Mechanics of Materials
Engr. Khadija Saliha
Department of Civil Engineering, NFC IEFR Faisalabad
Outline
• Bending stress
• Bending stress examples
• Shear stress
• Shear stress examples
• Shear flow
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Shear stresses in beams
VQ
=−
Ib
Q: the first moment.
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Shear stresses in beams
▪ Rectangular Cross
Section
- b = constant
▪ I beams
b = variable
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First moment Q
General equation:
Q = ydA
AQ
AQ is the cross sectional area above the level at which
is being evaluated. y is the distance to the centroid.
or divide
y
AQ into n parts
i
n
Q = yi A i
i =1
y i is the distance between the centroid of part (i) and the
centroid of the total cross-sectional area. Ai is the area of
part (i).
Unit: mm3 or m3
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Case 1: rectangular beam
y • First moment (Q) for position y1.
The area (A1) of the shaded part is the area
A1
h/2
above position y1.
y1
y1 h
z A1 = b − y1
2
Centroid
(neutral axis)
h/2
The distance between the centroid of the
shaded area and the cross-sectional area is y i .
b
h / 2 + y1
y1 =
2
So
b h2 2
Q = y1A1 = − y1
2 4
Q decreases with y1. Therefore, Q has the maximum
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value at y1=0, namely the neutral axis, for rectangle beam
Case 2: wide-flange beam
• First moment (Q) for position y1.
The shaded part is divided into two parts: A1
y
and A2.
h h h1
A1 A1 = b − 1 A2 = t − y1
h/2
2 2 2
h1/2 A2
y2
y1
The distances between the centroid of each
y1
z part and the cross-sectional area are
h + h1 h1 / 2 + y1
(neutral axis)
t
y1 = y2 =
4 2
So
Q = y1A1 + y2 A2 = (h − h1 ) + (h1 − 4y12 )
b
b 2 2 t 2
8 8
Maximum: Q=
8
(
b 2
h − h12 ) + h12
t
8
at y1=0
Q = (h − h12 )
b 2
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Minimum: at y1= h1/2
8
Shear flow
Shear flow is the horizontal shear force per unit
distance along the longitudinal axis of the beam.
VQ
f = b = −
I
The strength of a weld is the
shear flow at the connection
divided by 2 since there are two
welds at the connection.
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Example #1
Calculate the maximum shear stress τ max and the maximum bending stress σ max
in a wood beam (see figure) carrying a uniform load of 22.5 kN/m (which
includes the weight of the beam) if the length is 1.95 m and the cross section is
rectangular with width 150 mm and height 300 mm, and the beam is simply
supported as in the figure.
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Solution
q=22.5kN/m
A B
FAx
FAy FBy
L=1.95m
(1) ΣFx=0 FAx=0
(2) ΣMB=0
L
FAy L − q L = 0
2
L
FAy = q = 21 .94 kN
2
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q=22.5kN/m (3) Shear force and bending moment
A
M
V
FAy x
ΣFy=0
V
FAy − q( x ) − V = 0
Vmax=21.94kNm
V = FAy − q( x ) = (21 .94 − 22 .5x )kN
x
Vmin=-21.94kNm
ΣMA=0
x
M
M − q( x )( ) − V( x ) = 0
Mmax=10.7kNm 2
x
M = q( x )( ) + V( x ) = (21 .94 x − 11 .25x 2 )kNm
2
x
The maximum shear force Vmax is 21.94kNm at the supports.
The maximum bending moment Mmax is 10.7kNm at the middle point of beam
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where the shear force is zero.
(3) Maximum shear stress
bh 3 150 300 3
I= = = 3.375 10 8 mm 4
12 12
b h2 2
Q = − y1
2 4
The maximum shear stress occurs at the neutral axis y1=0
bh 2 150 300 2
Q max = = = 1.688 10 6 mm 3
8 8
The maximum shear stress is
Vmax Q max (21 .94 10 3 N)(1.688 10 6 mm 3 )
max =− =
Ib (3.375 10 8 mm 4 )(150 mm )
= −0.731 N / mm 2 = −0.731MPa
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(4) Maximum bending stresses
The maximum bending stress occurs at the top surface y=h/2
M max M max h (10 .7 10 6 Nmm ) (300 mm )
max =− y=− =− = −4.76 MPa
I I 2 (3.375 10 mm )
8 4
2
The maximum stress occurs at the bottom surface y=-h/2
M max M max h (10 .7 10 6 Nmm ) (300 mm )
max =− y=− (− ) = = 4.76 MPa
I I 2 (3.375 10 mm )
8 4
2
The maximum compression occurs at the top surface,
the maximum tension occurs at the bottom surface.
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Example #2
For the beam shown below, determine:
— The maximum shear stress
— The stress at the section (determine τ at a point 1” from the top of the beam and
1 ft from the left of the beam)
• The cross-section of the beam shown is 2" wide x 4" tall.
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Example #3
A cantilever beam of length L = 2 m supports a load P = 8.0 kN (see figure). The
beam is made of wood with cross-sectional dimensions 120 mm x 200 mm.
Calculate the shear stresses due to the load P at points located 25 mm, 50 mm, 75
mm, and 100 mm from the top surface of the beam. From these results, plot a
graph showing the distribution of shear stresses from top to bottom of the beam.
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Solution
P
MA
A B
FAx
FAy
L=2.0m
(1) ΣFx=0 FAx=0
(2) ΣMB=0
− M − P( L) = 0
M = − P( L) = −16kNm
(3) ΣFB=0
FAy − P = 0
FAy = P = 8.0kN
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(3) Shear force and bending moment
ΣFy=0
MA FAy − V = 0
A
V = FAy = 8kN
M
V
FAy x
ΣMA=0
− M A + M − V( x ) = 0
M = M A + V( x ) = (8x − 16 ) kNm
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(3) Shear stress
bh 3 120 200 3
I= = = 8 10 7 mm 4
12 12
b h2 2
Q = − y1 = 1 10 4 − 60 y12
2 4
The shear stress is
VQ (8 10 3 N)((60 10 4 − 60 y12 )mm 3 )
=− =
Ib (8 10 7 mm 4 )(120 mm )
1 y12
= − (1 − 4 )MPa
100
2 10
50 τ=-0.5MPa for y1=0mm
y1 τ=-0.469MPa for y1=25mm
0
τ=-0.375MPa for y1=50mm
-50
τ=-0.219MPa for y1=75mm
-100
τ=-0MPa for y1=100mm
-0.5 -0.4 -0.3 -0.2 -0.1 0.0
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Example #4
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Solution
y
bh 3 (b − t )h1
3
h
I= −
h1
z
12 12
(neutral axis)
t
450 (1600 + 2 24 )3 ( 450 − 10 ) 1600 3
I= −
12 12
= 1.766 1010 mm 4
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y A1 = 450 mm 24 mm = 10800 mm 2
A1
h/2
1600 + 24
h1/2 y1 y1 = ( )mm = 812 mm
2
z
(neutral axis)
t
Q = y1A1 = 8.77 10 6 mm 3
b
shear flow (f):
VQ (1300 10 3 N)(8.77 10 6 mm 3 )
f =− =
I (1.766 1010 mm 4 )
= −645 .6 N / mm
The strength of weld is f / 2 = 322.8N / mm
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