chapter 4 stresses in beam equation of bending stress
TRANSCRIPT
Darshan Institute of Engineering & Technology Civil Engineering Department
Equation of bending stress (Flexure eqn)
(i) Before bending (ii) C/S of Beam Stress diagram (iii) Before bending
Consider a layer PQ at a distance Y from the neural axis.
after bending layer of PQ compressed to PQ’
Decrease in length of PQ layer
l PQ - P'Q'
l PQ - P'Q' Strain ( ) _______________(1)
original length PQ
Now from geometry of the fig two section OP’Q’ & OR’S’ are similar.
n
P'Q' P'Q' 1 1
R'S' R'S'
R'S'-P'Q' R'S' RS PQ Neutral layer
R'S'
PQ-P'Q' ______________________________(2)
PQ
from eq (1) & (2)
stre
R y R y
R R
R R Y Y
R R
Y
R
Ybending
R
ss ( ) E
E
____________________________________(3)Y
Y
R E
E
R
Now for fig. i.e. C/S of beam, in which PQ small layer is at a distance Y from NA.
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
δa= Area of PQ layer
As we know that from qun (3)
Algebraic sum of all such moment about NA
Now, from qun (3)
Example 1 A simply supported beam of span 4.0 has a cross-section 200mm x 300mm. If the permissible
stress in the material of the beam is 20n/mm2, determine maximum udl it can carry.
Answer:
3 36 4
xx
2
200 300I 450 10
12 12
20 /
300150
2 2
bdmm
N mm
dy mm
nflexure eq M
I Y
RY
E
aa
NAabout force thisofMoment
a a layer PQ oflength in Decrease
2
R
EYY
R
EY
R
EY
a I
a
a M
2
2
2
YIR
E
YR
E
YR
E
R
Y
E
I
M
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
6620 450 10
M 600 10150
IN mm
Y
Now, for simply supported beam with U.D.L on entire span
26
max
2
6
M 600 108
4000 600 10
8
300
300
wl
w
w N mm
w kN m
Example 2 A cast iron test beam 30 mm square in cross-section 500 mm long is simply supported at end.
It fails at central point load at 4.32 KN. What load at free end will cause the failure of cantilever beam of
1 m span made of same material 30 mm x 60 mm in cross section?
Answer: (1) For simply supported beam
4 4 4
4
xx
3
n
2
30I 67500
12 12 12
3015
2 2
4.32 10 500540000
4 4
Mflexure eq
I
540000 120
67500 15
d bmm
dy mm
wlM N mm
Y
N mm
(2) For cantilever beam
3 33 4
xx
30 60I 540 10
12 12
3015
2 2
bdmm
dy mm
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
n
3
6
6
M M 120flexure eq
I 540 10 30
M 2.16 10
For cantilever M w
2.16 10 w 2160
1000
Y
N mm
l
N
w 2.16 KN
Example 3: A section of beam as shown in fig. is subjected to a bending stress of 10 kN.m about the major axis. Draw bending stress distribution across the section.
Answer:
(1) Center of a given beam
Sr.no Shape Area (mm2) Y (mm) AxY (mm
3)
1
A1=100x10
A1=1000
Y1=90+𝟏𝟎
𝟐
Y1=95
A1Y1=
95000
2
A2=90x10
A2=900
Y2=𝟗𝟎
𝟐
Y2=45
A1Y1=
405000
∑ A = 1900 ∑ AY = 135500
b
t
t
135500Y
1900
Y 71.32
Y 71.32
Y 100-71.32
Y 28.68
AY
A
mm
mm
mm
(2) Moment of inertia about centroidal horizontal axis
Sr.no. Area h (mm) Ah2 (mm4) IG or Iself (mm4) Ixx = IG+ Ah2(mm4)
1 A1=1000 h1=Yt- 𝒅𝟏
𝟐 =23.68 A1h1
2= 5607242.34 IG1=𝟏𝟎𝟎×𝟏𝟎𝟑
𝟏𝟐 = 8333.34 I1 = 569075.7
2 A2=900 h2=Yb- 𝒅𝟐
𝟐 = 26.32 A2h2
2= 623468.16 IG2=𝟏𝟎×𝟗𝟎𝟑
𝟏𝟐 = 607500 I2 = 1230968.16
C2 Y2
Y1
C1
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
(3) Bending stress
M = 10 KN = 10 X 103 X 10
3N·mm = 10 X 10
6N·mm
Ixx = 1.8 X 106 mm
4
Ytop = 100 – 71.32 = 28.36 mm
Ybottom = 71.32 mm Bending stress at top
M
I tY
6
6
10 10
1.8 10 28.68
2
top 159.34 N mm
Bending stress at bottom
M
I bY
6
6
10 10
1.8 10 71.32
2
bottom 392.22 N mm
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
Shear stress distribution across beam section (equation of shear stress)
Consider a small portion ABCD of length δx of abeam loaded with U.D.L as shown in fig (a).
Shear force (S.F) & Bending moment (B.M) will change at every point along the length of the
beam.
Let, M = B.M at AB
M+ δM = B.M at CD
F = S.F at AB
F+ δF = S.F at CD
I = M.I. of the section about the neutral axis.
Now consider an element strip EF at distance Y from the neutral axis.
a = cross section area of strip EF.
Σ = bending stress at AB.
Now, using
Y
Y
I
M
I
M eq flexure n
similarly, Y across CD section
M
I
acting across CD acting across AB
M Mbending stress
I
Y
Force Force
a
____ i a
stress area
M MY a
I
M
Y a _____ iiI
M
A C
E F
B D
Y Y N.A
M+δM
δX
d
b Fig.(a)
Fig.(b)
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
Net unbalanced force on the strip
Now, total unbalance force above neutral axis may be found out by integration above equation
between 0 &2
d
Ya
dYYadYYa
dYaY
I
M F
I
M
I
M
I
MF
2
d
0
2
d
0
2
d
0
Where, A = area of beam above N.A
Y = distance between the centroid of the area & N.A
bI
bI
YA
x
M
Area
forceshear total stressshear ,
YAF
bx
YAM
M
Now
FSF
ngsubstituti
.x
M
aY
aYaY
I
M
I
M -
I
MM
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
Prove that the maximum shear stress in a rectangular section of beam is 1.5 times of average
shear stress.
Consider rectangular beam as shown in fig.
b = width, d = depth
(i) Average shear stress:-
We know that average shear stress is given as
τavg = 𝐹
𝑏×𝑑
(ii) Maximum shear stress:-
Maximum shear stress occurs at neutral axis which lies at 𝑑
2 form top & bottom.
avg
db
F
db
F
bd
F
bbd
ddbF
bddY
bddhear
Ib
YFA
5.1
5.12
32
3
2
3
12
42
12 I ,
2
2
2
d A ,
max
avg
max
3max
3
max
d
𝑑
2
b
𝑌 =𝑑
4
𝑑
2
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
Prove that the maximum shear stress in a circular section of beam is 4/3 times of average
shear stress.
Consider a plane circular lamina having diameter‘d’
(i) Average shear stress:-
τavg = 𝐹
𝐴=
𝐹𝜋
4𝑑2=
4𝐹
𝜋𝑑2
(ii) Maximum shear stress:-
avg
d
Fd
F
d
F
dd
ddF
bd
ddhear
Ib
YFA
3
4
33.13
4
43
16
Now,
3
16
64
3
2
8
db , 12
I
3
2d
3
4rY ,
8 )
4(
2
1 A ,
max
2
2
avg
max
24
2
max
3
22
max
Maximum shear stress lies at neutral axis which
passes through its centroid.
The centroid of circle lies at 𝑑
2 form top & bottom.
A
A
d = 2r = dia
𝑌 =4𝑟
3𝜋
N
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
Example . A section of beam as shown in fig. is subjected to a shear force of 20KN. Draw
shear stress distribution across the section.
Ans:-
Shear force on beam = 20 KN
(i) shear stress at junction of flange & web.
Ytop = 28.68 mm, Ybottom = 71.32 mm
A Y = 100 10 × 28.68−10
2 = 23680 mm
3
(ii) Shear stress at C.G.
mm 100 b
1.8 I ,bI
YFA
2
2
6
3
N/mm 26.3 10
1002.63
stress.shear in the increasesudden is therejunction, the
/ 63.2100108.1
236801020
bI
At
mmNYFA
3 712.25424
71.174423680
2
68.181068.1855.2810100
mm
YA
2
6
3
max / 25.28100108.1
712.254241020
bImmN
YFA
100 mm
100 mm
10 mm
10 mm
τ = 0 τ = 2.63
N/mm2 26.3 N/mm
2
τmax = 28.25
N/mm2
Yt=
28.68
mm
Yb=
71.32
mm
10
mm
100 mm
28.68
mm
N A
shear stress diagram
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
Ex- Fig. shows a beam cross section subjected to a shear force of 200 KN. Determine the
shearing stress at neutral axis & sketch the shear stress distribution diagram.
Ans. (1) Center of a given section
Sr.no Shape Area (mm2) Y (mm) A*Y (mm3)
1
A1=10*120 A1=1200
Y1=
10+120+𝟏𝟎
𝟐
Y1= 135
A1Y1=
162000
2
A2=10*120 A2=1200
Y2 =10+𝟏𝟐𝟎
𝟐
Y2 =70
A1Y1= 84000
3
A2=10*120 A2=1200
Y2 =𝟏𝟎
𝟐
Y2 =5
A1Y1= 6000
∑ A = 3600 ∑ AY = 252000
(2) Moment of inertia about centroidal horizontal axis.
Sr.no. Area h (mm) Ah2 (mm4) IG or Iself (mm4) Ixx = IG + Ah2 (mm4)
1 A1=1200 h1=Yt- 𝒅𝟏
𝟐 =65 A1h1
2=5.07X106 IG1=
𝒃𝟏×𝒅𝟏𝟑
𝟏𝟐=1X104
I1 = 5.08X106
2 A2=1200 h2=Yt- Y2 = 0 A2h22=0
IG2=𝒃𝟐×𝒅𝟐
𝟑
𝟏𝟐=1.44X106
I2 = 1.44X106
3 A3=1200 H3=Yb- 𝒅𝟑
𝟐 = 65 A3h3
2=5.07X106 IG2=𝟏𝟎×𝟗𝟎𝟑
𝟏𝟐=1X104 I2 = 5.08X106
Ixx= I1 + I2+ I3=11.8 x 106 mm4
mm
A
AY
70
2
Y Y Y Y
3600
252000 Y
tb
120 mm
120 mm
10 mm
10 mm
120 mm
C2 Y2
Y1
C1
Y3 C3
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
(3) Shear stress
Shear stress at the junction of flange & web
Now, Shear stress at C point.
Shear stress at neutral axis.
From geometry of a given section is symmetric about centroidal axis (Horizontal &Vertical)
2
3
3
1
46
3
N/mm 8125.0
120106.11
78000105.14
Ib
YFA
mm 120b ,106.11
780002
1010120YA
Ib
YFA KN, 14.5F
B
B
xx
t
mmI
mmY
2
2
12
3
3
2
2
463
N/mm 75.910
120
b
b N/mm 75.9
elyAlternativ 10106.11
78000105.14
Ib
YFA
changes only width in mm 10bb
106.11 ,78000YA KN, 14.5F
BBCB
B
B
xx mmImm
2
3
3
3
1
221
11
2
46
N/mm 12 10106.11
96000105.14
mm 96000YA
2
60
2
10120
2
107010120
2YA
mm 10bb ,106.11 KN, 14.5F
D
xx
YA
YA
mmI
EcFBGA , ,
Yt=
70mm
A
C B
D
E
G
F
𝑌
1
2
3
𝑌1′
Yt=
70mm
C1
D
C2 𝑌2′
A
C B
D
E
G
F
τ = 0
0.815 N/mm2
9.75 N/mm2
12 N/mm2
τ = 0
0.815 N/mm2
9.75 N/mm2
shear stress distributiondiagram
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
Example As shows figure a cross section subjected to a shearing force of 200 kN. Determine the
shearing stress at neutral axis& sketch the shear stress distribution diagram across the section.
Ans. Given section is symmetric about Horizontal & Vertical axis
Moment of inertia about centroidal horizontal axis.
Now, Ixx = I1 + I2 + I3= 129.167 x 106 mm4
→ Stress at point B (shear)
→Stress at point C
S
n Area h (mm) Ah
2 (mm
4) IG or Iself (mm
4) Ixx = IG + Ah
2
1 A1=
50X100
=5000
h1=Yt- 𝟏𝟎𝟎
𝟐
= 100
A1h12=
5X107
IG1=𝒃𝟏×𝒅𝟏
𝟑
𝟏𝟐
=4.167X106
I1 = 5.4167x107
2 A2=
100X250
=25000
h2=0 A2h22= 0 IG2=
𝒃𝟐×𝒅𝟐𝟑
𝟏𝟐
=2.0834X107
I2 = 2.0834x107
3 A3= 5000
H3=Yb- 𝟏𝟎𝟎
𝟐
= 100
A3h32=
5X107
IG2=
4.167X106
I2 = 5.4167x107
mm 150 Y Y Y bottomtop
2
6
3
11
2
N/mm 48.15
5010167.129
100500010200
50bb ,1002
Y
KN 200F ,mm 500010050
B
B
t
Ib
YFA
mmmmd
Y
A
2N/mm 09.3250
50 BC
2
6
3
1
2
22
11
222
2111
N/mm 03.525010167.129
81250010200
50bb
mm 812500'2
'
42b
22db
Ib
YFA
mm
YA
YA
ddddA
G
𝑑
2
100mm
100mm
100mm
100
mm
100
mm
50
mm
b1= 50mm, d1= 100mm
b2= 250mm, d2= 100mm b3= 50mm, d3= 100mm
C
D E
A
B
G
F
Yt YB
𝑌1′
𝑌2′
C
A
B
G
C
A
B
G
τ = 0
3.09 N/mm2 15.48 N/mm
2
5.03 N/mm2
3.09 N/mm2
15.48 N/mm2
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam
Darshan Institute of Engineering & Technology Civil Engineering Department
Draw shear stress distribution diagram for standard section
(1) Rectangular section (2) Solid circular section
(3) Triangular section (4) I - section
(5) T - section (6) T - section
(7) Hollow rectangular section (8) Solid square with diagonal horizontal
(Rhombus)
τNA =4
3τavg
τmax = 1.5 τavg τmax =
4
3τavg
τmax = 1.5 τavg
A N
τmax
τmax
τ= 0
τmax
h
4
3h
τmax = 1.125 τavg
C C
Mechanics of Solids (3130608)
Chapter – 4 Stresses in beam