CAP_07_CISALHAMENTO_TRANSVERSAL

Prévia do material em texto

Problem 7-1
If the beam is subjected to a shear of V = 15 kN, determine the web's shear stress at A and B. Indicate
the shear-stress components on a volume element located at these points. Set w = 125 mm. Show that
the neutral axis is located at y = 0.1747 m from the bottom and INA = 0.2182(10-3) m4.
Given: bf 200mm:= b'f 125mm:= tf 30mm:=
tw 25mm:= dw 250mm:=
V 15kN:=
Solution:
Section Property : D dw 2tf+:=
A bf tf⋅ dw tw⋅+ b'f tf⋅+:= A 16000 mm
2=
y
⎯ Σ yi
⎯
Ai⋅( )⋅
Σ Ai( )⋅
= yc
b'f tf⋅( ) 0.5tf( )⋅ dw tw⋅( ) 0.5dw tf+( )⋅+ bf tf⋅( ) D 0.5tf−( )⋅+
A
:=
yc 174.69 mm=
I'f
1
12
b'f⋅ tf
3⋅ b'f tf⋅( ) 0.5tf yc−( )2⋅+:=
Iw
1
12
tw⋅ dw
3⋅ tw dw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
If
1
12
bf⋅ tf
3⋅ bf tf⋅( ) D 0.5tf− yc−( )2⋅+:=
I If Iw+ I'f+:= I 218.18 10
6−× m4=
Q Σ yi
⎯
⋅ A1⋅= QA D 0.5tf− yc−( ) bf tf⋅( )⋅:= QA 721875.00 mm3=
QB yc 0.5tf−( ) b'f tf⋅( )⋅:= QB 598828.12 mm3=
Shear Stress: τ
V Q⋅
I t⋅
=
τA
V QA⋅
I tw⋅
:= τA 1.99 MPa= Ans
τB
V QB⋅
I tw⋅
:= τB 1.65 MPa= Ans
Problem 7-2
If the wide-flange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in
the beam. Set w = 200 mm.
Given: bf 200mm:= tf 30mm:=
tw 25mm:= dw 250mm:=
V 30kN:=
Solution:
Section Property : D dw 2tf+:=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:=
I 268.65 10 6−× m4=
Q Σ yi
⎯
⋅ A1⋅= Qmax bf tf⋅( )
D
2
tf
2
−
⎛
⎜
⎝
⎞
⎠
⋅
dw
2
tw⋅
⎛
⎜
⎝
⎞
⎠
dw
4
⎛
⎜
⎝
⎞
⎠
⋅+:=
Qmax 1035312.50 mm
3=
Shear Stress: τ
V Q⋅
I t⋅
=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τmax
V Qmax⋅
I tw⋅
:= τmax 4.62 MPa= Ans
Problem 7-3
If the wide-flange beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the
web of the beam. Set w = 200 mm.
Given: bf 200mm:= dw 250mm:=
tf 30mm:= tw 25mm:=
V 30kN:=
Solution:
Section Property : D dw 2tf+:=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:=
I 268.65 10 6−× m4=
Af 0.5D y−( ) bf⋅= yf 0.5 0.5D y−( ) y+=
yf 0.5 0.5D y+( )=
Qf Af yf⋅=
Qf 0.5 0.5D y−( ) bf⋅ 0.5D y+( )⋅=
Qf 0.5bf 0.25D
2 y2−( )=
Shear Stress: τ
V Q⋅
I t⋅
=
τf
V
I bf⋅
⎛
⎜
⎝
⎞
⎠
Qf⋅=
τf
V
I bf⋅
⎛
⎜
⎝
⎞
⎠
0.5bf 0.25D
2 y2−( )⎡⎣ ⎤⎦⋅=
∫ == A 0dAV τResultant Shear Force:
For the flange,
Vf
V
I bf⋅ 0.5dw
0.5D
y0.5 bf⋅ 0.25 D
2⋅ y2−( )⋅⎡⎣ ⎤⎦ bf⋅
⌠
⎮
⌡
d
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅:=
Vf 1.46 kN=
Vw V 2Vf−:=
Vw 27.09 kN= Ans
Problem 7-4
If the wide-flange beam is subjected to a shear of V = 125 kN, determine the maximum shear stress in
the beam.
Given: bf 200mm:= dw 250mm:=
tf 25mm:= tw 25mm:=
V 125kN:=
Solution:
Section Property : D dw 2tf+:=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:=
Qmax bf tf⋅( )
D
2
tf
2
−
⎛
⎜
⎝
⎞
⎠
⋅
dw
2
tw⋅
⎛
⎜
⎝
⎞
⎠
dw
4
⎛
⎜
⎝
⎞
⎠
⋅+:=
Maximum Shear Stress: τ
V Q⋅
I t⋅
=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τmax
V Qmax⋅
I tw⋅
:=
τmax 19.87 MPa= Ans
Problem 7-5
If the wide-flange beam is subjected to a shear of V = 125 kN, determine the shear force resisted by
the web of the beam.
Given: bf 200mm:= dw 250mm:=
tf 25mm:= tw 25mm:=
V 125kN:=
Solution:
Section Property : D dw 2tf+:=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:=
A1 bf tf⋅= y1c 0.5 D tf−( )=
A2 0.5dw y−( ) tw⋅= y2c 0.5 0.5dw y−( ) y+=
y2c 0.5 0.5dw y+( )=
Qw A1 y1c( )⋅ A2 y2c( )⋅+=
Qw 0.5bf tf⋅ D tf−( )⋅ 0.5 0.5dw y−( ) tw⋅ 0.5dw. y+( )⋅+=
Qw 0.5bf tf⋅ D tf−( )⋅ 0.5tw 0.25dw2 y2−⎛⎝ ⎞⎠+=
Shear Stress: τ
V Q⋅
I t⋅
=
τw
V
I tw⋅
⎛
⎜
⎝
⎞
⎠
Qw⋅=
τw
V
I tw⋅
⎛
⎜
⎝
⎞
⎠
0.5 bf⋅ tf⋅ D tf−( )⋅ 0.5 tw⋅ 0.25 dw2⋅ y2−⎛⎝ ⎞⎠⋅+⎡⎣ ⎤⎦⋅=
∫ == A 0dAV τResultant Shear Force: For the web.
Vw
0.5− dw
0.5dw
y
V
I tw⋅
⎛
⎜
⎝
⎞
⎠
0.5 bf⋅ tf⋅ D tf−( )⋅ 0.5 tw⋅ 0.25 dw2⋅ y2−⎛⎝ ⎞⎠⋅+⎡⎣ ⎤⎦⋅ tw⋅
⌠
⎮
⎮
⌡
d:=
Vw 115.04 kN= Ans
Problem 7-6
The beam has a rectangular cross section and is made of wood having an allowable shear stress of
τallow = 11.2 MPa. If it is subjected to a shear of V = 20 kN, determine the smallest dimension a of its
bottom and 1.5a of its sides.
Given: V 20kN:= τallow 11.2MPa:=
Solution:
Section Property :
I
1
12
a⋅ 1.5a( )3⋅= t a=
Qmax
1.5a
2
a⋅⎛⎜
⎝
⎞
⎠
1.5a
4
⎛⎜
⎝
⎞
⎠
⋅=
Allowablwe Shear Stress: τ
V Q⋅
I t⋅
=
I t⋅
V Qmax⋅
τallow
=
1
12
a⋅ 1.5a( )3⋅ a⋅
V
τallow
⎛
⎜
⎝
⎞
⎠
1.5a
2
a⋅⎛⎜
⎝
⎞
⎠
⋅
1.5a
4
⎛⎜
⎝
⎞
⎠
⋅=
a
V
τallow
:=
a 42.26 mm= Ans
Problem 7-7
The beam has a rectangular cross section and is made of wood. If it is subjected to a shear of V = 20
kN, and a = 250 mm, determine the maximum shear stress and plot the shearstress variation over the
cross section. Sketch the result in three dimensions.
Given: a 250mm:= V 20kN:=
Solution:
Section Property : b a:= d 1.5a:=
I
1
12
b⋅ d3⋅:=
Qmax
d
2
b⋅⎛⎜
⎝
⎞
⎠
d
4
⎛⎜
⎝
⎞
⎠
⋅:=
Maximum Shear Stress: τ
V Q⋅
I b⋅
=
Maximum shear stress occurs at the point where
the neutral axis passes through the section.
τmax
V Qmax⋅
I b⋅
:=
τmax 0.320 MPa= Ans
Problem 7-8
Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 20 kN.
Given: bf 120mm:= tf 12mm:=
tw 80mm:= dw 60mm:=
V 20kN:=
Solution:
Section Property : D dw 2tf+:=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:=
I 5.21 10 6−× m4=
Q Σ yi
⎯
⋅ A1⋅= Qmax bf tf⋅( )
D
2
tf
2
−
⎛
⎜
⎝
⎞
⎠
⋅
dw
2
tw⋅
⎛
⎜
⎝
⎞
⎠
dw
4
⎛
⎜
⎝
⎞
⎠
⋅+:=
Qmax 87840.00 mm
3=
Shear Stress: τ
V Q⋅
I t⋅
=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τmax
V Qmax⋅
I tw⋅
:=
τmax 4.22 MPa= Ans
Problem 7-9
Determine the maximum shear force V that the strut can support if the allowable shear stress for the
material is τallow = 40 MPa.
Given: bf 120mm:= tf 12mm:=
tw 80mm:= dw 60mm:=
τallow 40MPa:=
Solution:
Section Property : D dw 2tf+:=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:=
I 5.21 10 6−× m4=
Q Σ yi
⎯
⋅ A1⋅= Qmax bf tf⋅( )
D
2
tf
2
−
⎛
⎜
⎝
⎞
⎠
⋅
dw
2
tw⋅
⎛
⎜
⎝
⎞
⎠
dw
4
⎛
⎜
⎝
⎞
⎠
⋅+:=
Qmax 87840.00 mm
3=
Shear Stress: τ
V Q⋅
I t⋅
=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τallow
V Qmax⋅
I tw⋅
= V
I tw⋅ τallow⋅
Qmax
:=
V 189.69 kN= Ans
Problem 7-10
Plot the intensity of the shear stress distributed over the cross section of the strut if it is subjected to a
shear force of V = 15 kN.
Given: bf 120mm:= tf 12mm:=
tw 80mm:= dw 60mm:=
V 15kN:=
Solution:
Section Property : D dw 2tf+:=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:=
I 5.21 10 6−× m4=
Q Σ yi
⎯
⋅ A1⋅=
QA bf tf⋅( )
D
2
tf
2
−
⎛
⎜
⎝
⎞
⎠
⋅:= QA 51840.00 mm
3=
Qmax bf tf⋅( )
D
2
tf
2
−
⎛
⎜
⎝
⎞
⎠
⋅
dw
2
tw⋅
⎛
⎜
⎝
⎞
⎠
dw
4
⎛
⎜
⎝
⎞
⎠
⋅+:=
Qmax 87840.00 mm
3=
Shear Stress: τ
V Q⋅
I t⋅
=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τmax
V Qmax⋅
I tw⋅
:= τmax 3.16 MPa= Ans
τw_A
V QA⋅
I tw⋅
:= τw_A 1.87 MPa= Ans
τf_A
V QA⋅
I bf⋅
:= τf_A 1.24 MPa= Ans
Problem 7-11
If the pipe is subjected to a shear of V = 75 kN, determine the maximum shear stress in the pipe.
Given: ro 60mm:= ri 50mm:= V 75kN:=
Solution:
Section Property : t ro ri−:=
I
π
4
ro
4 ri
4−⎛⎝
⎞
⎠⋅:=
Qmax
4ro
3π
π ro
2⋅
2
⎛
⎜
⎝
⎞
⎠
⋅
4ri
3π
π ri
2⋅
2
⎛
⎜
⎝
⎞
⎠
⋅−:=
Maximum Shear Stress: τ
V Q⋅
I b⋅
=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τmax
V Qmax⋅
I 2t( )⋅
:=
τmax 43.17 MPa= Ans
Problem 7-12
The strut is subjected to a vertical shear of V = 130 kN. Plot the intensity of the shear-stress
distribution acting over the cross-sectional area, and compute the resultant shear force developed in the
vertical segment AB.
Given: bf 350mm:= tf 50mm:=
tw 50mm:= dw 350mm:=
V 130kN:=
Solution:
Section Property : a 0.5 bf tw−( ):=
I
1
12
bf tf
3⋅ tw dw
3⋅+ tw tf
3⋅−⎛⎝
⎞
⎠⋅:=
I 181.77 10 6−× m4=
Q Σ yi
⎯
⋅ A1⋅=
QC a tw⋅( )
a
2
tf
2
+
⎛
⎜
⎝
⎞
⎠
⋅:= QC 750000 mm
3=
QD a tw⋅( )
a
2
tf
2
+
⎛
⎜
⎝
⎞
⎠
⋅
tf
2
bf⋅
⎛
⎜
⎝
⎞
⎠
tf
4
⎛
⎜
⎝
⎞
⎠
⋅+:= QD 859375 mm
3=
τ
V Q⋅
I t⋅
=Shear Stress:
τw_C
V QC⋅
I tw⋅
:= τf_C
V QC⋅
I bf⋅
:= τD
V QD⋅
I bf⋅
:=
τw_C 10.73 MPa= τf_C 1.53 MPa=τD 1.76 MPa=
∫ == A 0dAV τResultant Shear Force:
Aw 0.5dw y−( ) tw⋅= yw 0.5 0.5dw y−( ) y+=
yw 0.5 0.5dw y+( )=
Qw Aw yw⋅=
Qw 0.5 0.5dw y−( ) tw⋅ 0.5dw y+( )⋅=
Q 0.5tw 0.25dw
2 y2−⎛⎝
⎞
⎠=
τw
V
I tw⋅
⎛
⎜
⎝
⎞
⎠
Qw⋅= τw
V
I tw⋅
⎛
⎜
⎝
⎞
⎠
0.5tw 0.25dw
2 y2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅=
VAB
V
I tw⋅ 0.5tf
0.5dw
y0.5 tw⋅ 0.25 dw
2⋅ y2−⎛⎝
⎞
⎠⋅
⎡
⎣
⎤
⎦ tw⋅
⌠⎮
⎮⌡
d
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅:= VAB 50.29 kN= Ans
Problem 7-13
The steel rod has a radius of 30 mm. If it is subjected to a shear of V = 25 kN, determine the
maximum shear stress.
Given: r 30mm:= V 25kN:=
Solution:
Section Property :
I
π
4
r4⋅:=
Qmax
4r
3π
π r2⋅
2
⎛
⎜
⎝
⎞
⎠
⋅:=
Maximum Shear Stress: τ
V Q⋅
I b⋅
=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τmax
V Qmax⋅
I 2r( )⋅
:=
τmax 11.79 MPa= Ans
Problem 7-14
If the T-beam is subjected to a vertical shear of V = 60 kN, determine the maximum shear stress in the
beam. Also, compute the shear-stress jump at the flange-web junction AB. Sketch the variation of the
shear-stress intensity over the entire cross section.
Given: bf 300mm:= dw 150mm:=
tf 75mm:= tw 100mm:=
V 60kN:=
Solution:
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) dw tw⋅( )+
bf tf⋅ dw tw⋅+
:= yc 82.50 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
tw⋅ dw
3⋅ dw tw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I I1 I2+:=
Qmax tw D yc−( )⋅
D yc−
2
⋅:=
QAB bf tf⋅( ) yc 0.5tf−( )⋅:=
Shear Stress: τ
V Q⋅
I b⋅
=
τmax
V Qmax⋅
I tw⋅
:= τmax 3.993 MPa= Ans
τf_AB
V QAB⋅
I bf⋅
:= τf_AB 1.327 MPa= Ans
τw_AB
V QAB⋅
I tw⋅
:= τw_AB 3.982 MPa= Ans
Problem 7-15
If the T-beam is subjected to a vertical shear of V = 60 kN, determine the vertical shear force resisted
by the flange.
Given: bf 300mm:= dw 150mm:=
tf 75mm:= tw 100mm:=
V 60kN:=
Solution:
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) dw tw⋅( )+
bf tf⋅ dw tw⋅+
:= yc 82.50 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
tw⋅ dw
3⋅ dw tw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I I1 I2+:=
A'f yc y−( ) bf⋅= yfc 0.5 yc y−( ) y+=
yfc 0.5 yc y+( )=
Q A'f yfc⋅=
Q 0.5 yc y−( ) bf⋅ yc y+( )=
Q 0.5bf yc
2 y2−⎛⎝
⎞
⎠=
Shear Stress: τ
V Q⋅
I b⋅
=
τf
V
I bf⋅
⎛
⎜
⎝
⎞
⎠
Q⋅=
τf
V
I bf⋅
⎛
⎜
⎝
⎞
⎠
0.5bf yc
2 y2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅=
∫ == A 0dAV τResultant Shear Force: For the flange. yo yc tf−:=
Vf
yo
yc
y
V
I bf⋅
⎛
⎜
⎝
⎞
⎠
0.5bf yc
2 y2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅ bf⋅
⌠
⎮
⎮
⌡
d:=
Vf 19.08 kN= Ans
Problem 7-16
The T-beam is subjected to the loading shown. Determine the maximum transverse shear stress in the
beam at the critical section.
Given: L1 2m:= L2 2m:= L3 3m:=
bf 100mm:= dw 100mm:=
tf 20mm:= tw 20mm:=
P 20kN:= w 8
kN
m
:=
Solution: L L1 L2+ L3+:=
Support Reaction :
 Equilibrium : Given
+ ΣFy=0; A P− w L3⋅− B+ 0=
 
ΣΜB=0; A L P L2 L3+( )⋅− w L3⋅( ) 0.5L3( )⋅− 0=
Guess A 1kN:= B 1kN:=
A
B
⎛
⎜
⎝
⎞
⎠
Find A B,( ):=
A
B
⎛
⎜
⎝
⎞
⎠
19.43
24.57
⎛
⎜
⎝
⎞
⎠
kN=
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) dw tw⋅( )+
bf tf⋅ dw tw⋅+
:= yc 40.00 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
tw⋅ dw
3⋅ dw tw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I I1 I2+:= I 5333333.33 mm4=
Qmax tw D yc−( )⋅
D yc−
2
⋅:= Qmax 64000 mm
3=
Maximum Shear Stress: τ
V Q⋅
I b⋅
= Vmax B:=
τmax
Vmax Qmax⋅
I( ) tw⋅
:=
τmax 14.74 MPa= Ans
Shear Force Diagram:
x1 0 0.01 L1⋅, L1..:= x2 L1 1.01 L1⋅, L1 L2+( )..:= x3 L1 L2+( ) 1.01 L1 L2+( )⋅, L( )..:=
V1 x1( )
A
kN
:= V2 x2( ) A P−( )
1
kN
⋅:= V3 x3( ) A P− w x3 L1− L2−( )⋅−⎡⎣ ⎤⎦
1
kN
⋅:=
0 2 4 6
20
0
20
Distance (m)
Sh
ea
r (
kN
) V1 x1( )
V2 x2( )
V3 x3( )
x1 x2, x3,
Problem 7-17
Determine the largest end forces P that the member can support if the allowable shear stress is τallow =
70 MPa. The supports at A and B only exert vertical reactions on the beam.
Given: L1 1m:= τallow 70MPa:=
L2 2m:=
w 3
kN
m
:=L3 1m:=
do 100mm:= bo 160mm:=
di 60mm:= bi 80mm:=
Solution:
Section Property :
yc
0.5do bo do⋅( ) 0.5di( ) bi di⋅( )−
bo do⋅ bi di⋅−
:= yc 58.57 mm=
I1
1
12
bo⋅ do
3⋅ bo do⋅( ) 0.5do yc−( )2⋅+:=
I2
1
12
bi⋅ di
3⋅ bi di⋅( ) 0.5di yc−( )2⋅+:= I I1 I2−:=
Qmax A'f yfc⋅= A'f yc bo bi−( )⋅= y'c 0.5yc= Qmax 0.5yc2 bo bi−( )×=
Maximum Shear Stress: τ
V Q⋅
I b⋅
= Vmax P=
τallow
P
I bo bi−( )⋅
⎡
⎢
⎣
⎤
⎥
⎦
0.5yc
2 bo bi−( )⎡⎣ ⎤⎦⋅=
τallow
P
I
⎛⎜
⎝
⎞
⎠
0.5yc
2⎛
⎝
⎞
⎠⋅=
P
I
0.5yc
2
⎛
⎜
⎝
⎞
⎠
τallow( )⋅:=
P 373.42 kN= Ans
Shear Force Diagram: L L1 L2+ L3+:=
 Equilibrium : Given
+ ΣFy=0; A w L2⋅− B+ 2P− 0=
 
ΣΜB=0; P− L1 L2+( )⋅ A L2⋅+ w L2⋅( ) 0.5L2( )⋅− P L3⋅+ 0=
Guess A 1kN:= B 1kN:=
A
B
⎛
⎜
⎝
⎞
⎠
Find A B,( ):=
A
B
⎛
⎜
⎝
⎞
⎠
376.42
376.42
⎛
⎜
⎝
⎞
⎠
kN=
x1 0 0.01 L1⋅, L1..:= x2 L1 1.01 L1⋅, L1 L2+( )..:= x3 L1 L2+( ) 1.01 L1 L2+( )⋅, L( )..:=
V1 x1( )
P−
kN
:= V2 x2( ) P− A+ w x2 L1−( )⋅−⎡⎣ ⎤⎦
1
kN
⋅:= V3 x3( ) P− A+ w L2⋅− B+( )
1
kN
⋅:=
0 1 2 3 4
0
Distance (m)
Sh
ea
r (
kN
) V1 x1( )
V2 x2( )
V3 x3( )
x1 x2, x3,
Problem 7-18
If the force P = 4 kN, determine the maximum shear stress in the beam at the critical section. The
supports at A and B only exert vertical reactions on the beam.
Given: L1 1m:= P 4kN:=
L2 2m:=
w 3
kN
m
:=L3 1m:=
do 100mm:= bo 160mm:=
di 60mm:= bi 80mm:=
Solution:
Section Property :
yc
0.5do bo do⋅( ) 0.5di( ) bi di⋅( )−
bo do⋅ bi di⋅−
:= yc 58.57 mm=
I1
1
12
bo⋅ do
3⋅ bo do⋅( ) 0.5do yc−( )2⋅+:=
I2
1
12
bi⋅ di
3⋅ bi di⋅( ) 0.5di yc−( )2⋅+:= I I1 I2−:=
Qmax A'f yfc⋅= A'f yc bo bi−( )⋅= y'c 0.5yc= Qmax 0.5yc2 bo bi−( )×=
Maximum Shear Stress: τ
V Q⋅
I b⋅
= Vmax P=
τmax
P
I bo bi−( )⋅
⎡
⎢
⎣
⎤
⎥
⎦
0.5yc
2 bo bi−( )⎡⎣ ⎤⎦⋅=
τmax
P
I
⎛⎜
⎝
⎞
⎠
0.5yc
2⎛
⎝
⎞
⎠⋅:=
τmax 0.750 MPa= Ans
Shear Force Diagram: L L1 L2+ L3+:=
 Equilibrium : Given
+ ΣFy=0; A w L2⋅− B+ 2P− 0=
 
ΣΜB=0; P− L1 L2+( )⋅ A L2⋅+ w L2⋅( ) 0.5L2( )⋅− P L3⋅+ 0=
Guess A 1kN:= B 1kN:=
A
B
⎛
⎜
⎝
⎞
⎠
Find A B,( ):=
A
B
⎛
⎜
⎝
⎞
⎠
7.00
7.00
⎛
⎜
⎝
⎞
⎠
kN=
x1 0 0.01 L1⋅, L1..:= x2 L1 1.01 L1⋅, L1 L2+( )..:= x3 L1 L2+( ) 1.01 L1 L2+( )⋅, L( )..:=
V1 x1( )
P−
kN
:= V2 x2( ) P− A+ w x2 L1−( )⋅−⎡⎣ ⎤⎦
1
kN
⋅:= V3 x3( ) P− A+ w L2⋅− B+( )
1
kN
⋅:=
0 1 2 3 4
10
0
10
Distance (m)
Sh
ea
r (
kN
) V1 x1( )
V2 x2( )
V3 x3( )
x1 x2, x3,
Problem 7-19
Plot the shear-stress distribution over the cross section of a rod that has a radius c. By what factor is
the maximum shear stress greater than the average shear stress acting over the cross section?
Problem 7-20
Develop an expression for the average vertical component of shear stress acting on the horizontal plane
through the shaft, located a distance y from the neutral axis.
Problem 7-21
Railroad ties must be designed to resist large shear loadings. If the tie is subjected to the 150-kN rail
loadings and the gravel bed exerts a distributed reaction as shown, determine the intensity w for
equilibrium, and find the maximum shear stress in the tie.
Given: L1 0.45m:= P 150kN:=
L2 0.90m:=
w 3
kN
m
:=L3 0.45m:=
d 150mm:= b 200mm:=
Solution:
 Equilibrium :
+ ΣFy=0; 0.5w L1⋅ w L2⋅+ 0.5w L3⋅+ 2P− 0=
w
2P
0.5L1 L2+ 0.5L3+
:=
w 222.22
kN
m
=
Section Property : I
1
12
b⋅ d3⋅:= Qmax 0.5 b⋅ d⋅( ) 0.25⋅ d:=
Shear Force Diagram: L L1 L2+ L3+:=
x1 0 0.01 L1⋅, L1..:= x2 L1 1.01 L1⋅, L1 L2+( )..:= x3 L1 L2+( ) 1.01 L1 L2+( )⋅, L( )..:=
V1 x1( ) 0.5 x1⋅
w x1⋅
L1
⎛
⎜
⎝
⎞
⎠
1
kN
⋅:= V2 x2( ) 0.5w L1⋅ P− w x2 L1−( )⋅+⎡⎣ ⎤⎦
1
kN
⋅:=
V3 x3( ) 0.5 w⋅ L1⋅ 2 P⋅− w L2⋅+ w x3 L1− L2−( )⋅ 1
x3 L1− L2−
2L3
−
⎛
⎜
⎝
⎞
⎠
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
1
kN
⋅:=
0 0.5 1 1.5
100
0
100
Distance (m)
Sh
ea
r (
kN
) V1 x1( )
V2 x2( )
V3 x3( )
x1 x2, x3,
Maximum Shear Stress:
τ
V Q⋅
I b⋅
=
Vmax V2 L1( ) kN⋅:=
τmax
Vmax
I b⋅
⎛
⎜
⎝
⎞
⎠
Qmax⋅:=
τmax 5 MPa= Ans
Problem 7-22
The beam is subjected to a uniform load w. Determine the placement a of the supports so that the shear
stress in the beam is as small as possible. What is this stress?
Set w
kN
m
:= a m:= L 5m:=
Given: L1 a:= L3 a:=
L2 L 2a−:=
Solution:
 Equilibrium : By equilibrium, A = B = R
+
ΣFy=0; 2R w L⋅− 0=
R 0.5wL⋅:=
Shear Force Diagram:
x1 0 0.01 L1⋅, L1..:= x2 L1 1.01 L1⋅, L1 L2+( )..:= x3 L1 L2+( ) 1.01 L1 L2+( )⋅, L( )..:=
V1 x1( ) w x1⋅( )
1
kN
⋅:= V2 x2( ) w x2⋅ R−( )
1
kN
⋅:= V3 x3( ) w x3⋅ 2 R⋅−( )
1
kN
⋅:=
0 1 2 3 4 5
0
Distance (m)
Sh
ea
r (
kN
) V1 x1( )
V2 x2( )
V3 x3( )
x1 x2, x3,
Require,
V1 a( ) V2 a( )−=
w a⋅ w a⋅ R−( )−=
2w a⋅ 0.5w L⋅=
a 0.25L=
Vmax w a⋅= Vmax 0.25w L⋅:=
Section Property : I
1
12
b⋅ d3⋅= Qmax b
d
2
⋅⎛⎜
⎝
⎞
⎠
d
4
⋅=
Maximum Shear Stress: τ
V Q⋅
I b⋅
=
τmax
Vmax
I b⋅
⎛
⎜
⎝
⎞
⎠
Qmax⋅= τmax
0.25w L⋅
1
12
b⋅ d3⋅ b⋅
⎛
⎜
⎜
⎝
⎞
⎠
b
d
2
⋅⎛⎜
⎝
⎞
⎠
d
4
⋅⎡⎢
⎣
⎤⎥
⎦
⋅=
τmax
3w L⋅
8 b⋅ d⋅
= Ans
Problem 7-23
The timber beam is to be notched at its ends as shown. If it is to support the loading shown, determine
the smallest depth d of the beam at the notch if the allowable shear stress is τallow = 3 MPa. The beam
has a width of 200 mm. 
Given: L1 1.2m:= P1 12.5kN:=
L2 1.8m:= P2 25.0kN:=
L3 1.8m:= P3 12.5kN:=
L4 1.2m:= b 200mm:=
τallow 3MPa:=
Solution:
 Equilibrium : By symmetry, R1=R , R2=R 
+ ΣFy=0; P1 P2+ P3+ 2R− 0=
R 0.5 P1 P2+ P3+( ):= R 25.00 kN=
Section Property : I
1
12
b⋅ d3⋅=
Qmax 0.5 b⋅ d⋅( ) 0.25⋅ d= Qmax 0.125 b⋅ d
2⋅=
Maximum Shear Stress:
τ
V Q⋅
I b⋅
= Vmax R:=
τallow
R 0.125b d2⋅( )⋅
1
12
b⋅ d3⋅⎛⎜
⎝
⎞
⎠
b⋅
=
d
12 0.125( )⋅ R⋅
τallow( ) b⋅
:=
d 62.5 mm= Ans
Problem 7-24
The beam is made from three boards glued together at the seams A and B. If it is subjected to the
loading shown, determine the shear stress developed in the glued joints at section a-a. The supports at
C and D exert only vertical reactions on the beam.
Given: bf 150mm:= dw 200mm:=
tf 40mm:= tw 50mm:=
P 25kN:=
Solution:
 Equilibrium : By symmetry, RC=R , RD=R 
+ ΣFy=0; 3P 2R− 0=
R 1.5P:=
R 37.50 kN=
Section Property : D dw 2tf+:=
I
1
12
bf⋅ D
3⋅
1
12
tw⋅ dw
3⋅−:=
QA bf tf⋅( ) 0.5D 0.5tf−( )⋅:=
QB QA:=
Shear Stress: τ
V Q⋅
I b⋅
= Vaa R P−:=
τA
Vaa QA⋅
I tw⋅
:= τA 0.747 MPa= Ans
τB
Vaa QB⋅
I tw⋅
:= τB 0.747 MPa= Ans
Problem 7-25
The beam is made from three boards glued together at the seams A and B. If it is subjected to the
loading shown, determine the maximum shear stress developed in the glued joints. The supports at C
and D exert only vertical reactions on the beam.
Given: bf 150mm:= dw 200mm:=
tf 40mm:= tw 50mm:=
P 25kN:=
Solution:
 Equilibrium : By symmetry, RC=R , RD=R 
+ ΣFy=0; 3P 2R− 0=
R 1.5P:=
R 37.50 kN=
Section Property : D dw 2tf+:=
I
1
12
bf⋅ D
3⋅
1
12
tw⋅ dw
3⋅−:=
QA bf tf⋅( ) 0.5D 0.5tf−( )⋅:=
QB QA:=
Shear Stress: τ
V Q⋅
I b⋅
= Vmax R:=
τA
Vmax QA⋅
I tw⋅
:= τA 2.24 MPa= Ans
τB
Vmax QB⋅
I tw⋅
:= τB 2.24 MPa= Ans
Problem 7-26
The beam is made from three boards glued together at the seams A and B. If it is subjected to the
loading shown, determine the maximum vertical shear force resisted by the top flange of the beam. The
supports at C and D exert only vertical reactions on the beam.
Given: bf 150mm:= dw 200mm:=
tf 40mm:= tw 50mm:= P 25kN:=
Solution:
 Equilibrium : By symmetry, RC=R , RD=R 
+ ΣFy=0; 3P 2R− 0=
R 1.5P:=
Section Property : D dw 2tf+:=
I
1
12
bf⋅ D
3⋅
1
12
tw⋅ dw
3⋅−:=
yc 0.5D:=
A'f yc y−( ) bf⋅= yfc 0.5 yc y−( ) y+=
yfc 0.5 yc y+( )=
Q A'f yfc⋅=
Q 0.5 yc y−( ) bf⋅ yc y+( )=
Q 0.5bf yc
2 y2−⎛⎝
⎞
⎠=
Shear Stress in flange: τ
V Q⋅
I b⋅
= V R=
τ
R
I bf⋅
⎛
⎜
⎝
⎞
⎠
Q⋅=
τ
R
I
⎛⎜
⎝
⎞
⎠
0.5 yc
2 y2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅=
Resultant Shear Force: For the flange. yo yc tf−:= Vf
A
Aτ
⌠
⎮
⌡
d=
Vf
yo
yc
y
R
I
⎛⎜
⎝
⎞
⎠
0.5 yc
2 y2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅ bf⋅
⌠
⎮
⎮
⌡
d:=
Vf 2.36 kN= Ans
Problem 7-27
Determine the shear stress at points B and C located on the web of the fiberglass beam.
Given: bf 100mm:= dw 150mm:=
tf 18mm:= tw 12mm:=
L1 2m:= L2 0.6m:=
L3 2m:=
wo 2.5
kN
m
:= w1 3
kN
m
:=
Solution: L L1 L2+ L3+:=
 Equilibrium : Given
+ ΣFy=0; A wo L1⋅− 0.5 w1⋅ L3⋅− D+ 0=
 
ΣΜD=0; A L⋅ wo L1⋅ L 0.5L1−( )⋅− 0.5 w1⋅ L3⋅
2L3
3
⎛
⎜
⎝
⎞
⎠
⋅− 0=
Guess A 1kN:= D 1kN:=
A
D
⎛
⎜
⎝
⎞
⎠
Find A D,( ):=
A
D
⎛
⎜
⎝
⎞
⎠
4.783
3.217
⎛
⎜
⎝
⎞
⎠
kN=
Section Property : D dw 2tf+:=
I
1
12
bf⋅ D
3⋅
1
12
tw⋅ dw
3⋅−:=
QB bf tf⋅( ) 0.5D 0.5tf−( )⋅:=
QC QB:=
Shear Stress: τ
V Q⋅
I b⋅
= VBC A wo 0.5L1( )⋅−:=
τB
VBC QB⋅
I tw⋅
:= τB 0.572 MPa= Ans
τC
VBC QC⋅
I tw⋅
:= τC 0.572 MPa= Ans
100mm
84mm
18mm
18mm
12mm
75mm
75mm
Problem 7-28
Determine the maximum shear stress acting in the fiberglass beam at the critical section.
Given: bf 100mm:= dw 150mm:=
tf 18mm:= tw 12mm:=
L1 2m:= L2 0.6m:=
L3 2m:=
wo 2.5
kN
m
:= w1 3
kN
m
:=
Solution: L L1 L2+ L3+:=
 Equilibrium : Given
+ ΣFy=0; A wo L1⋅− 0.5 w1⋅ L3⋅− D+ 0=
 
ΣΜD=0; A L⋅ wo L1⋅ L 0.5L1−( )⋅− 0.5 w1⋅ L3⋅
2L3
3
⎛
⎜
⎝
⎞
⎠
⋅− 0=
Guess A 1kN:= D 1kN:=
A
D
⎛
⎜
⎝
⎞
⎠
Find A D,( ):=
A
D
⎛
⎜
⎝
⎞
⎠
4.783
3.217
⎛
⎜
⎝
⎞
⎠
kN=
Section Property : D dw 2tf+:=
I
1
12
bf⋅ D
3⋅
1
12
tw⋅ dw
3⋅−:=
Qmax bf tf⋅( ) 0.5D 0.5tf−( )⋅ 0.5dw tw⋅( ) 0.25dw( )⋅+:=
Shear Stress: τ
V Q⋅
I b⋅
= Vmax A:=
τmax
Vmax Qmax⋅
I tw⋅
:= τmax 1.467 MPa= Ans
Shear Force Diagram: L L1 L2+ L3+:=
x1 0 0.01 L1⋅, L1..:= x2 L1 1.01 L1⋅, L1 L2+( )..:= x3 L1 L2+( ) 1.01 L1 L2+( )⋅, L( )..:=
V1 x1( ) A wo x1⋅−( )
1
kN
⋅:= V2 x2( ) A wo L1⋅−( )
1
kN
⋅:=
V3 x3( ) A wo L1⋅− w1 x3 L1− L2−( )⋅ 1
x3 L1− L2−
2L3
−
⎛
⎜
⎝
⎞
⎠
⋅−
⎡
⎢
⎣
⎤
⎥
⎦
1
kN
⋅:=
0 1 2 3 4
5
0
5
Distance (m)
Sh
ea
r (
kN
) V1 x1( )
V2 x2( )
V3 x3( )
x1 x2, x3,
Problem 7-29
The beam is made from three plastic pieces glued together at the seams A and B. If it is subjected to
the loading shown, determine the shear stress developed in the glued joints at the critical section. The
supports at C and D exert only vertical reactions on the beam.
Given: bf 200mm:= dw 200mm:=
tf 50mm:= tw 50mm:=
wo 3
kN
m
:= L 2.5m:=
Solution:
 Equilibrium : By symmetry, RC=R , RD=R 
+ ΣFy=0; wo L⋅ 2R− 0=
R 0.5 wo L⋅( ):=
R 3.75 kN=
Section Property : D dw 2tf+:=
I
1
12
bf⋅ D
3⋅
1
12
tw⋅ dw
3⋅−:=
QA bf tf⋅( ) 0.5D 0.5tf−( )⋅:=
QB QA:=
Shear Stress: τ
V Q⋅
I b⋅
= Vmax R:=
τA
Vmax QA⋅
I tw⋅
:= τA 0.225 MPa= Ans
τB
Vmax QB⋅
I tw⋅
:= τB 0.225 MPa= Ans
Problem 7-30
The beam is made from three plastic pieces glued together at the seams A and B. If it is subjected to
the loading shown, determine the vertical shear force resisted by the top flange of the beam at the
critical section. The supports at C and D exert only vertical reactions on the beam.
Given: bf 200mm:= dw 200mm:=
tf 50mm:= tw 50mm:=
L 2.5m:= wo 3
kN
m
:=
Solution:
 Equilibrium : By symmetry, RC=R , RD=R 
+ ΣFy=0; wo L⋅ 2R− 0=
R 0.5 wo L⋅( ):=
R 3.75 kN=
Section Property : D dw 2tf+:=
I
1
12
bf⋅ D
3⋅
1
12
tw⋅ dw
3⋅−:=
yc 0.5D:=
A'f yc y−( ) bf⋅= yfc 0.5 yc y−( ) y+=
yfc 0.5 yc y+( )=
Q A'f yfc⋅=
Q 0.5 yc y−( ) bf⋅ yc y+( )=
Q 0.5bf yc
2 y2−⎛⎝
⎞
⎠=
Shear Stress in flange: τ
V Q⋅
I b⋅
= Vmax R=
τ
R
I bf⋅
⎛
⎜
⎝
⎞
⎠
Q⋅=
τ
R
I
⎛⎜
⎝
⎞
⎠
0.5 yc
2 y2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅=
Resultant Shear Force: For the flange. yo yc tf−:= Vf
A
Aτ
⌠
⎮
⌡
d=
Vf
yo
yc
y
R
I
⎛⎜
⎝
⎞
⎠
0.5 yc
2 y2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅ bf⋅
⌠
⎮
⎮
⌡
d:=
Vf 0.30 kN= Ans
Problem 7-31
Determine the variation of the shear stress over the cross section of a hollow rivet. What is the
maximum shear stress in the rivet? Also, show that if then τ max = 2( V/A ).01 rr →
Problem 7-32
The beam has a square cross section and is subjected to the shear force V. Sketch the shear-stress
distribution over the cross section and specify the maximum shear stress. Also, from the neutral axis,
locate where a crack along the member will first start to appear due to shear.
Problem 7-33
Write a computer program that can be used to determine the maximum shear stress in the beam that
has the cross section shown, and is subjected to a specified constant distributed load w and
concentrated force P. Show an application of the program using the values L = 4 m, a = 2 m, P = 1.5
kN, d1 = 0, d2 = 2 m, w = 400 N/m, t1 = 15 mm, t2 = 20 mm, b = 50 mm, and h = 150 mm.
Problem 7-34
The beam has arectangular cross section and is subjected to a load P that is just large enough to
develop a fully plastic moment Mp = PL at the fixed support. If the material is elastic-plastic, then at a
distance x < L the moment M = P x creates a region of plastic yielding with an associated elastic core
having a height 2y'. This situation has been described by Eq. 6-30 and the moment M is distributed
over the cross section as shown in Fig. 6-54e. Prove that the maximum shear stress developed in the
beam is given by τ max = 3/2 (P/A'),where A' = 2y'b, the cross-sectional area of the elastic core.
Problem 7-35
The beam in Fig. 6-54f is subjected to a fully plastic moment Mp. Prove that the longitudinal and
transverse shear stresses in the beam are zero. Hint: Consider an element of the beam as shown in Fig.
7-4d.
Problem 7-36
The beam is constructed from two boards fastened together at the top and bottom with two rows of
nails spaced every 150 mm. If each nail can support a 2.5-kN shear force, determine the maximum
shear force V that can be applied to the beam.
Given: b 150mm:= d1 50mm:=
d2 50mm:= sn 150mm:=
Fallow 2.5kN:=
Solution:
Section Property : D d1 d2+:=
I
1
12
b⋅ D3⋅:=
Q b d1⋅( ) 0.5d1( )⋅:=
Shear Flow : q
V Q⋅
I
=
There are two rows of nails. Hence, the allowable shear flow is qallow
2Fallow
sn
:=
2Fallow
sn
Vmax Q⋅
I
=
Vmax
2 I⋅ Fallow
Q sn⋅
:=
Vmax 2.222 kN= Ans
Problem 7-37
The beam is constructed from two boards fastened together at the top and bottom with two rows of
nails spaced every 150 mm. If an internal shear force of V = 3 kN is applied to the boards, determine
the shear force resisted by each nail.
Given: b 150mm:= d1 50mm:=
d2 50mm:= sn 150mm:=
V 3kN:=
Solution:
Section Property : D d1 d2+:=
I
1
12
b⋅ D3⋅:=
Q b d1⋅( ) 0.5d1( )⋅:=
Shear Flow : q
V Q⋅
I
:= q 45.00
kN
m
=
There are two rows of nails. Hence, the shear force resisted by each nail is F
q
2
sn⋅:=
F 3.37 kN= Ans
Problem 7-38
A beam is constructed from five boards bolted together as shown. Determine the maximum shear
force developed in each bolt if the bolts are spaced s = 250 mm apart and the applied shear is V = 35
kN.
Given: d1 250mm:= d2 350mm:=
t 25mm:= a 100mm:=
s 250mm:= V 35kN:=
Solution: a' d2 d1 a−( )−:= a' 200 mm=
h a' 0.5d1+:= h 325 mm=
Section Property :
yc
2 d1 t⋅( ) 0.5d1 a'+( )⋅ 3 d2 t⋅( ) 0.5d2( )⋅+
2 d1 t⋅( ) 3 d2 t⋅( )+
:=
yc 223.39 mm=
I1
1
12
2t( )⋅ d1
3⋅ 2t d1⋅( ) 0.5d1 a'+ yc−( )2⋅+:=
I2
1
12
3 t⋅( )⋅ d2
3⋅ 3 t⋅ d2⋅( ) 0.5d2 yc−( )2⋅+:=
I I1 I2+:= I 523597110.22 mm
4=
Q d1 2t( )⋅ h yc−( )⋅:= Q 1270161.29 mm3=
Shear Flow :
q
V Q⋅
I
:= q 84.90
kN
m
=
There are four planes on the bolt. Hencs, the shear
force resisted by each shear plane of the bolt is
F
q
4
s⋅⎛⎜
⎝
⎞
⎠
:= F 5.31 kN= Ans
Problem 7-39
A beam is constructed from five boards bolted together as shown. Determine the maximum spacing s
of the bolts if they can each resist a shear of 20 kN and the applied shear is V = 45 kN.
Given: d1 250mm:= d2 350mm:=
t 25mm:= a 100mm:=
V 45kN:= Fallow 20kN:=
Solution: a' d2 d1 a−( )−:= a' 200 mm=
h a' 0.5d1+:= h 325 mm=
Section Property :
yc
2 d1 t⋅( ) 0.5d1 a'+( )⋅ 3 d2 t⋅( ) 0.5d2( )⋅+
2 d1 t⋅( ) 3 d2 t⋅( )+
:=
yc 223.39 mm=
I1
1
12
2t( )⋅ d1
3⋅ 2t d1⋅( ) 0.5d1 a'+ yc−( )2⋅+:=
I2
1
12
3 t⋅( )⋅ d2
3⋅ 3 t⋅ d2⋅( ) 0.5d2 yc−( )2⋅+:=
I I1 I2+:= I 523597110.22 mm
4=
Q d1 2t( )⋅ h yc−( )⋅:= Q 1270161.29 mm3=
Shear Flow : q
V Q⋅
I
:= q 109.16
kN
m
=
Since there are four planes on the bolt, the allowable shear flow is qallow
4Fallow
s
=
s
4Fallow
q
:=
s 732.9 mm= Ans
Problem 7-40
The beam is subjected to a shear of V = 800 N. Determine the average shear stress developed in the
nails along the sides A and B if the nails are spaced s = 100 mm apart. Each nail has a diameter of 2
mm.
Given: bf 250mm:= dw 150mm:=
t 30mm:= a 100mm:=
s 100mm:= V 800N:=
do 2mm:=
Solution: h' 0.5t:= h' 15 mm=
Section Property :
yc
bf t⋅ 0.5t( )⋅ dw 2t( )⋅ 0.5dw( )⋅+
bf t⋅ 2 dw t⋅( )+
:=
yc 47.73 mm=
I1
1
12
bf⋅ t
3⋅ bf t⋅( ) 0.5t yc−( )2⋅+:=
I2
1
12
2 t⋅( )⋅ dw
3⋅ 2 t⋅ dw⋅( ) 0.5dw yc−( )2⋅+:=
I I1 I2+:= I 32164772.73 mm
4=
Q bf t⋅ yc h'−( )⋅:= Q 245454.55 mm3=
Shear Flow :
q
V Q⋅
I
:= q 6.105
kN
m
=
F q s⋅:= F 0.6105 kN=
Since each side of the beam resists this shear force, then
Ao
π
4
do
2⋅:= τavg
F
2Ao
:=
τavg 97.16 MPa= Ans
Problem 7-41
The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of
150 mm and a thickness of 12 mm. If a shear of V = 250 kN is applied to the cross section, determine
the maximum spacing of the bolts. Each bolt can resist a shear force of 75 kN.
Given: bf 75mm:= dw 75mm:= t 12mm:=
dp 150mm:= dg 50mm:=
V 250kN:= Fallow 75kN:=
Solution:
Section Property : D 2dw 2t+ dg+:=
IT
1
12
bf⋅ D
3⋅
1
12
t⋅ dg
3⋅−
1
12
bf t−( )⋅ D 2t−( )3⋅−:=
IP
1
12
2t( )⋅ dp
3⋅:=
I IT IP+:=
Q bf t⋅( ) 0.5D 0.5t−( )⋅ t dw⋅( ) 0.5dw 0.5dg+( )⋅+:=
Shear Flow : q
V Q⋅
I
=
Since there are two shear planes on the bolt, the allowable shear flow is q
2F
sn
=
2Fallow
sn
V Q⋅
I
=
sn
2 I⋅ Fallow
V Q⋅
:=
sn 138.0 mm= Ans
Problem 7-42
The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of
150 mm and a thickness of 12 mm. If the bolts are spaced at s = 200 mm, determine the maximum
shear force V that can be applied to the cross section. Each bolt can resist a shear force of 75 kN.
Given: bf 75mm:= dw 75mm:= t 12mm:=
dp 150mm:= dg 50mm:=
sn 200mm:= Fallow 75kN:=
Solution:
Section Property : D 2dw 2t+ dg+:=
IT
1
12
bf⋅ D
3⋅
1
12
t⋅ dg
3⋅−
1
12
bf t−( )⋅ D 2t−( )3⋅−:=
IP
1
12
2t( )⋅ dp
3⋅:=
I IT IP+:=
q
V Q⋅
I
Q bf t⋅( ) 0.5D 0.5t−( )⋅ t dw⋅( ) 0.5dw 0.5dg+( )⋅+:=
Shear Flow : =
Since there are two shear planes on the bolt, the allowable shear flow is q
2F
sn
=
2Fallow
sn
V Q⋅
I
=
V
2 I⋅ Fallow
sn Q⋅
:=
V 172.5 kN= Ans
Problem 7-43
The double-web girder is constructed from two plywood sheets that are secured to wood members at
its top and bottom. If each fastener can support 3 kN in single shear, determine the required spacing s
of the fasteners needed to support the loading P = 15 kN. Assume A is pinned and B is a roller.
Given: bi 150mm:= di 250mm:=
t 12mm:= do 350mm:=
P 15kN:= Fallow 3kN:=
Solution:
 Equilibrium : By symmetry, A=R , B=R 
+ ΣFy=0; 2R P− 0=
R 0.5P:=
Section Property : bo bi 2t+:= d' 0.5 do di−( ):=
I
1
12
bo⋅ do
3⋅
1
12
bi⋅ di
3⋅−:=
Q bi d'⋅( ) 0.5do 0.5d'−( )⋅:=
Shear Flow : q
V Q⋅
I
= Vmax R:=
Since there are two shear planes on the bolt, the allowable shear flow is q
2F
sn
=
2Fallow
sn
Vmax Q⋅
I
=
sn
2 I⋅ Fallow
Vmax Q⋅
:=
sn 303.2 mm= Ans
Problem 7-44
The double-web girder is constructed from two plywood sheets that are secured to wood members at
its top and bottom. The allowable bending stress for the wood is σallow = 56 MPa and the allowable
shear stress is τallow = 21 MPa. If the fasteners are spaced s = 150 mm and each fastener can support
3 kN in single shear, determine the maximum load P that can be applied to the beam. 
Given: bi 150mm:= di 250mm:=
t 12mm:= do 350mm:=
sn 150mm:= σallow 56MPa:=
Fallow 3kN:= τallow 21MPa:=
Solution:
 Equilibrium : By symmetry, A=R , B=R 
+ ΣFy=0; 2R P− 0=
R 0.5P=
Section Property : bo bi 2t+:=
d' 0.5 do di−( ):=
I
1
12
bo⋅ do
3⋅
1
12
bi⋅ di
3⋅−:=
Q bi d'⋅( ) 0.5do 0.5d'−( )⋅:=
Shear Flow : q
V Q⋅
I
= Vmax R= q
2F
sn
=
Since there are two shear planes on the bolt, the allowable shear flow is
2Fallow
sn
Vmax Q⋅
I
=
2Fallow
sn
0.5P Q⋅
I
=
P
4 I⋅ Fallow
sn Q⋅
:=
P 30.32 kN= Ans
Problem 7-45
The beam is made from three polystyrene strips that are glued together as shown. If the glue has a
shear strength of 80 kPa, determine the maximum load P that can be applied without causing the glue
to lose its bond.
Given: bf 30mm:= tf 40mm:=
tw 20mm:= dw 60mm:=
τallow 0.080MPa:=
Solution:
 Equilibrium : By equilibrium, A = B = R
+
ΣFy=0; 2R P− 2
1
4
P⋅⎛⎜
⎝
⎞
⎠
− 0=
R 0.75P=Maximum Shear : Vmax R= Vmax
3
4
P=
Section Property : D dw 2tf+:=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:=
I 6.68 10 6−× m4=
Q Σ yi
⎯
⋅ A1⋅= Q bf tf⋅( )
D
2
tf
2
−
⎛
⎜
⎝
⎞
⎠
⋅:= Q 60000 mm3=
Shear Stress: τ
V Q⋅
I t⋅
=
τallow
Vmax Q⋅
I tw⋅
= Vmax
τallow I⋅ tw⋅
Q
=
3
4
P
τallow I⋅ tw⋅
Q
=
P
4τallow I⋅ tw⋅
3Q
:=
P 0.238 kN= Ans
Problem 7-46
The beam is made from four boards nailed together as shown. If the nails can each support a shear
force of 500 N., determine their required spacings s' and s if the beam is subjected to a shear of V =
3.5 kN.
Given: bf 250mm:= dw 250mm:=
tf 25mm:= tw 40mm:=
tb 25mm:= db 75mm:=
V 3.5kN:= Fallow 0.5kN:=
Solution:
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) dw tw⋅( )+ 0.5db 2tb db⋅( )+
bf tf⋅ dw tw⋅+ 2 tb db⋅( )+
:=
yc 85.94 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
tw⋅ dw
3⋅ dw tw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I3
1
12
2tb( )⋅ db3⋅ 2tb db⋅( ) 0.5db yc−( )2⋅+:=
I I1 I2+ I3+:=
QC tb db⋅( ) yc 0.5db−( )⋅:=
QD dw tw⋅( ) D yc− 0.5dw−( )⋅:=
Shear Flow : q
V Q⋅
I
=
The allowable shear flow at points C and D are : qC
F
sn
= qD
F
s'n
=
Fallow
sn
V QC⋅
I
=
Fallow
s'n
V QD⋅
I
=
sn
I Fallow
V QC⋅
:= s'n
I Fallow
V QD⋅
:=
sn 216.6 mm= s'n 30.7 mm= Ans
Problem 7-47
The beam is fabricated from two equivalent channels and two plates. Each plate has a height of 150
mm and a thickness of 12 mm. If a shear of V = 250 kN is applied to the cross section, determine the
maximum spacing of the bolts. Each bolt can resist a shear force of 75 kN.
Given: bf 300mm:= dw 88mm:= t 12mm:=
dp 150mm:= dg 50mm:=
V 250kN:= Fallow 75kN:=
Solution:
Section Property : D 2dw 2t+ dg+:=
IU
1
12
bf⋅ D
3⋅
1
12
2t( )⋅ dg
3⋅−
1
12
bf 2t−( )⋅ D 2t−( )3⋅−:=
IP
1
12
2t( )⋅ dp
3⋅:=
I IU IP+:=
Q bf t⋅( ) 0.5D 0.5t−( )⋅ 2t dw⋅( ) 0.5dw 0.5dg+( )⋅+:=
Shear Flow : q
V Q⋅
I
=
Since there are two rows of bolts, the allowable shear flow is q
2F
sn
=
2Fallow
sn
V Q⋅
I
=
sn
2 I⋅ Fallow
V Q⋅
:=
sn 137.6 mm= Ans
Problem 7-48
A built-up timber beam is made from n boards, each having a rectangular cross section. Write a
computer program that can be used to determine the maximum shear stress in the beam when it is
subjected to any shear V. Show an application of the program using a cross section that is in
the form of a “T” and a box.
Problem 7-49
The timber T-beam is subjected to a load consisting of n concentrated forces Pn ., If the allowable
shear Vnail for each of the nails is known, write a computer program that will specify the nail spacing
between each load. Show an application of the program using the values L = 4.5 m, a1 = 1.2 m, P1 = 3
kN, a2 = 2.4 m, P2 = 7.5 kN, b1 = 37.5 mm, h1 = 250 mm, b2 = 200 mm, h2 = 25 mm, and Vnail = 1
kN.
Problem 7-50
The strut is constructed from three pieces of plastic that are glued together as shown. If the allowable
shear stress for the plastic is τallow = 5.6 MPa and each glue joint can withstand 50 kN/m, determine
the largest allowable distributed loading w that can be applied to the strut.
Given: L1 1m:= τallow 5.6MPa:=
L2 2m:= qallow 50
kN
m
:=
L3 1m:=
bf 74mm:= tf 25mm:=
dw 75mm:= tw 12mm:=
Solution:
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) 2tw dw⋅( )+
bf tf⋅ 2tw dw⋅+
:=
yc 37.16 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
2tw( )⋅ dw3⋅ 2tw dw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I I1 I2+:=
Qmax 2tw( ) D yc−( )⋅
D yc−
2
⋅:=
QA bf tf⋅( ) yc 0.5tf−( )⋅:=
Allowable Shear Stress: τ
V Q⋅
I t⋅
= Vmax w L1⋅=
τallow
w L1⋅
I 2tw( )⋅
⎡
⎢
⎣
⎤
⎥
⎦
2tw( ) D yc−( )⋅
D yc−
2
⋅
⎡
⎢
⎣
⎤
⎥
⎦
⋅=
τallow
w L1⋅
2I
D yc−( )2⋅= w
2I
L1 D yc−( )2⋅
τallow( )⋅:= w 9.13 kNm=
Shear Flow : Assume the beam fails at the glue joint and the allowable shear flow is 2 qallow⋅
2qallow
V Q⋅
I
= 2qallow
w L1⋅( ) QA⋅
I
=
w
2 I⋅ qallow
L1 QA⋅
:= w 7.06
kN
m
= (Controls !) Ans
Shear Force Diagram: L L1 L2+ L3+:=
 Equilibrium : By symmetry, R1 R= R2 R=
+ ΣFy=0; 2R w L⋅− 0=
R 0.5w L⋅:=
x1 0 0.01 L1⋅, L1..:= x2 L1 1.01 L1⋅, L1 L2+( )..:= x3 L1 L2+( ) 1.01 L1 L2+( )⋅, L( )..:=
V1 x1( )
w− x1⋅
kN
:= V2 x2( ) w− x2⋅ R+( )
1
kN
⋅:= V3 x3( ) w− x3⋅ 2R+( )
1
kN
⋅:=
0 1 2 3 4
10
0
10
Distance (m)
Sh
ea
r (
kN
) V1 x1( )
V2 x2( )
V3 x3( )
x1 x2, x3,
Problem 7-51
The strut is constructed from three pieces of plastic that are glued together as shown. If the distributed
load w = 3 kN/m, determine the shear flow that must be resisted by each glue joint.
Given: L1 1m:=
L2 2m:= w 3
kN
m
:=
L3 1m:=
bf 74mm:= tf 25mm:=
dw 75mm:= tw 12mm:=
Solution:
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) 2tw dw⋅( )+
bf tf⋅ 2tw dw⋅+
:=
yc 37.16 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
2tw( )⋅ dw3⋅ 2tw dw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I I1 I2+:=
QA bf tf⋅( ) yc 0.5tf−( )⋅:=
Shear Flow : Since there are two glue joints, hence 2q
V Q⋅
I
=
Vmax w L1⋅:=
q
Vmax QA⋅
2I
:=
q 21.24
kN
m
= Ans
Shear Force Diagram: L L1 L2+ L3+:=
 Equilibrium : By symmetry, R1 R= R2 R=
+ ΣFy=0; 2R w L⋅− 0=
R 0.5w L⋅:=
x1 0 0.01 L1⋅, L1..:= x2 L1 1.01 L1⋅, L1 L2+( )..:= x3 L1 L2+( ) 1.01 L1 L2+( )⋅, L( )..:=
V1 x1( )
w− x1⋅
kN
:= V2 x2( ) w− x2⋅ R+( )
1
kN
⋅:= V3 x3( ) w− x3⋅ 2R+( )
1
kN
⋅:=
0 1 2 3 4
10
0
10
Distance (m)
Sh
ea
r (
kN
) V1 x1( )
V2 x2( )
V3 x3( )
x1 x2, x3,
Problem 7-52
The beam is subjected to the loading shown, where P = 7 kN. Determine the average shear stress
developed in the nails within region AB of the beam. The nails are located on each side of the beam and
are spaced 100 mm apart. Each nail has a diameter of 5 mm.
Given: bf 250mm:= dw 150mm:=
t 30mm:= a 2m:=
s 100mm:= do 5mm:=
P' 3kN:= P 7kN:=
Solution:
Section Property :
I
1
12
bf 2 t⋅+( ) dw3⋅ bf dw 2 t⋅−( )3⋅−⎡⎣ ⎤⎦⋅:=
I 72000000 mm4=
Q bf t⋅ 0.5dw 0.5t−( )⋅:=
Q 450000 mm3=
Maximum Shear : Vmax P' P+:= Vmax 10 kN=
Shear Flow :
q
Vmax Q⋅
I
:= q 62.500
kN
m
=
There are two rows of nails. Hence, the sher force
resisted by each nail is
F
q
2
s⋅:= F 3.125 kN=
Ao
π
4
do
2⋅:= τavg
F
Ao
:=
τavg 159.2 MPa= Ans
Problem 7-53
The beam is constructed from four boards which are nailed together. If the nails are on both sides of
the beam and each can resist a shear of 3 kN, determine the maximum load P that can be applied to the
end of the beam.
Given: bf 250mm:= dw 150mm:=
t 30mm:= a 2m:=
s 100mm:= P' 3kN:=
Fallow 3kN:=
Solution:
Section Property :
I
1
12
bf 2 t⋅+( ) dw3⋅ bf dw 2 t⋅−( )3⋅−⎡⎣ ⎤⎦⋅:=
I 72000000 mm4=
Q bf t⋅ 0.5dw 0.5t−( )⋅:=
Q 450000 mm3=
Maximum Shear : Vmax P' P+=
There are two rows of nails. Hence, the allowable sher flow is
qallow
2Fallow
s
:=
qallow 60.00
kN
m
=
Shear Flow :
qallow
Vmax Q⋅
I
= qallow
P' P+( ) Q⋅
I
=
P
qallow I⋅
Q
⎛
⎜
⎝
⎞
⎠
P'−:=
P 6.60 kN= Ans
Problem 7-54
The member consists of two plastic channel strips 12 mm thick, bonded together at A and B. If the
glue can support an allowable shear stress of τallow = 4.2 MPa, determine the maximum intensity w0 of
the triangular distributed loading that can be applied to the member based on the strength of the glue.
Given: bo 150mm:= t 12mm:=
do 150mm:= L 4m:=
τallow 4.2MPa:=
Solution:
 Equilibrium : By symmetry, A=R , B=R 
+ ΣFy=0; 2R 0.5wo L⋅− 0=
R 0.25wo L⋅=
Section Property : bi bo 2t−:= di do 2t−:=
I
1
12
bo⋅ do
3⋅
1
12
bi⋅ di
3⋅−:=
Q bo t⋅( ) 0.5do 0.5t−( )⋅ 2 t⋅ 0.5 di⋅( )⋅⎡⎣ ⎤⎦ 0.5di( )⋅+:=
Shear Flow : q
V Q⋅
I
= Vmax R=
Since there are two planes of glue, the allowable shear flow is 2t τallow⋅
2t( ) τallow⋅
Vmax Q⋅
I
=
2 t⋅( )τallow
0.25wo L⋅( ) Q⋅
I
=
wo
8t I⋅ τallow
L Q⋅
:=
wo 9.73
kN
m
= Ans
Problem 7-55
The member consists of two plastic channel strips 12 mm thick, glued together at A and B. If the
distributed load has a maximum intensity of w0 = 50 kN/m, determine the maximum shear stress
resisted by the glue.
Given: bo 150mm:= do 150mm:= L 4m:=
t 12mm:= wo 50
kN
m
:=
Solution:
 Equilibrium : By symmetry, A=R , B=R 
+ ΣFy=0; 2R 0.5wo L⋅− 0=
R 0.25wo L⋅:=Section Property : bi bo 2t−:= di do 2t−:=
I
1
12
bo⋅ do
3⋅
1
12
bi⋅ di
3⋅−:=
Q bo t⋅( ) 0.5do 0.5t−( )⋅ 2 t⋅ 0.5 di⋅( )⋅⎡⎣ ⎤⎦ 0.5di( )⋅+:=
Allowable Shear Stress: τ
V Q⋅
I b⋅
= Vmax R:=
τmax
Vmax Q⋅
I 2t( )⋅
:=
τmax 21.58 MPa= Ans
Problem 7-56
A shear force of V = 18 kN is applied to the symmetric box girder. Determine the shear flow at A and
B.
Given: bf 125mm:= dw 300mm:= t 10mm:=
dm 200mm:= dg 30mm:= V 18kN:=
Solution:
Section Property :
I1
1
12
bf⋅ t
3⋅ bf t⋅( ) 0.5dw 0.5t−( )2⋅+:=
I2
1
12
bf⋅ t
3⋅ bf t⋅( ) 0.5t 0.5dm+( )2⋅+:=
I3
1
12
t⋅ dw
3⋅:=
I 2I1 2I2+ 2I3+:= I 125166666.67 mm
4=
QA bf t⋅( ) 0.5dw 0.5t−( )⋅:= QA 181250 mm3=
QB bf t⋅( ) 0.5t 0.5dm+( )⋅:= QB 131250 mm3=
Shear Flow :
qA
1
2
V QA⋅
I
:= qA 13.03
kN
m
= Ans
qB
1
2
V QB⋅
I
:= qB 9.44
kN
m
= Ans
Problem 7-57
A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at C.
Given: bf 125mm:= dw 300mm:= t 10mm:=
dm 200mm:= dg 30mm:= V 18kN:=
Solution:
Section Property :
I1
1
12
bf⋅ t
3⋅ bf t⋅( ) 0.5dw 0.5t−( )2⋅+:=
I2
1
12
bf⋅ t
3⋅ bf t⋅( ) 0.5t 0.5dm+( )2⋅+:=
I3
1
12
t⋅ dw
3⋅:=
I 2I1 2I2+ 2I3+:=
I 125166666.67 mm4=
QC bf t⋅( ) 0.5dw 0.5t−( )⋅ bf t⋅( ) 0.5t 0.5dm+( )⋅+ 2 0.5dw t⋅( ) 0.25dw( )⋅+:=
QC 537500 mm
3=
Shear Flow :
qC
1
2
V QC⋅
I
:= qC 38.65
kN
m
= Ans
Problem 7-58
The channel is subjected to a shear of V = 75 kN. Determine the shear flow developed at point A.
Given: bf 400mm:= dw 200mm:=
tf 30mm:= tw 30mm:=
V 75kN:=
Solution:
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) 2tw dw⋅( )+
bf tf⋅ 2tw dw⋅+
:=
yc 72.50 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
2tw( )⋅ dw3⋅ 2tw dw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I I1 I2+:=
I 120250000 mm4=
QA bf tf⋅( ) yc 0.5tf−( )⋅:=
QA 690000 mm
3=
Shear Flow :
qA
1
2
V QA⋅
I
:= qA 215.2
kN
m
= Ans
Problem 7-59
The channel is subjected to a shear of V = 75 kN. Determine the maximum shear flow in the channel.
Given: bf 400mm:= dw 200mm:=
tf 30mm:= tw 30mm:=
V 75kN:=
Solution:
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) 2tw dw⋅( )+
bf tf⋅ 2tw dw⋅+
:=
yc 72.50 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
2tw( )⋅ dw3⋅ 2tw dw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I I1 I2+:=
I 120250000 mm4=
Qmax tw D yc−( )⋅
1
2
⋅ D yc−( )⋅:=
Qmax 372093.75 mm
3=
Shear Flow :
qmax
V Qmax⋅
I
:= qmax 232.1
kN
m
= Ans
Problem 7-60
The beam supports a vertical shear of V = 35 kN. Determine the resultant force developed in segment
AB of the beam.
Given: bf 125mm:= dw 250mm:=
tf 12mm:= tw 12mm:=
V 35kN:=
Solution:
Section Property : D dw 2tf+:=
I
1
12
2tf( )⋅ bf3⋅
1
12
dw⋅ tw
3⋅+:=
Q A'f yfc⋅=
Q tf 0.5bf y−( )⋅ 0.5 0.5bf y−( ) y+⎡⎣ ⎤⎦⋅=
Q 0.5tf 0.25bf
2 y2−⎛⎝
⎞
⎠=
Shear Flow: q
V Q⋅
I
=
q
V tf⋅
2I
0.25bf
2 y2−⎛⎝
⎞
⎠⋅=
Resultant Shear Force: For AB: yo 0.5tw:= VAB
A
yq
⌠
⎮
⌡
d=
VAB
yo
0.5bf
y
V tf⋅
2I
0.25bf
2 y2−⎛⎝
⎞
⎠⋅
⌠
⎮
⎮
⌡
d:=
VAB 7.43 kN= Ans
Problem 7-61
The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V
= 150 N, determine the shear flow at points A and B.
Given: bf 60mm:= b'f 80mm:=
tf 10mm:= dw 40mm:=
tw 10mm:= V 150N:=
Solution:
Section Property : D dw 2tf+:=
A bf tf⋅ 2dw tw⋅+ b'f tf⋅+:=
A 2200 mm2=
y
⎯ Σ yi
⎯
Ai⋅( )⋅
Σ Ai( )⋅
= yc
b'f tf⋅( ) 0.5tf( )⋅ 2 dw tw⋅( ) 0.5dw tf+( )⋅+ bf tf⋅( ) D 0.5tf−( )⋅+
A
:=
yc 27.73 mm=
I'f
1
12
b'f⋅ tf
3⋅ b'f tf⋅( ) 0.5tf yc−( )2⋅+:=
Iw
1
12
tw⋅ dw
3⋅ tw dw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
If
1
12
bf⋅ tf
3⋅ bf tf⋅( ) D 0.5tf− yc−( )2⋅+:=
I If 2Iw+ I'f+:=
I 981.9697 10 9−× m4=
QA 0.5b'f tf⋅( ) yc 0.5tf−( )⋅:= QA 9090.91 mm3=
QB bf tf⋅( ) D 0.5tf− yc−( )⋅:= QB 16363.64 mm3=
Shear Flow :
qA
V QA⋅
I
:= qA 1.39
kN
m
= Ans
qB
1
2
V QB⋅
I
:= qB 1.25
kN
m
= Ans
Problem 7-62
The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V
= 150 N, determine the maximum shear flow in the strut.
Given: bf 60mm:= b'f 80mm:=
tf 10mm:= dw 40mm:=
tw 10mm:= V 150N:=
Solution:
Section Property : D dw 2tf+:=
A bf tf⋅ 2dw tw⋅+ b'f tf⋅+:=
A 2200 mm2=
y
⎯ Σ yi
⎯
Ai⋅( )⋅
Σ Ai( )⋅
= yc
b'f tf⋅( ) 0.5tf( )⋅ 2 dw tw⋅( ) 0.5dw tf+( )⋅+ bf tf⋅( ) D 0.5tf−( )⋅+
A
:=
yc 27.73 mm=
I'f
1
12
b'f⋅ tf
3⋅ b'f tf⋅( ) 0.5tf yc−( )2⋅+:=
Iw
1
12
tw⋅ dw
3⋅ tw dw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
If
1
12
bf⋅ tf
3⋅ bf tf⋅( ) D 0.5tf− yc−( )2⋅+:=
I If 2Iw+ I'f+:=
I 981.9697 10 9−× m4=
Qmax bf tf⋅( ) D 0.5tf− yc−( )⋅ 2 tw⋅ D yc− tf−( )⋅
1
2
⋅ D yc− tf−( )⋅+:=
Qmax 21324.38 mm
3=
Shear Flow :
qmax
1
2
V Qmax⋅
I
:= qmax 1.63
kN
m
= Ans
Problem 7-63
The angle is subjected to a shear of V = 10 kN. Sketch the distribution of shear flow along the leg AB.
Indicate numerical values at all peaks.
Given: L 125mm:= t 6mm:=
θ 45deg:= V 10kN:=
Solution:
Section Property :
h L cos θ( )⋅:= b t
sin θ( )
:=
I
1
12
2b( )⋅ h3⋅:=
Q A' y'c⋅=
Q t
0.5h y−
sin θ( )
⎛⎜
⎝
⎞
⎠
⋅ 0.5 0.5h y−( ) y+[ ]⋅=
Q
t
2 sin θ( )
0.25h2 y2−( )=
Shear Flow: q
V Q⋅
I
=
q
V t⋅
2I sin θ( )⋅
0.25h2 y2−( )⋅=
At y = 0, q qmax=
qmax
V t⋅
2I sin θ( )⋅
0.25h2( )⋅:=
qmax 84.85
kN
m
= Ans
Problem 7-64
The beam is subjected to a shear force of V = 25 kN. Determine the shear flow at points A and B.
Given: bf 274mm:= tf 12mm:=
b'f 250mm:= t'f 12mm:=
dw 200mm:= tw 12mm:=
d'w 50mm:= V 25kN:=
Solution:
Section Property : D dw tf+:=
D' D d'w−:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) 2tw dw⋅( )+ D' 0.5t'f−( ) b'f t'f⋅( )+
bf tf⋅ 2tw dw⋅+ b'f t'f⋅+
:=
yc 92.47 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
2tw( )⋅ dw3⋅ 2tw dw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I3
1
12
b'f⋅ t'f
3⋅ b'f t'f⋅( ) D' 0.5t'f− yc−( )2⋅+:=
I I1 I2+ I3+:=
QA bf tf⋅( ) yc 0.5tf−( )⋅:=
QB b'f t'f⋅( ) D' 0.5t'f− yc−( )⋅:=
Shear Flow :
qA
V QA⋅
2I
:= qA 65.09
kN
m
= Ans
qB
V QB⋅
2I
:= qB 43.63
kN
m
= Ans
Problem 7-65
The beam is constructed from four plates and is subjected to a shear force of V = 25 kN. Determine
the maximum shear flow in the cross section.
Given: bf 274mm:= tf 12mm:=
b'f 250mm:= t'f 12mm:=
dw 200mm:= tw 12mm:=
d'w 50mm:= V 25kN:=
Solution:
Section Property : D dw tf+:=
D' D d'w−:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) 2tw dw⋅( )+ D' 0.5t'f−( ) b'f t'f⋅( )+
bf tf⋅ 2tw dw⋅+ b'f t'f⋅+
:=
yc 92.47 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
2tw( )⋅ dw3⋅ 2tw dw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I3
1
12
b'f⋅ t'f
3⋅ b'f t'f⋅( ) D' 0.5t'f− yc−( )2⋅+:=
I I1 I2+ I3+:=
Qmax bf tf⋅( ) yc 0.5tf−( )⋅ 2tw
yc tf−
2
⎛
⎜
⎝
⎞
⎠
⋅ yc tf−( )⋅+:=
Maximum Shear Flow :
qmax
V Qmax⋅
2I
:= qmax 82.88
kN
m
= Ans
Problem 7-66
A shear force of V = 18 kN is applied to the box girder. Determine the position d of the stiffener plates
BE and FG so that the shear flow at A is twice as great as the shear flow at B. Use the centerline
dimensions for the calculation. All plates are 10 mm thick.
Given: t 10mm:= bf 135mm t−:=
V 18kN:= dw t 290mm+:=
Solution:
Section Property :
QA bf t⋅( ) 0.5dw 0.5t−( )⋅:= QA 181250 mm3=
QB bf t⋅( ) d( )⋅=
Shear Flow :
qA
1
2
V QA⋅
I
= qB
1
2
V QB⋅
I
=
Require, qA 2qB=
1
2
V QA⋅
I
2
1
2
V QB⋅
I
⎛
⎜
⎝
⎞
⎠
⋅=
QA 2QB=
QA 2 bf t⋅( ) d( )⋅=
d
QA
2bf t⋅
:=
d 72.50 mm= Ans
Problem 7-67
The pipe is subjected to a shear force of V = 40 kN. Determine the shear flow in the pipe at points A
and B.
Given: ri 150mm:= t 5mm:=
V 40kN:=
Solution:
Section Property : ro ri t+:=
I
π
4
ro
4 ri
4−⎛⎝
⎞
⎠⋅:=
Since a' -> 0, then QA 0:=
QB
4ro
3π
π ro
2⋅
2
⎛
⎜
⎝
⎞
⎠
4ri
3π
π ri
2⋅
2
⎛
⎜
⎝
⎞
⎠
−:=
Shear Flow :
qA
V QA⋅
I
:= qA 0.00
kN
m
= Ans
qB
V QB⋅
2I
:= qB 83.48
kN
m
= Ans
Problem 7-68
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown where b2 > b1. The member segments have the same thickness t.
Problem 7-69
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.The member segments have the same thickness t.Problem 7-70
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-71
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-72
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-73
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-74
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Given: dθ 150mm:= dv 150mm:=
θ 30deg:=
Solution: Set t 1mm:=
Section Property :
hθ dθ sin θ( )⋅:= bθ
t
sin θ( )
:=
Iθ
1
12
bθ⋅ hθ
3⋅ dθ t⋅( )
dv hθ+
2
⎛
⎜
⎝
⎞
⎠
2
⋅+:=
Iv
1
12
t⋅ dv
3⋅:=
I 2Iθ Iv+:=
y'c 0.5dv hθ+ 0.5x sin θ( )⋅−=
Q A' y'c⋅= Q x t⋅( ) 0.5dv hθ+ 0.5x sin θ( )⋅−( )⋅=
Shear Flow Resultant:
q
V Q⋅
I
= V P=
q
P
x t⋅
I
0.5dv hθ+ 0.5x sin θ( )⋅−( )⋅=
F1
P
0
dθ
x
x t⋅
I
0.5dv hθ+ 0.5x sin θ( )⋅−( )⋅
⌠
⎮
⎮⌡
d=
Shear Center: Summing moment about point A
P e⋅ F1 dv⋅ cos θ( )⋅=
e
F1
P
dv⋅ cos θ( )⋅=
e
0
dθ
xdv cos θ( )⋅
x t⋅
I
⎛⎜
⎝
⎞
⎠
⋅ 0.5dv hθ+ 0.5x sin θ( )⋅−( )⋅
⌠
⎮
⎮⌡
d:=
e 43.30 mm= Ans
Problem 7-75
Determine the location e of the shear center, point O, for the thin-walled member having a slit along its
side.
Given: a 100mm:= b 100mm:=
Solution:
Set t mm:= P kN:=
Section Property :
h 2a:=
I
1
12
2t( )⋅ h3⋅ 2 b t⋅( ) a2⋅+:= I 3.3333 a3t=
Q1 y t⋅( )
1
2
⋅ y( )= Q1
t
2
y2=
Q2 a t⋅( )
1
2
⋅ a( ) x t⋅( ) a( )⋅+= Q1
a t⋅
2
a 2x+( )=
Shear Flow Resultant:
q1
V Q1⋅
I
= q1
P t⋅ y2⋅
2I
=
q2
V Q2⋅
I
= q2
P a t⋅( )⋅ a 2x+( )⋅
2I
=
Fw
0
a
yq1
⌠
⎮
⌡
d= Fw
0
a
y
P t⋅ y2⋅
2I
⌠
⎮
⎮
⌡
d:= Fw 0.05 P=
Ff
0
b
xq2
⌠
⎮
⌡
d= Ff
0
b
x
P a t⋅( )⋅ a 2x+( )⋅
2I
⌠
⎮
⎮
⌡
d:= Ff 0.3 P=
Shear Center: Summing moment about point A
P e⋅ 2Fw b⋅ Ff h⋅+=
e
1
P
2Fw b⋅ Ff h⋅+( )⋅:=
e 70 mm= Ans
Problem 7-76
Determine the location e of the shear center, point O, for the thin-walled member having a slit along its
side. Each element has aconstant thickness t.
Given: a mm:= b a:=
Solution:
Set t mm:= P kN:=
Section Property :
h 2a:=
I
1
12
2t( )⋅ h3⋅ 2 b t⋅( ) a2⋅+:= I 3.3333 a3t=
Q1 y t⋅( )
1
2
⋅ y( )= Q1
t
2
y2=
Q2 a t⋅( )
1
2
⋅ a( ) x t⋅( ) a( )⋅+= Q1
a t⋅
2
a 2x+( )=
Shear Flow Resultant:
q1
V Q1⋅
I
= q1
P t⋅ y2⋅
2I
=
q2
V Q2⋅
I
= q2
P a t⋅( )⋅ a 2x+( )⋅
2I
=
Fw
0
a
yq1
⌠
⎮
⌡
d= Fw
0
a
y
P t⋅ y2⋅
2I
⌠
⎮
⎮
⌡
d:=
Fw 0.05 P=
Ff
0
b
xq2
⌠
⎮
⌡
d= Ff
0
b
x
P a t⋅( )⋅ a 2x+( )⋅
2I
⌠
⎮
⎮
⌡
d:=
Ff 0.3 P=
Shear Center: Summing moment about point A
P e⋅ 2Fw b⋅ Ff h⋅+=
e
1
P
2Fw b⋅ Ff h⋅+( )⋅:=
e 0.7 a= Ans
Problem 7-77
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.
Given: a mm:= θ 60deg:=
Solution:
Set t mm:= P kN:=
Section Property :
b a sin θ( )⋅:= b 0.86603 a=
t'
t
cos θ( )
:= t' 2 t= h
a
2
:=
I
1
12
t( )⋅ a3⋅
1
12
t'( ) a3⋅+:= I 0.25 a3t=
Q1 y' t'⋅( )
1
2
⋅ y'( )= Q1
t'
2
y'2=
Q2 h t'⋅( )
1
2
⋅ h( ) h y−( ) t⋅ y
1
2
h y−( )+⎡⎢
⎣
⎤⎥
⎦
+=
Q2
1
2
t'h2 t h2 y2−( )+⎡⎣ ⎤⎦=
Shear Flow Resultant:
q1
V Q1⋅
I
= q1
P t'⋅ y'2⋅
2I
=
q2
V Q2⋅
I
= q2
P t'h2 t h2 y2−( )+⎡⎣ ⎤⎦⋅
2I
=
F'
0
h
yq1
⌠
⎮
⌡
d= F'
0
h
y'
P t'⋅ y'2⋅
2I
⌠
⎮
⎮
⌡
d:= F' 0.1667 P=
F
0
h
yq2
⌠
⎮
⌡
d= F
0
h
y
P t'h2 t h2 y2−( )+⎡⎣ ⎤⎦⋅
2I
⌠
⎮
⎮
⌡
d:= F 0.6667 P=
Shear Center: Summing moment about point A
P e⋅ 2 F b⋅ F' 0⋅+( )=
e
1
P
2F b⋅( )⋅:=
e 1.1547 a= Ans
Problem 7-78
If the angle has a thickness of 3 mm, a height h = 100 mm, and it is subjected to a shear of V = 50 N,
determine the shear flow at point A and the maximum shear flow in the angle.
Given: h 100mm:= t 3mm:=
θ 45deg:= V 50N:=
Solution:
Section Property :
t'
t
sin θ( )
:= t' 4.2426 mm=
I
1
12
2t'( )⋅ h3⋅:=
Q A' y'c⋅=
Q t' 0.5h y−( )⋅
1
2
0.5h y−( ) y+⎡⎢
⎣
⎤⎥
⎦
⋅=
Q
t'
2
0.25h2 y2−( )=
Shear Flow: q
V Q⋅
I
=
q
V t'⋅
2I
0.25h2 y2−( )⋅=
At A, yA 0.5 h⋅:= qA 0:= Ans
At y = 0, q qmax=
qmax
V t'⋅
2I
0.25h2( )⋅:=
qmax 375
N
m
= Ans
Problem 7-79
The angle is subjected to a shear of V = 10 kN. Sketch the distribution of shear flow along the leg AB.
Indicate numerical values at all peaks. The thickness is 6 mm and the legs (AB) are 125 mm.
Given: L 125mm:= t 6mm:=
θ 45deg:= V 10kN:=
Solution:
Section Property :
h L cos θ( )⋅:= t' t
sin θ( )
:= t' 8.4853 mm=
I
1
12
2t'( )⋅ h3⋅:=
Q A' y'c⋅=
Q t' 0.5h y−( )⋅
1
2
0.5h y−( ) y+⎡⎢
⎣
⎤⎥
⎦
⋅=
Q
t'
2
0.25h2 y2−( )=
Shear Flow: q
V Q⋅
I
=
q
V t'⋅
2I
0.25h2 y2−( )⋅= Ans
At y = 0, q qmax=
qmax
V t'⋅
2I
0.25h2( )⋅:=
qmax 84.85
kN
m
= Ans
Problem 7-80
Determine the placement e for the force P so that the beam bends downward without twisting. Take
h = 200 mm.
Given: h1 100mm:= bf 300mm:=
h2 200mm:=
Solution:
Set t mm:= P kN:=
Section Property :
I
1
12
bf t
3⋅ t h1
3⋅+ t h2
3⋅+⎛⎝
⎞
⎠⋅:=
Qw2 t 0.5h2 y−( )⋅ y
1
2
0.5h2 y−( )+⎡⎢⎣
⎤⎥
⎦
⋅=
Qw2
t
2
0.25h2
2 y2−⎛⎝
⎞
⎠⋅=
Shear Flow Resultant:
qw2
V Qw2⋅
I
= qw2
P t⋅ 0.25h2
2 y2−⎛⎝
⎞
⎠⋅
2I
=
Fw2
0.5− h2
0.5h2
xqw2
⌠⎮
⎮⌡
d=
Fw2
0.5− h2
0.5h2
y
P t⋅ 0.25h2
2 y2−⎛⎝
⎞
⎠⋅
2I
⌠⎮
⎮
⎮⌡
d:=
Fw2 0.8889 P=
Shear Center: Summing moment about point A
P e⋅ Fw2 bf t+( )⋅=
e
1
P
Fw2 bf t+( )⋅⎡⎣ ⎤⎦⋅:=
e 267.5 mm= Ans
Problem 7-81
A force P is applied to the web of the beam as shown. If e = 250 mm, determine the height h of the
right flange so that the beam will deflect downward without twisting. The member segments have the
same thickness t.
Given: h1 100mm:= bf 300mm:=
e 250mm:=
Solution:
Set t mm:= P kN:=
Shear Center: Summing moment about point A
P e⋅ Fw2 bf t+( )⋅= Fw2
e
bf t+
P⋅:=
Assume bf+t equal to bf : Fw2
e
bf
P⋅:=
Section Property : Assume bf t3 negligible.
I
1
12
t h1
3⋅ t h2
3⋅+⎛⎝
⎞
⎠⋅=
Qw2 t 0.5h2 y−( )⋅ y
1
2
0.5h2 y−( )+⎡⎢⎣
⎤⎥
⎦
⋅=
Qw2
t
2
0.25h2
2 y2−⎛⎝
⎞
⎠⋅=
Shear Flow Resultant:
qw2
V Qw2⋅
I
= qw2
P t⋅ 0.25h2
2 y2−⎛⎝
⎞
⎠⋅
2I
=
Fw2
0.5− h2
0.5h2
xqw2
⌠⎮
⎮⌡
d=
e
bf
P⋅
0.5− h2
0.5h2
y
6P t⋅ 0.25h2
2 y2−⎛⎝
⎞
⎠⋅
t h1
3⋅ t h2
3⋅+
⌠
⎮
⎮
⎮
⌡
d=
Given
e h1
3 h2
3+⎛⎝
⎞
⎠⋅
6bf 0.5− h2
0.5h2
y0.25h2
2 y2−⎛⎝
⎞
⎠
⌠⎮
⎮⌡
d=
Guess h2 10mm:= h2 Find h2( ):=
h2 171.0 mm= Ans
Problem 7-82
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.
Problem 7-83
Determine the location e of the shear center, point O, for the tube having a slit along its length.
Problem 7-84
The beam is fabricated from four boards nailed together as shown. Determine the shear force each nail
along the sides C and the top D must resist if the nails are uniformly spaced s = 75 mm. The beam is
subjected to a shear of V = 22.5 kN.
Given: bf 250mm:= dw 300mm:=
tf 25mm:= tw 25mm:=
tb 25mm:= db 100mm:=
V 22.5kN:= sn 75mm:=
Solution:
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) dw tw⋅( )+ 0.5db 2tb db⋅( )+
bf tf⋅ dw tw⋅+ 2 tb db⋅( )+
:=
yc 87.50 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
tw⋅ dw
3⋅ dw tw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I3
1
12
2tb( )⋅ db3⋅ 2tb db⋅( ) 0.5db yc−( )2⋅+:=
I I1 I2+ I3+:=
QC tb db⋅( ) yc 0.5db−( )⋅:=
QD dw tw⋅( ) D yc− 0.5dw−( )⋅:=
Shear Flow : q
V Q⋅
I
=
The allowable shear flow at points C and D are : qC
FC
sn
= qD
FD
sn
=
FC
sn
V QC⋅
I
=
FD
sn
V QD⋅
I
=
FC
V QC⋅ sn⋅
I
:= FD
V QD⋅ sn⋅
I
:=
FC 0.987 kN= FD 6.906 kN= Ans
Problem 7-85
The beam is constructed from four boards glued together at their seams. If the glue can withstand 15
kN/m, what is the maximum vertical shear V that the beam can support?
Given: bf 100mm:= dw 249mm:=
tf 12mm:= tw 12mm:= di 75mm:=
qallow 15
kN
m:=
Solution:
Section Property :
I
2
12
tw⋅ dw
3⋅
2
12
bf⋅ tf
3⋅+ 2bf tf⋅
di tf+
2
⎛
⎜
⎝
⎞
⎠
2
⋅+:=
Q bf tf⋅( )
di tf+
2
⋅:=
Shear Flow : Since there are two glue joints, hence 2q
V Q⋅
I
=
Vmax
2I qallow⋅
Q
:=
Vmax 20.37 kN= Ans
Problem 7-86
Solve Prob. 7-85 if the beam is rotated 90° from the position shown.
Given: bf 100mm:= dw 249mm:=
tf 12mm:= tw 12mm:= di 75mm:=
qallow 15
kN
m
:=
Solution:
Section Property :
I
2
12
tf⋅ bf
3⋅
2
12
dw⋅ tw
3⋅+ 2dw tw⋅
bf tw+
2
⎛
⎜
⎝
⎞
⎠
2
⋅+:=
Q dw tw⋅( )
bf tw+
2
⋅:=
Shear Flow : Since there are two glue joints, hence 2q
V Q⋅
I
=
Vmax
2I qallow⋅
Q
:=
Vmax 3.731 kN= Ans
Problem 7-87
The member is subjected to a shear force of V = 2 kN. Determine the shear flow at points A, B, and C.
The thickness of each thin-walled segment is 15 mm.
Given: bf 200mm:= dw 300mm:=
tf 15mm:= tw 15mm:=
tb 15mm:= db 115mm:=
V 2kN:=
Solution:
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) dw tw⋅( )+ 0.5db 2tb db⋅( )+
bf tf⋅ dw tw⋅+ 2 tb db⋅( )+
:=
yc 87.98 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
tw⋅ dw
3⋅ dw tw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I3
1
12
2tb( )⋅ db3⋅ 2tb db⋅( ) 0.5db yc−( )2⋅+:=
I I1 I2+ I3+:=
I 86939045.38 mm4=
QA 0:=
QB tb db⋅( ) yc 0.5db−( )⋅:= QB 52577.05 mm3=
QC QB tf 0.5bf 0.5tw−( )⋅ yc 0.5tf−( )⋅+:= QC 164242.29 mm3=
Shear Flow : q
V Q⋅
I
=
qA 0:= Ans
qB
V QB⋅
I
:= qB 1.210
kN
m
= Ans
qC
V QC⋅
I
:= qC 3.778
kN
m
= Ans
Problem 7-88
The member is subjected to a shear force of V = 2 kN. Determine the maximum shear flow in the
member. All segments of the cross section are 15 mm thick.
Given: bf 200mm:= dw 300mm:=
tf 15mm:= tw 15mm:=
tb 15mm:= db 115mm:=
V 2kN:=
Solution:
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) dw tw⋅( )+ 0.5db 2tb db⋅( )+
bf tf⋅ dw tw⋅+ 2 tb db⋅( )+
:=
yc 87.98 mm=
I1
1
12
bf⋅ tf
3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
tw⋅ dw
3⋅ dw tw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I3
1
12
2tb( )⋅ db3⋅ 2tb db⋅( ) 0.5db yc−( )2⋅+:=
I I1 I2+ I3+:=
I 86939045.38 mm4=
Qmax tw D yc−( )⋅
1
2
⋅ D yc−( )⋅:=
Qmax 386537.47 mm
3=
Shear Flow : q
V Q⋅
I
=
Maximum shear flow occurs at the point where the neutral axis passes through the section.
qmax
V Qmax⋅
I
:= qmax 8.892
kN
m
= Ans
Problem 7-89
The beam is made from three thin plates welded together as shown. If it is subjected to a shear of
V = 48 kN, determine the shear flow at points A and B. Also, calculate the maximum shear stress in
the beam.
Given: bf 215mm:= tf 15mm:=
tw 15mm:= dw 315mm:=
V 48kN:= h 200mm:=
Solution:
Section Property : a 0.5 bf tw−( ):=
yc
bf tw−( ) tf⋅⎡⎣ ⎤⎦ h 0.5tf+( ) dw tw⋅( ) 0.5dw( )+
bf tw−( ) tf⋅ dw tw⋅+
:=
yc 176.92 mm=
If
1
12
bf tw−( )⋅ tf3⋅ bf tw−( ) tf⋅ h 0.5 tf⋅+ yc−( )2⋅+:=
Iw
1
12
tw⋅ dw
3⋅ tw dw⋅( ) yc 0.5dw−( )2⋅+:=
I If Iw+:= I 43.71347 10
6−× m4=
QA a tw⋅( ) dw yc− 0.5a−( )⋅:= QA 132123.79 mm3=
QB a tf⋅( ) h 0.5 tf⋅+ yc−( )⋅:= QB 45873.79 mm3=
Qmax tw yc( )⋅
1
2
⋅ yc( )⋅:= Qmax 234748.45 mm3=
Shear Flow : q
V Q⋅
I
=
qA
V QA⋅
I
:= qA 145.1
kN
m
= Ans
qB
V QB⋅
I
:= qB 50.37
kN
m
= Ans
Maximum Shear Stress: τ
V Q⋅
I b⋅
=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τmax
V Qmax⋅
I tw( )⋅
:= τmax 17.18 MPa= Ans
Problem 7-90
A steel plate having a thickness of 6 mm is formed into the thin-walled section shown. If it is subjected
to a shear force of V = 1.25 kN, determine the shear stress at points A and C. Indicate the results on
volume elements located at these points.
Given: bf 100mm:= dw 50mm:= b'f 25mm:=
t 6mm:= V 1.25kN:=
Solution:
Section Property : D dw 2t+:=
mm
mm
yc
0.5t 2b'f t⋅( ) 0.5dw t+( ) 2dw t⋅( )+ D 0.5t−( ) bf t⋅( )+
2b'f t⋅ 2dw t⋅+ bf t⋅+
:=
yc 36.60 mm=
I1
1
12
b'f⋅ t
3⋅ b'f t⋅( ) 0.5t yc−( )2⋅+:=
I2
1
12
t⋅ dw
3⋅ dw t⋅( ) 0.5dw t+ yc−( )2⋅+:=
I3
1
12
bf⋅ t
3⋅ bf t⋅( ) D 0.5t− yc−( )2⋅+:=
I 2I1 2I2+ I3+:=
QA b'f t⋅( ) yc 0.5t−( )⋅:=
QC b'f t⋅( ) yc 0.5t−( )⋅ dw t⋅( ) 0.5dw t+ yc−⋅+ 0.5bf t⋅( ) D 0.5t− yc−( )⋅−:=
QC 0.00 mm
3= (since A' = 0)
Shear Stress : τ
V Q⋅
I t⋅
=
τA
V QA⋅
I t⋅
:= τA 1.335 MPa= Ans
τC
V QC⋅
I t⋅
:= τC 0 MPa= Ans
Problem 7-91
A steel plate having a thickness of 6 mm is formed into the thin-walled section shown. If it is subjected
to a shear force of V = 1.25 kN, determine the shear stress at point B.
Given: bf 100mm:= dw 50mm:= b'f 25mm:=
t 6mm:= V 1.25kN:=
Solution:
Section Property : D dw 2t+:=
yc
0.5t 2b'f t⋅( ) 0.5dw t+( ) 2dw t⋅( )+ D 0.5t−( ) bf t⋅( )+
2b'f t⋅ 2dw t⋅+ bf t⋅+
:=
yc 36.60 mm=
I1
1
12
b'f⋅ t
3⋅ b'f t⋅( ) 0.5t yc−( )2⋅+:=
I2
1
12
t⋅ dw
3⋅ dw t⋅( ) 0.5dw t+ yc−( )2⋅+:=
I3
1
12
bf⋅ t
3⋅ bf t⋅( ) D 0.5t− yc−( )2⋅+:=
I 2I1 2I2+ I3+:=
QB bf t⋅( ) D 0.5t− yc−( )⋅:=
Shear Stress : τ
V Q⋅
I b⋅
=
τB
V QB⋅
I 2t( )⋅
:= τB 1.781 MPa= Ans
Problem 7-92
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.
Problem 7-93
Sketch the intensity of the shear-stress distribution acting over the beam's cross-sectional area, and
determine the resultant shear force acting on the segment AB. The shear acting at the section is V =
175 kN. Show that INA = 340.82(106) mm4.
Given: b1 200mm:= d1 200mm:= V 175kN:=
b2 50mm:= d2 150mm:=
Solution:
Section Property : D d1 d2+:=
yc
0.5d1 b1 d1⋅( ) 0.5d2 d1+( ) b2 d2⋅( )+
b1 d1⋅ b2 d2⋅+
:= yc 127.63 mm=
y'c D yc−:=
I1
1
12
b1⋅ d1
3⋅ b1 d1⋅( ) 0.5d1 yc−( )2⋅+:=
I2
1
12
b2⋅ d2
3⋅ b2 d2⋅( ) D 0.5d2− yc−( )2⋅+:=
I I1 I2+:= I 340.82 10
6× mm4= (Q.E.D)
A'1 yc y1−( ) b1⋅= y1c 0.5 yc y1−( ) y1+= y1c 0.5 yc y1+( )=
Q1 A'1 y'1c⋅= Q1 0.5 yc y1−( ) b1⋅ yc y1+( )= Q1 0.5b1 yc2 y12−⎛⎝ ⎞⎠=
A'2 y'c y2−( ) b2⋅= y2c 0.5 y'c y2−( ) y2+= y2c 0.5 y'c y2+( )=
Q2 A'2 y'2c⋅= Q2 0.5 y'c y2−( ) b2⋅ y'c y2+( )= Q2 0.5b2 y'c2 y22−⎛⎝ ⎞⎠=
Shear Stress : τ
V Q⋅
I b⋅
= τCB
V
I
⎛⎜
⎝
⎞
⎠
0.5 yc
2 y1
2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅=
τ1B
V
I
⎛⎜
⎝
⎞
⎠
0.5yc
2⎛
⎝
⎞
⎠⋅:=At B: y1 0=
τ1B 4.18 MPa=
At C: y1 yc d1−:= τ1C
V
I
⎛⎜
⎝
⎞
⎠
0.5 yc
2 y1
2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅:=
τ1C 2.84 MPa=
τAB
V
INA
⎛
⎜
⎝
⎞
⎠
0.5 y'c
2 y2
2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅=
At C: y2 y'c d2−:= τ2C
V
I
⎛⎜
⎝
⎞
⎠
0.5 y'c
2 y2
2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅:=
τ2C 11.35 MPa=
Resultant Shear Force: For segment AB. yo y'c d2−:= VAB
A
AτAB
⌠
⎮
⌡
d=
VAB
yo
y'c
y
V
I
⎛⎜
⎝
⎞
⎠
0.5 y'c
2 y2−⎛⎝
⎞
⎠
⎡
⎣
⎤
⎦⋅ b2⋅
⌠
⎮
⎮
⌡
d:=
VAB 49.78 kN= Ans
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