Now R1 = 300 mm; R2 = 320 mm; R= 327.14 mm; R3 = 380 mm
Using the Relation:
2
2
31
1
22
32 log.log. R
R
Rt
R
Rb
A
Rh ee
23
2 )14.327()320
380(log*20)
300
320(log*80
2800
)14.327(
eeh
= 12503.8(5.16+3.44) – 107020.6 = 512.08
y = )(56.108.512)14.327(
08.512*14.327222
2
mm
hR
Rh
where y = e (eccentricity) = distance of the neutral axis from the centroidal axis. Negative sign indicates that neutral axis is locates below the centroidal axis.
10. Fig. shows a C- frame subjected to a load of 120 kN. Determine the stresses at A and B.
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 136 Strength of Materials
Solution:
Load (P) = 120 kN Area of cross – section = b1t1 +b2t2+ b3t3
= 120*30 + 150*30 +180*30 = 0.0135 mm2 To find c.g of the section about the edge LL,
21
2211
AA
xAxAx
)30*180()30*150()30*120(
)120*30*180()15*30*150()225*20*120(1
y =113 mm=0.113 m
y2 = 240 – 113 = 127 mm = 0.127 m
R1 = 225 mm = 0.225 m
R2 = 225 + 30 = 255 mm = 0.255 m R = 225 + 113 = 338 mm = 0.338 m
R3 = 225 +210 = 435 mm = 0.435 m R4= 225 + 240 = 465 mm = 0.465 m
2
3
41
2
33
1
22
32 logloglog R
R
Rb
R
Rt
R
Rb
A
Rh eee
23
2 338.0435.0
465.0log12.0
255.0
435.0log03.0
225.0
255.0log15.0
0135.0
)338.0(
eeeh
= 2.86 (0.01877 +0.016 +0.008) – 0.1142 = 0.008122 m2
Direct stress, σd = )(/89.810*0135.0
10*120 263
compmMNA
P
Bending moment, M = P*R
Bending stress at A due to the bending moment,
2
2
2
2
1)(yR
y
h
R
AR
MAb
127.0338.0
127.0
008122.0
338.01
*)(
2
2
AR
RPAb
= 8.89 (1+3.842) = 43.04 MN/m2 (tensile) Bending stress at B due to the bending moment:
1
1
2
2
1)(yR
y
h
R
AR
MAb
113.0338.0
113.0
008122.0
338.01
*)(
2
AR
RPAb
= 8.89 ( 1- 7.064) = -53.9 MN /m2 = 53.9 MN/m2 (comp) Stress at A, σA = σd + (σb)A
= -8.89 + 43.04 = 34.15 MN/m2 (tensile) Stress at B, σB = σd + (σb)B
= -8.89 – 53.9 = 62.79 MN/m2 (comp)
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 137 Strength of Materials
11. Derive the formula for the deflection of beams due to unsymmetrical bending.
Solution:
Fig. shows the transverse section of the beam with centroid G. XX and YY are two rectangular co-ordinate axes and UU and VV are the principal axes inclined at an angle θ to
the XY set of co-ordinates axes. W is the load acting along the line YY on the section of the beam. The load W can be resolved into the following two components:
(i) W sin θ …… along UG
(ii) W cos θ …… along VG
Let, δu = Deflection caused by the component W sin θ along the line GU for its bending about VV axis, and
Δv = Deflection caused by the component W cos θ along the line GV due to bending abodt UU axis. Then depending upon the end conditions of the beam, the values of δu and δv are given by
VVu
EI
lWK 3sin
UUv
EI
lWK 3cos
where, K = A constant depending on the end conditions of the beam and position of the load along the beam, and
l = length of the beam The total or resultant deflection δ can then be found as follows:
22vu
223 cossin
UUVV I
W
I
W
E
Kl
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 138 Strength of Materials
UUVV IIE
Kl2
2
2
23 cossin
The inclination β of the deflection δ, with the line GV is given by:
tantan
VV
UU
I
I
v
u
12. A 80 mm x 80 mm x 10 mm angle section shown in fig. is used as a simply supported
beam over a span of 2.4 m. It carries a load of 400 kN along the line YG, where G is the
centroid of the section. Calculate (i) Stresses at the points A, B and C of the mid – section of
the beam (ii) Deflection of the beam at the mid-section and its direction with the load line
(iii) Position of the neutral axis. Take E = 200 GN/m2
Solution:
Let (X,Y) be the co-ordinate of centroid G, with respect to the rectangular axes BX1 and BY1.
Now X = Y = mm66.23700800
350032000
10*7010*80
5*10*7040*10*80
Moment of inertia about XX axis:
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 139 Strength of Materials
2
32
3
)66.2345(*10*7012
70*10)566.23(*10*80
12
10*80XXI
= (6666.66 + 278556) + (285833.33 + 318777) = 889833 mm4
= 8.898 * 105 mm4 = IYY (since it is an equal angle section) Co-ordinates of G1 = + (40-23.66), - (23.66-5) = (16.34,- 18.66) Co-ordinates of G2 = -(23.66-5). + (45 – 23.66) = (-18.66, + 21.34)
(Product of inertia about the centroid axes is zero because portions 1 and 2 are rectangular strips)
If θ is the inclination of principal axes with GX, passing through G then,
90tan2
2tan
XXXY
XY
II
I (since Ixx =Iyy)
2θ = 90º
i.e. θ1 = 45º and θ2 = 90º + 45º = 135º are the inclinations of the principal axes GU and GV respectively.
Principal moment of inertia:
IUU = 22 )()2
()(2
1XY
XXYYYYXX I
IIII
= 25255
55 )10*226.5()2
10*898.810*895.8()10*898.810*895.8(
2
1
= (8.898 + 5.2266) *105 = 14.1245*105 mm4
IUU + IVV = IXX + IYY IVV = IXX IYY – IUU
= 2*8.898 x 105 – 14.1246 x 105 = 3.67 x 105 mm4
(i) Stresses at the points A, B and C: Bending moment at the mid-section,
Nmm
WlM 5
3
10*4.24
10*4.2*400
4
The components of the bending moments are; M’ = M sin θ = 2.4 x 105 sin 45º = 1.697 x 105 Nmm
M’’ = M cos θ = 2.4 x 105 cos 45º = 1.697 x 105 Nmm u,v co-ordinates:
Point A: x = -23.66, y = 80-23.66 = 56.34 mm u = x cos θ + y sin θ = -23.66 x cos 45º + 56.34 x sin 45º = 23.1 mm
v = y cosθ + x sin θ = 56.34 cos 45º - (-23.66 x sin 45º) = 56.56 mm
Point B: x = -23.66, y = -23.66
u = x cos θ + y sin θ = -23.66 x cos 45º + (-23.66 x sin 45º ) = - 33.45 mm
v = y cosθ + x sin θ = -23.66 cos 45º - (-23.66 x sin 45º) = 0
S.K.P. Engineering College, Tiruvannamalai IV SEM
Civil Engineering Department 140 Strength of Materials
Point C ; x = 80 – 23.66 = 56.34, y = -23.66 u = x cos θ + y sin θ = 56.34 cos 45º -23.66 x sin 45º = 23.1 mm
v = y cosθ + x sin θ = -23.66 cos 45º - 56.34 sin 45º) =- 56.56 mm
UUVV
AI
vM
I
uM "'
2
5
5
5
5
/47.17101246.14
)56.56(10*697.1
1067.3
)1.23(10*697.1mmN
xxA
2
55
5
/47.15101246.14
0
1067.3
)45.33(10*697.1mmN
xxB
2
55
5
/788.3101246.14
56.56
1067.3
)1.23(10*697.1mmN
xxB
(ii) Deflection of the beam, δ: The deflection δ is given by:
UUVV IIE
KWl2
2
2
23 cossin
where K = 1/48 for a beam with simply supported ends and carrying a point load at the centre. Load , W = 400 N
Length l = 2.4 m E = 200 x 103 N/mm2
IUU = 14.1246 x 105 mm4 IVV = 3.67 x 105 mm4 Substituting the values, we get
25
2
25
233
)101246.14(
45cos
)1067.3(
45sin)104.2(400
48
1
xxE
xx
δ = 1.1466 mm
The deflection δ will be inclined at an angle β clockwise with the kine GV, given by
848.345tan1067.3
101246.14tantan
5
5
x
x
I
I
VV
UU
β = 75.43º - 45º = 30.43º clockwise with the load line GY’. (iii) Position of the neutral axis:
The neutral axis will be at 90º - 30.43º = 59.57º anti-clockwise with the load line, because the neutral axis is perpendicular to the line of deflection.