1

MECHANICS OF MATERIALS REVIEW

Notation:

a = normal stress (psi or Pa)= shearing stress (psi or Pa)

€ = strain (in/in or rn/rn)

o = deformation (in. or m)= moment of inertia

of area (in4 or m4)

J = polar moment of

inertia of area (in4 or m4)N = revolutions per mm.

E = modulus of elasticity (psi or Pa)G = shearing modulus (psi or Pa)p = Poisson’s ratioa = coefficient of thermal

expansion (1°F or 1°C)M = bending moment in beamsT = torque

= temperature change (°F or °C)H.P. = horsepower (1 H.P. = 550 ft-lb/sec)

General Information:factor of safety = (failure-producing load)1(actual load or allowed load)thermal strain = = att

Section 1: Centric Loads

a. Axial Loaded MembersIf the load, P, passes through the centroid of the resisting

section:

axial stress = a =A

axial strain = =L

P

-f

_-j

lithe material is also elastic, then,

stress-strain relation: c = 2.E

alL PLaxial deformation: 8 = — = —

- E AE

2

b. Average shear stress in pins (or rivets)

double shear:

single shear: r =

—* P/2

P

P / 2

P

; P

Section 2: Torsion of Circular SectionsIi a shaft has a circular cross section and thematerial is elastic, the shearing stress in theshaft varies linearly with the distance from thecenter of the shaft (p) and is given by

shearing stress: t =

J

Also the angle of twist () of one end relative to theother is given by

. tL TLangle of twist: 0 = — = —

pG JG

The maximum shearing stress in the shaft is at theoutside surface and equals:

Tr3tmax

=

The oolar moment of inertia is

icr4solid section: J =

hollv section: J = (r -r14)

Also the shaft has maximum and minimumnormal stresses which are tensile andcompressive and act at 450 to the shaft axis.Their values are

Tr3lens 0camp max

—

\\

1_•

0c0mp Tm0,

/•fl TmQ

226

Section 3: Beam Behavior

a. Bending Stresses (elastic)

3

If the loads on a beam act in its plane of a

symmetry and the beam is elastic, thebending stresses acting normal to the L._._Licross section vary linearly with the

S..rfoc,

distance from the neutral axis (N.A.) andhave the value

)

bending stresses: a =

In the absence of axial loads

centrodial axis = neutral axJJ

In the sketch the cross section is shown rectangular. However, the crosssection, in general, can be circular, triangular, etc. Many sections such asT-, I-, H-sections can be found in handbooks. lithe section is notstandard, you must be prepared to derive the centroidal location as wellas the value for 1NA• The maximum bending stress occurs at the fiberwhere (My) is maximum.

maximum bending stress: a = Z = section modulus =max z

b. Shearing Stresses in Beams

The transverse and longitudinalshearing stress in a beam is

iven by

V

4JA1

where Q is the first moment of the shaded area with respect th thecentroidal axis if the shearing stress is evaluated at line 1-2

Now Q = [(c-a)t][(c+a)12] {for rectangular section}

The maximum shearing stress will occur where Q/t is maximum. 0 isalways maximum at the centroidal surface. However, Q/t may or may notbe maximum at the centroidal surface. You check all possibilities.

Shear flaw = q = .2 (lb/in.)

4

You must be able to draw shear

__________

and moment diagrams for beamsin order to determine themaximum values of M and V.

b. Beam Deflectioris.

__________

The deflections of beams isdetermined from the equation

Ely”

where primes denote derivatives with respect to x and M is the bendingmoment equation. Integration of this equation determines the slopeequation.

E[Y’=fMx+Cij

Additional integration gives the deflection equation or the equation of theelastic curve

Ely = f[fMdE)dx + C1x + C2

The constants C1 and C2 are found from the slope and deflectionboundary conditions.

The solution to these equations for many different types of supports andloadings can be found in handbooks. Using the known handbooksolutions and the principle of superposition many other problems can besolved.

For slender columns L’Ir> 100 (steel) the buckling loads andstresses are given by

2EJ it2Ep = 0 =Cr

(L’)2 (L’Ir)2

where L’ is the effective length of a pinned-pinned columnand r is the radius of gyration r = /(IIA)

fixed-fixed L’ = L/2fixed-free L’ = 2L

fixed-pinned L’ = 0. 7L

L

Section 4: Bucklino

PC,

(

P_

See handbooks for codes for columns of intermediate length.

m26

Section 5a: Stresses at a Point: Mohr’s Circle for Stresses

STRESS

5

0+00

=

a 2

2I 1 —

oYJ%J I, 2 ry0 +0x y +

2

0 +0cos 20 - t sin 20

2 = 0a + R

= ax - at,,1 sin 26 + ti,, cos 20

2

tan 20 =

(a — a,,)

2

= 0a - R

Rotation on Mohr1scircle is twice the rotation on the element and in the same direction

The top half of the plane corresponds to shear stresses trying to rotate the elementclockwise; the lower half to shear stresses which try to rotate the elementcounterclockwise

For plane stress the third principal stress is zero= !4(o - an); where o, a, and a are labled right to left along the horizontalaxis.

Radius = R = = R

2226

Section 5b: Strains at a Point: Mohr’s Circle for Strains

+ € ya 2

V STAIN

2

2

6

€ +Ex y +

2

€ +€ycos

220 - sin 20

2€ =€ +Rmax a

YxY — €x — Ey

2 2sin 20 + cos 20

2- R

Yy

tan26= 2

-

2

Rotation on Mohr’s circle is twice the rotation on the element and in the same direction

The top half of the plane corresponds to shear strains trying to rotate the directionclockwise; the lower half to shear strains which try to rotate the directioncounterclockwise

For plane strain the third principal stress is zero

Ymax = (e - c); where , c, and c are tabled right to left along the horizontalaxis

Radius=R=2

I -

N 2y=2R

22Z6

C(1)

a,C)

0,

a,04.-

4-a

Va,E(I,(I)(1

a,

a,

VC(U.c

VCD

(I)CD

U)

U)

(0000.2

(I)0

(U)

;;a,

CDCD

CDQ

E.9

.9.9

a)U

)(0

(0a,

(U>

-

++

•0t

>‘-

CDG

)CD

—4-a

U)

c’ac

U)

(E.9

..9

0C

U)

U)

U)

QU

)a,

++

4-

‘-Q

Ua)L

Cl)

CD

CDE

-Cw

L_

fi

c,ac’i

qU

)U

)U

)C

)..—

00

0a0,

£I0

0C

)4-

aC

))(

0w

c)(a)

‘-6

rII

tiI)

-oC

)

iø

(a)(a)

Ø

U)

C0(Ua,.E

>—

f

Q.

EU)

-cC)

I)>

—-—

-—

——

-—

U)

0a.

0a.-a,EE0C.)

>‘

C(U

(a)+0

Ci)+(a)

oC

%J

0(i)’

ftft

Itx

(a)(a)

>-

(I)0(a)

RC

a)(a)

Ca)C

’J

IIII

Ca)Ci)

>-

x

0CDtJ

>N

(a)

(a)

II

0(a)

0(a)(a)

><

0CDL

)

8

Section 6: Stress-Strain Relations

If the material is linear-elastic (Le, stress proportional to strain) and isotropic, then

ç = —

+= (1 +v)(j _2v)[(1 — v)E + +

=— v(a÷a)j

= (l+v)(1_2v)[(1 )€ +

=—

= (1÷v)(1 _2v)[(1 —v)€ +

y=t,JG

y=t/G t=Gy

y=tG t,1=Gy

If we have plane stress where: o = = = 0, then

ç = vci)= (i V2)(€x

+ v€)

=— var)

= (i _v2)’+ v€)

= —_(a+o)

V oz=o

txy = GYXY

where p is the value of Poisson’s ratio determined from a uniaxial tensile or

compression test= €lateral

G is the shearing modulus of elasticity G = tly, and related to E and p by E = 2(1 + IL)G.

1226

9

Pressure Vessels

p = pressure r = radius t = wall thicknessRelationships are valid for thin-walled vessels (hr < 0.1)

For any coordinate system on the outside surface

= (1 )(€X ÷

pr—÷p

t

€ = (€ ÷c)r(1—!.t)

Spherical:

o = pr/2t; (In any direction)

= o = pr/2t

= -p; (on the inside surface)

pr—+p2t

= 2(on the inside surface)

I fpr€ = = —l—(I

-

E[21

1

/

A ,

Hoop

Cylindrical: (A=axial; H=hoop)

= pr/2t ; 0H = pr/t

GPIOH °P20A

= -p (on the inside surface)

tmx2

iIpr(1 —

=

rpr€ = _l_(

A E121- 2k)]

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10

Statically Determinant 2-force Members

P4 f -

Equilibrium — - F1 + F2 - F3 + F4 = 0P = F1 (Tension) = F4 - F3 + F2P = F1 - F2 (Tension) = F4 - F3

= F1 - F2 + F3 (Tension) = F4PLL’ A1-JAB “AB

Thermal— r

UAB- 1AB 1AB

Statically Indeterm inant 2-Force Members

p p

.AU

gid i....[

L = LengthE= Modulus of ElasticityA = Areaa = Coefficient of Thermal expansion

= Temperature change

•19

-

Stiffness:

=

____

LAaAM)

Stiffness:AE

(8- LA UA 7)

A8E8(8 - Cl ÷ LB a8 7)

Equilibrium:A + P8 = P

- Cl + La8tT)

Equilibrium:PA(a) + PB(b) = P(b+c)

Member A was assumed in tension,therefore subtract LaSTMember B was assumed incompression, therefore add LatT

11

Torsion

Stress= Tr/J

Itt 4 4 Itt 4 4I = —d - d, = —ir —

32’ 0 2’°T = Torquer = radiusL = Length

Rotation

JGG = Shear Modulus (Modulus of Rigidity)

Power Transmission

Power = (Torque)(Angular Velocity); (Angular Velocity must be in radians/seconds)1 hp = 550 ft-lb/sec1 RPM = (2rT/60) rad/sec

Transmitting 100 hp at 3000 RPM, what is the torque?

= lOOhp (550fl_ThIsec3O0ORPM hp

Indtrminant Torsion

RPM= 175fl-lb

2it—md/sec60

TA+TB=T

TB =

___

T1 + T11 = TJGr =

= JI!!e

Brass sleeveAluminum Care

z:zzzz®

b + °

I

U

0

I—

T,e

12

Beams

- bh3

12- hb3

12= bh3

+ Ad2a12

I-Beam

= (1.5x3)3÷

____

+ (4X1X2)2J 39375j,4

Qcentro,d (1 5X1 .5)( 1) + (4)(1X2) = 9.6875 jpi3

QA = (4)(1)(2) = 8 in

The width (t) at the centroid or A is 1.5’. At the bottom or

______________________

top the width is 4’.

Failure is most likely where V or M is a maximum asobtained from the shear and moment diagrams.

a) Top or Bottom, Pure Bending, a = My/Ib) Centroid, Pure Shear, t = VQ/Itc) Wide-Flange Intersection (Point A), combination of bending and shear.

Find principal stresses and maximum shear stress.

o = My/It = VQ/ItM = Bending Momenty = distance from centroidal axis of bending

= 2’ Area Moment Of InertiaQ = 1 Area Moment = A9t = width

r -/ A

fl

//

D

0—- - -

12Z6

13

Beam Deflections

1. Find the support reactions usingStatics.

2. Find the moment as a piecewise 325 k

continuous function of x.

3. Integrate twice to get the deflection. ‘V.

_______

325kip D-xH -10---

Region 1: 0 x 5’ Region 2: 5’ x 10’

M1 = 3.25x - M2 = 7.5 - 0.75x2 ,,3

E101 = 3.25 .i - .i+ El02 = 7.5x - 0.75.i_ + C2

2 6 23 4 2 3

Ely1 = 3.25 - .-. + C1 x + D, Ely2 = 7.5 .-. - 0.75 + C2x + D26 24 2 6

4. Solve for the constants of integration using the boundary and continuity conditions.Boundary: y(0) = 0; y2(10) = 0Continuity: ø(5) = 02(5); y(5) = Y2(5)

5. The above equations can then be used to solve for the slope or deflection at anypoint along the beam.

Superposition - useful for getting the deflection at a point.

1 kp/ft 1 /ftIII H HH H 11111 H 11TH 11111 HIll HI H H Ill

5- 5- 5. 5, 5 5,1 <p 1 kip. 6 1 kip, 62

Center Deflection: = ö1 + Y2ö2 L Same cemter deflection I

1 kip/ft I kip/ft I kip/ftHHIIIIIIHHIII!IInhIIITII wit +

.

14

Indeterminant Beams

1. Remove enough supports to make the beam statically determinant, and replacethem with unknown loads.

2. Require that the unknown loads beand slopes.

or

such that they give the appropriate deflections

P must be such that the deflection of theright end is zero.

M must be such that the slope atthe left end is zero.

orM2

M1 and M2 must make the slopesat both ends zero.

P and M must make both the slope anddeflection at the right end zero.

p1

8 = wL48 El

- PL3-

3E1

/ //

P = 18.75 kipsWall Reactions

= 0 = 18.75 + R - 50R=31.25kips I

= 0 = -(50)(5) + (18.25)(10) + MM = 62.5 kip-ft (CCW)

5 Hft

U

2i22€

15

REVIEW PROBLEMS

1. An aluminum bar having a constant cross sectional area of ¼ in2 carries the axialloads applied at the positions shown. Find the total deformation of the bar.

a. _0.0192 in.

b. 0.2880 in. 80001b 6ft 60001b— 2000th

c. _0.3264 in.E 10x108d. _0.3840 in.

e. _None of these.

2. A steel rod with a cross sectional area of ½ in2 is stretched between two rigid walls.The temperature coefficient of strain is 6.5 x lO4in.Iin./°F and E is 30 x lO6psi. Ifthe tensile load is 2000 lb at 80° F, find the tensile load at 0°F.

a. _5800 lb.

b. _7800 lb.

c. _8800 lb.

d. ..9800 lb.

e. _19,600 lb.

3. The composite bar shown is firmly attached to unyielding supports at the ends andis subjected to the axial load P shown. lithe aluminum is stresses to 10,000 psi,find the stress in the steel.

a. _1000 lb.1511.

k ‘)ññf’i IA

______________

Id. _.Id’.JW JId.

frJ. St. -

c. _5000 lb.-i A1.5 11 A—20 112

lnrwri1.. 6 —

U. — I V,VVV Ii). E lOx 10 P& E = 30 x lo6p&

e. _20,000 lb.

2i2Z6

16

4. Find the length L to make the relative total angle of twist of both ends of the shaftequal zero.

a._3ft.

b._4ft.

6. A hollow aluminum shaft and a solidsteel shaft are rigidly connected ateach end. This compound shaft is thenloaded as shown. Determine themaximum shearing stress in eachmaterial and the angle of twist of thefree end. Assume elastic behavior.Gaijm = 4 X 106 psiG =l2xlO6psi

HL

fl

/.

e

I /I I

. I IA’.d‘ /

t. 62800 lb

b

1000 lb

E•

c._6ft.

d. 9 ft.

e. _12 ft.

5. Determine the shearing stress atpoints A and B which are at theinside and outside surface of thehollow shaft. Assume elasticbehavior.

__6+T_

1. 62800 lb

2/2Z6

17

7. Determine the maximum bending moment in the beam.

a._3600 ft-lb.

b. — 5400 ft-lb.

c. — 7200 ft-lb.

d. _8100 ft-lb.

4Ø lb/Fl

uwwwI— 6Fi 6-ri

e._4050 ft-lb.

8. Find the maximum transverse shearing force in the beam shown.

a. — 450 lb.

b. — 1800 lb.

C. — 2250 lb.

d._3600lb.

e. — 4050 lb.

gee b zr1

5-ri

9. By means of strain gages, the flexural stresses are found to be -12,000 psi at Aand +4000 psi at B. Assuming the elastic limit of the material has not beenexceeded, find the flexural stress at the bottom of the beam.

a. — 6000 psi.

b._8000 psi.

c. — 9000 psi.

d. — 10,000 psi.

e. — 12,000 psi.

10. For the cast iron beam shown, the maximum permissible compressive stress is12,000 psi and the maximum permissible tensile stress is 5000 psi. Find themaximum safe load P applied to the beam as shown.

A $e-ci r 4-ft.

a._220lb.

b. — 333 lb.

c....12501b. [d._3000lb.

S.c,o.’ A

e. _7500 lb.

18

11. A 12-inch, 35-lb I-beam 30 ft. long is supported at 5 ft. from each end and carries auniformily distributed load of 1600 lbs per ft including its own weight. Determinethe maximum flexural stress in the beam.

12. Find the maximum vertical shearing force which may be applied to a box beamhaving the cross section shown without exceeding a horizontal shearing stress of500 psi.

a._3065lb.—

b._4000lb.

c. 6000 lb.

d._61301b.

e._63001b.

13. Find the reaction at the rightend of the beam shown.

a._wLJS

b.wU4

c. 3wLI8

_______________________

14. Two beams, simply supported at their ends, jointly support a load P = 3500 lb.applied to the upper 6-ft. beam at its midpoint. The beams are identical except forlength and cross at their midpoints. Find the load carried by the lower 9-ft. beam.

a. 700 lb

b._8001b

c. — 1000 lb

d. 1750 lb

e. _2700 lb

NA

‘—4- IN—

6-IN

o orino +L

Z226

19

15. Determine the deflection at the end of this L 1 0 ftbeam. 200x0 b—in

— I ft’fTh

i

____ ____

P=iQ,000 b

16. The critical Euler load for the pin-endedslender column restrained at the midpoint asshown in Fig. A is 1000 lb. What is the criticalEuler load for the same column with themidpoint restraint removed as shown in Fig.B.

a.

________

250 lb.

b.

______

5001b.

C.

______

7501b. Fig. A Fg.d.

______

l000lb.

e.

_______

4000 lb.

17. A rectangular bar is loaded as shown. Find the maximum tensile stress developedover section A-A.

4ft 6ft —I12,000 b A r’4 12,000 b

2Z6

20

18. For stress conditions on the element shown, findthe principal stresses and the plane on which themaximum principal stress acts.

19. A circular shaft of brittle material subjected to torsion fractures along a 45° angle.Failure is due to what kind of stress?

a.

________

Shearing stress.

b.

_________

Compressive stress.

c.

_________

Tensile stress.

d. Combined stress.

e.

________

None of these.

20. Which has the higher shear stress for a given elastic torque?

a. a one-inch diameter rod, or b. a two-inch diameter rod.

21. Identical rods of aluminum and steel are each subjected to the same elastic torque.Which rod will have the higher shear stress?

a. steel b. aluminum c. both the same stress

22. If G represents the modulus of rigidity (or shear modulus of elasticity), E is modulusof elasticity, and p is Poisson’s ratio, which of the following statements is true forany homogeneous material?

a.

_______

G is independent of E.

b.

_______

G is 0.4E.

C.

_______

G is 0.5E.

d.

________

G depends upon both E and p.

e.

________

None of these.

3DOD DS

DS

2f226

21

ANSWERS EIT REVIEW

MECHANICS OF MATERIALS

1. C

2. d

3. e

4. d

5. T8=5,333.3 psi TA=2,6€6PS

6. Ts = 81,528 TA = 54,352 psi

8 = 1.086 rad.

7. e

8. c

9. b

10. d

11. ,=19,100psi

12. d

13. c

14. b

15. 19.8 in.

16. a

17. 8,000 psi

18. umax = 605 psi 0mãn = 6606 psi

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