9789810679941 sm 14

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Problem 14-01
A material is subjected to a general state of plane stress. Express the strain energy density in terms of
the elastic constants E, G, and ν and the stress components σx , σy , and τxy .
Problem 14-02
The strain-energy density must be the same whether the state of stress is represented by σx , σy , and
τxy , or by the principal stresses σ1 and σ2 . This being the case, equate the strain-energy expressions
for each of these two cases and show that G = E/[2(1 + ν)].
Problem 14-03
Determine the strain energy in the rod assembly. Portion AB is steel, BC is brass, and CD is aluminum.
Est = 200 GPa, Ebr = 101 GPa, and Eal = 73.1 GPa.
Given: Lst 0.3m:= dst 15mm:=
Lbr 0.4m:= dbr 20mm:=
Lal 0.2m:= dal 25mm:=
Est 200GPa:= PA 3kN:=
Ebr 101GPa:= PB 4kN:=
Eal 73.1GPa:= PC 10− kN:=
Solution:
Section Properties :
Ast
π dst2⋅
4
:= Ast 176.71 mm2=
Abr
π dbr2⋅
4
:= Abr 314.16 mm2=
Aal
π dal2⋅
4
:= Aal 490.87 mm2=
Internal Axial Forces : 
Pst PA:=
Pbr PA PB+:=
Pal PA PB+ PC+:=
∑= L/(2EA)NU 2iAxial Strain Energy:
Ui
Pst
2 Lst⋅
2 Est⋅ Ast⋅
Pbr
2 Lbr⋅
2 Ebr⋅ Abr⋅
+
Pal
2 Lal⋅
2 Eal⋅ Aal⋅
+:=
Ui 0.372 J= Ans
Problem 14-04
Determine the torsional strain energy in the A-36 steel shaft. The shaft has a radius of 40 mm.
Given: LAB 0.6m:= TA 8kN m⋅:= r 40mm:=
LBC 0.4m:= TB 6− kN m⋅:= G 75GPa:=
LCD 0.5m:= TC 12− kN m⋅:=
Solution:
Section Properties :
J
π r4⋅
2
:= J 4021238.60 mm4=
Internal Torsional Moment :
TAB TA:=
TBC TA TB+:=
TCD TA TB+ TC+:=
∑= L/(2GJ)TU 2iAxial Strain Energy:
Ui
TAB
2 LAB⋅
2 G⋅ J⋅
TBC
2 LBC⋅
2 G⋅ J⋅+
TCD
2 LCD⋅
2 G⋅ J⋅+:=
Ui 149.2 J= Ans
Problem 14-05
Determine the torsional strain energy in the A-36 steel shaft. The shaft has a radius of 30 mm.
Given: L 0.5m:= r 30mm:= G 75GPa:=
TA 0:= TB 3kN m⋅:= TC 4− kN m⋅:=
Solution:
Section Properties :
J
π r4⋅
2
:= J 1272345.02 mm4=
Internal Torsional Moment :
TAB TA:=
TBC TA TB+:=
TCD TA TB+ TC+:=
∑= L/(2GJ)TU 2iAxial Strain Energy:
Ui
TAB
2 L⋅
2 G⋅ J⋅
TBC
2 L⋅
2 G⋅ J⋅+
TCD
2 L⋅
2 G⋅ J⋅+:=
Ui 26.20 J= Ans
Problem 14-06
Determine the bending strain energy in the beam. EI is constant.
Solution:
Support Reactions :
+
By symmetry, A = B = R
ΣFy=0; 2R P− 0=
R 0.5P=
Internal Moment Function: 
M R x⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Applying Eq. 14-17:
Ui
1
2E I⋅ 0
L
xM2
⌠⎮⌡ d=
2
2E I⋅ 0
0.5L
xM2
⌠⎮⌡ d=
2
2E I⋅ 0
0.5L
xR x⋅( )2⌠⎮⌡ d=
Ui
1
4E I⋅ 0
0.5L
xP x⋅( )2⌠⎮⌡ d=
P2
4E I⋅ 0
0.5L
xx2
⌠⎮⌡ d=
Ui
P2 L3⋅
96E I⋅= Ans
P 30kN:=Given: a 4m:= b 2m:=
I 22.5 106( )⋅ mm4:=
E 200GPa:=
Solution: L a b+:=
Use W 250x18 :
Ans
Support Reactions :
+
 
ΣFy=0; A B+ P− 0= (1)
ΣΜB=0; A a⋅ P b⋅+ 0= (2)
Solving Eqs. (1) and (2): A
P− b⋅
a
:= B P L⋅
a
:=
A 15− kN= B 45 kN=
Internal Moment Function: 
M1 A x1⋅= M2 A a x2−( )⋅=
M4 P− x4⋅= M3 P− b x3−( )⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
a) Using coocrdinates x1 and x4 and applying Eq. 14-17: 
Ui
1
2E I⋅
0
a
x1M1
2⌠⎮⌡ d 0
b
x4M4
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ui
1
2E I⋅
0
a
x1A x1⋅( )2⌠⎮⌡ d 0
b
x4P x4⋅( )2⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ui 800 J= Ans
b) Using coocrdinates x2 and x3 and applying Eq. 14-17: 
Ui
1
2E I⋅
0
a
xM2
2⌠⎮⌡ d 0
b
x3M3
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ui
1
2E I⋅
0
a
x2A a x2−( )⋅⎡⎣ ⎤⎦2⌠⎮⌡ d 0
b
x3P− b x3−( )⋅⎡⎣ ⎤⎦2⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ui 800 J=
Problem 14-07
Determine the bending strain energy in the A-36 structural steel W250 X 18 beam. Obtain the answer
using the coordinates (a) x1 and x4 , and (b) x2 and x3.
Problem 14-08
Determine the total axial and bending strain energy in the A-36 steel beam. A = 2300 mm2, I = 9.5(106)
mm4.
Given: L 10m:= A 2300mm2:=
P 15kN:= I 9.5 106( )⋅ mm4:=
E 200GPa:= w 1.5 kN
m
:=
Solution:
Support Reactions :
+
 
ΣFy=0; Av Bv+ w L⋅− 0= (1)
ΣΜB=0; Av L⋅ w L⋅ 0.5L( )⋅− 0= (2)
Solving Eqs. (1) and (2): Av 0.5w L⋅:= Av 7.5 kN=
Bv 0.5w L⋅:= Bv 7.5 kN=
+ ΣFx=0; Ah P+ 0=
Ah P−:= Ah 15− kN=
Internal Moment and Force Function: 
M' Av x⋅ w x⋅ 0.5x( )⋅−= Av x⋅ 0.5w x2⋅−=
N' Ah−=
∫= L0 2i dx/(2EA)NUAxial Strain Energy:
Ua.i
1
2E A⋅ 0
L
xN'2
⌠⎮⌡ d=
Ua.i
1
2E A⋅
0
L
xAh−( )2⌠⎮⌡ d:=
Ua.i 2.4457 J=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Ub.i
1
2E I⋅ 0
L
xM'2
⌠⎮⌡ d=
Ub.i
1
2E I⋅
0
L
xAv x⋅ 0.5w x2⋅−⎛⎝ ⎞⎠
2⌠⎮
⎮⌡ d:=
Ub.i 493.4211 J=
Total Strain Energy: 
Ui Ua.i Ub.i+:= Ui 495.9 J= Ans
Problem 14-09
Determine the total axial and bending strain energy in the A-36 structural steel W200 X 86 beam.
Given: a 3m:= P1 25kN:=
θ 30deg:= P2 15kN:=
E 200GPa:=
Solution: L 2a:=
Use W 200x86 : I 94.7 106( )⋅ mm4:=
A 11000mm2:=
Support Reactions :
+
 
ΣFy=0; Av Bv+ P1− P2 sin θ( )⋅− 0= (1)
ΣΜB=0; Av 2a( )⋅ P1 a⋅− 0= (2)
Solving Eqs. (1) and (2): Av 0.5P1:= Av 12.5 kN=
Bv 0.5P1 0.5P2+:= Bv 20 kN=
+ ΣFx=0; Ah P2 cos θ( )⋅+ 0=
Ah P2 cos θ( )⋅:= Ah 12.99 kN=
Internal Moment and Force Function: 
M1 Av x1⋅= M2 Bv P2 sin θ( )⋅−( )− x2⋅= M2 Av− x2⋅=
N1 Ah= N2 Ah=
∫= L0 2i dx/(2EA)NUAxial Strain Energy:
Ua.i
1
2E A⋅
0
a
x1N1
2⌠⎮⌡ d 0
a
x2N2
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ua.i
1
2E A⋅
0
a
x1Ah
2⌠⎮⌡ d 0
a
x2Ah
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
:=
Ua.i 0.23 J=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Ub.i
1
2E I⋅
0
a
x1M1
2⌠⎮⌡ d 0
a
x2M2
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ub.i
1
2E I⋅
0
a
x1Av x1⋅( )2⌠⎮⌡ d 0
a
x2Av− x2⋅( )2⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ub.i 74.25 J=
Total Strain Energy: 
Ui Ua.i Ub.i+:= Ui 74.48 J= Ans
Problem 14-10
Determine the bending strain energy in the beam. EI is constant.
Solution:
Support Reactions :
By symmetry, A = B = R
ΣMB=0; R L⋅ Mo+ Mo− 0=
R 0=
Internal Moment Function: 
M Mo=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Applying Eq. 14-17:
Ui
1
2E I⋅ 0
L
xM2
⌠⎮⌡ d=
1
2E I⋅
0
L
xMo
2⌠⎮⌡ d=
Ui
Mo
2
4E I⋅ 0
L
x1
⌠⎮⌡ d=
Ui
Mo
2 L⋅
2E I⋅= Ans
Problem 14-11
Determine the bending strain energy in the A-36 steel beam due to the loading shown. Obtain the
answer using the coordinates (a) x1 and x4 , and (b) x2 and x3. I = 21(106) mm4.
Unit used: kJ 1000J:=
Given: a 3m:= b 1.5m:= w 120 kN
m
:=
E 200GPa:=
Solution: L a b+:=
Section Property : I 21 106( )⋅ mm4:=
Support Reactions :
+
 
ΣFy=0; A B+ w b⋅− 0= (1)
ΣΜB=0; A a⋅ w b⋅( ) 0.5b( )⋅+ 0= (2)
Solving Eqs. (1) and (2): A
w− b2⋅
2a
:= B w b⋅ b 2a+( )⋅
2a
:=
A 45− kN= B 225 kN=
Internal Moment Function: 
M1 A x1⋅= M2 A a x2−( )⋅=
M4 0.5− w x42⋅= M3 0.5− w b x3−( )2⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
a) Using coocrdinates x1 and x4 and applying Eq. 14-17: 
Ui
1
2E I⋅
0
a
x1M1
2⌠⎮⌡ d 0
b
x4M4
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ui
1
2E I⋅
0
a
x1A x1⋅( )2⌠⎮⌡ d 0
b
x40.5− w x42⋅⎛⎝ ⎞⎠
2⌠⎮
⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ui 2.821 kJ= Ans
b) Using coocrdinates x2 and x3 and applying Eq. 14-17: 
Ui
1
2E I⋅
0
a
x2M2
2⌠⎮⌡ d 0
b
x3M3
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ui
1
2E I⋅
0
a
x2A a x2−( )⋅⎡⎣ ⎤⎦2⌠⎮⌡ d 0
b
x30.5− w b x3−( )2⋅⎡⎣ ⎤⎦2
⌠⎮
⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ui 2.821 kJ= Ans
Problem 14-12
Determine the bending strain energy in the cantilevered beam due to a uniform load w. Solve the
problem two ways. (a) Apply Eq. 14-17. (b) The load w dx acting on a segment dx of the beam is
displaced a distance y, where y = w (-x4 + 4L3x - 3L4)/(24EI), the equation of the elastic curve. Hence
the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e.,
dUi = ½ (w dx)(-y). Integrate this
equation to obtain the total strain energy in the beam. EI is constant.
Problem 14-13
Determine the bending strain energy in the simply supported beam due to a uniform load w. Solve the
problem two ways. (a) Apply Eq. 14-17. (b) The load w dx acting on a segment dx of the beam is
displaced a distance y, where y = w (-x4 + 2Lx3 - L3x)/(24EI), the equation of the elastic curve. Hence
the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e.,
dUi = ½ (w dx)(-y). Integrate this equation to obtain the total strain energy in the beam. EI is constant.
Problem 14-14
Determine the shear strain energy in the beam. The beam has a rectangular cross section of area A, and
the shear modulus is G.
Problem 14-15
The concrete column contains six 25-mm-diameter steel reinforcing rods. If the column supports a
load of 1500 kN, determine the strain energy in the column. Est = 200 GPA, Ec = 25 GPa.
Given: L 1.5m:= rc 300mm:= dst 25mm:=
Est 200GPa:= Econc 25GPa:= P 1500kN:=
Solution:
Section Properties : Ast 6
π dst2⋅
4
:= Ast 2945.24 mm2=
Aconc π rc2⋅ Ast−:= Aconc 279798.10 mm2=
Given
Equilibrium : 
+ ΣFy=0; Pconc Pst+ P− 0= (1)
Compatibility : ∆conc = ∆st 
Pconc L⋅
Aconc Econc⋅
Pst L⋅
Ast Est⋅
= (2)
Solving Eqs. (1) and (2): Guess Pconc 1kN:= Pst 1kN:=
Pconc
Pst
⎛⎜⎜⎝
⎞
⎠
Find Pconc Pst,( ):= PconcPst
⎛⎜⎜⎝
⎞
⎠
1383.50
116.50
⎛⎜⎝
⎞
⎠ kN=
∑= L/(2EA)NU 2iAxial Strain Energy:
Ui
Pconc
2 L⋅
2 Econc⋅ Aconc⋅
Pst
2 L⋅
2 Est⋅ Ast⋅
+:=
Ui 222.51 J= Ans
Problem 14-16
Determine the bending strain energy in the beam and the axial strain energy in each of the two rods.
The beam is made of 2014-T6 aluminum and has a square cross section 50 mm by 50 mm. The rods
are made of A-36 steel and have a circular cross section with a 20-mm diameter.
Given: Lst 2m:= Est 200GPa:= dst 20mm:=
Lal 4m:= Eal 73.1GPa:= dal 50mm:=
a 1m:= b 2m:= P 8kN:=
Solution:
Section Properties :
Ast
π dst2⋅
4
:= Ial
dal
4
12
:=
Support Reactions :
+
By symmetry, A = B = Fst
ΣFy=0; 2Fst 2P− 0=
Fst P:= Fst 8 kN=
Internal Moment and Force Function: 
Nst Fst=
M1 Fst x1⋅=
M2 Fst a x2+( )⋅ P x2⋅−= P a⋅=
Axial Strain Energy:
Ua.i
1
2Est Ast⋅ 0
Lst
xNst
2⌠⎮⌡ d=
Ua.i
1
2Est Ast⋅ 0
Lst
xFst
2⌠⎮⌡ d:=
Ua.i 1.019 J= Ans
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Using coocrdinates x1 and x2 and applying Eq. 14-17: 
Ubi
1
2Eal Ial⋅
2
0
a
x1M1
2⌠⎮⌡ d 0
b
x2M2
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ubi
1
2Eal Ial⋅
2
0
a
x1Fst x1⋅( )2⌠⎮⌡ d 0
b
x2P a⋅( )2
⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ubi 2241.3 J= Ans
∫= L0 2i dx/(2EA)NU
Problem 14-17
Determine the bending strain energy in the beam due to the distributed load. EI is constant.
Solution:
Internal Moment Function: 
M
x
2
wo
x
L
⋅⎛⎜⎝
⎞
⎠⋅
x
3
⋅= wo
x3
6L
⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Applying Eq. 14-17:
Ui
1
2E I⋅ 0
L
xM2
⌠⎮⌡ d=
1
2E I⋅
0
L
xwo
x3
6L
⋅⎛⎜⎝
⎞
⎠
2⌠⎮
⎮
⎮⌡
d=
Ui
wo
2
72E I⋅ L2⋅ 0
L
xx6
⌠⎮⌡ d=
Ui
wo
2 L5⋅
504E I⋅= Ans
Problem 14-18
The beam shown is tapered along its width. If a force P is applied to its end, determine the strain
energy in the beam and compare this result with that of a beam that has a constant rectangular cross
section of width b and height h.
Problem 14-19
The bolt has a diameter of 10 mm, and the link AB has a rectangular cross section that is 12 mm wide
by 7 mm thick. Determine the strain energy in the link due to bending and in the bolt due to axial force.
The bolt is tightened so that it has a tension of 500 N. Both members are made of A-36 steel. Neglect
the hole in the link.
Given: Lo 60mm:= do 10mm:= d 12mm:=
a 50mm:= b 30mm:= t 7mm:=
E 200GPa:= P 0.5kN:=
Solution:
Section Properties : Ao
π do2⋅
4
:= I d t
3⋅
12
:=
Support Reactions :
Ans
 
ΣFx=0; P A− B− 0= (1)
ΣΜB=0; A a b+( )⋅ P b⋅− 0= (2)
Solving Eqs. (1) and (2): A
P b⋅
a b+:= B P A−:=
A 0.1875 kN= B 0.3125 kN=
Internal Moment and Force Function: 
No P:=
M1 A x1⋅=
M2 B x2⋅=
Axial Strain Energy:
Ua.i
1
2E Ao⋅ 0
Lo
xNo
2⌠⎮⌡ d=
Ua.i
1
2E Ao⋅ 0
Lo
xP2
⌠⎮⌡ d:=
Ua.i 4.77 10
4−× J= Ans
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Using coocrdinates x1 and x2 and applying Eq. 14-17: 
Ubi
1
2 E⋅ I⋅
0
a
x1M1
2⌠⎮⌡ d 0
b
x2M2
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
⋅=
Ubi
1
2 E⋅ I⋅
0
a
x1A x1⋅( )2⌠⎮⌡ d 0
b
x2B x2⋅( )2⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
⋅:=
Ubi 0.0171 J=
+
∫= L0 2i dx/(2EA)NU
Problem 14-20
The steel beam is supported on two springs, each having a stiffness of k = 8 MN/m. Determine the
strain energy in each of the springs and the bending strain energy in the beam. Est = 200 GPa, I =
5(106) mm4.
Given: L 4m:= E 200GPa:= I 5 106( )⋅ mm4:=
a 1m:= k 8000 kN
m
:= w 2 kN
m
:=
b 2m:=
Solution:
Support Reactions :
+
By symmetry, A = B = R
ΣFy=0; 2R w L⋅− 0=
R 0.5w L⋅:= R 4 kN=
Internal Moment Function:
M1
w
2
x1
2⋅=
M2
w
2
a x2+( )2⋅ R x2⋅−=
2/U 2i δk=Spring Strain Energy: 
The spring deforms δ R
k
:= δ 0.500 mm=
Usp.i
1
2
k δ2⋅:=
Usp.i 1.000 J= Ans
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Using coocrdinates x1 and x2 and applying Eq. 14-17: 
Ubi
1
2E I⋅ 2
0
a
x1M1
2⌠⎮⌡ d 0
b
x2M2
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ubi
1
2E I⋅ 2
0
a
x1
w
2
x1
2⋅⎛⎜⎝
⎞
⎠
2⌠⎮
⎮⌡
d
0
b
x2
w
2
a x2+( )2⋅ R x2⋅−⎡⎢⎣ ⎤⎥⎦
2⌠⎮
⎮⌡
d+
⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
:=
Ubi 0.400 J= Ans
Problem 14-21
Determine the bending strain energy in the 50-mm-diameter A-36 steel rod due to the loading shown.
Given: a 0.6m:= b 0.6m:= do 50mm:=
E 200GPa:= P 400N:=
Solution:
Section Properties : I
π do4⋅
64
:=
Internal Moment Function: 
M1 P x1⋅= M2 P a⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Ui
1
2E I⋅ 2
0
a
x1M1
2⌠⎮⌡ d 0
b
x2M2
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ui
1
2E I⋅ 2
0
a
x1P x1⋅( )2⌠⎮⌡ d 0
b
x2P a⋅( )2
⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ui 0.469 J= Ans
Problem 14-22
The A-36 steel bar consists of two segments, one of circular cross section of radius r, and one of
square cross section. If it is subjected to the axial loading of P, determine the dimensions a of the
square segment so that the strain energy within the square segment is the same as in the circular
segment.
Problem 14-23
Consider the thin-walled tube of Fig. 5-30. Use the formula for shear stress, τavg = T/2tAm, Eq. 5-18,
and the general equation of shear strain energy, Eq. 14-11, to show that the twist of the tube is given
by Eq. 5-20. Hint: Equate the work done by the torque T to the strain energy in the tube, determined
from integrating the strain energy for a differential element, Fig. 14-4, over the volume of material.
Problem 14-24
Determine the horizontal displacement of joint C. AE is constant.
Problem 14-25
Determine the horizontal displacement of joint A. Each bar is made of A-36 steel and has a cross-
sectional area of 950 mm2.
Given: a 0.9m:= b 1.2m:= A 950mm2:=
E 200GPa:= P 10kN:=
Solution: c a2 b2+:=
Member Forces: Applying the method of joints to Joint A.
Given
+ ΣFy=0; FAB FAD
a
c
⎛⎜⎝
⎞
⎠⋅− 0= (1)
+ ΣFx=0; FAD
b
c
⎛⎜⎝
⎞
⎠⋅ P− 0= (2)
Solving Eqs. (1) and (2): Guess FAB 1kN:= FAD 1kN:=
(C)FAB
FAD
⎛⎜⎜⎝
⎞
⎠
Find FAB FAD,( ):= FABFAD
⎛⎜⎜⎝
⎞
⎠
7.50
12.50
⎛⎜⎝
⎞
⎠ kN= (T)
Applying the method of joints to Joint D.
Given
+ ΣFy=0; FCD− FBD
a
c
⎛⎜⎝
⎞
⎠⋅+ FAD
a
c
⎛⎜⎝
⎞
⎠⋅+ 0= (3)
+ ΣFx=0; FBD
b
c
⎛⎜⎝
⎞
⎠⋅ FAD
b
c
⎛⎜⎝
⎞
⎠⋅− 0= (4)
Solving Eqs. (3) and (4): Guess FCD 1kN:= FBD 1kN:=
(C)FBD
FCD
⎛⎜⎜⎝
⎞
⎠
Find
FBD FCD,( ):= FBDFCD
⎛⎜⎜⎝
⎞
⎠
12.50
15.00
⎛⎜⎝
⎞
⎠ kN= (T)
∑= L/(2EA)NU 2iAxial Strain Energy:
Ui
1
2 E⋅ A⋅ FAB
2 2a( )⋅ FAD2 c( )⋅+ FBD2 c( )⋅+ FCD2 b( )⋅+⎡⎣ ⎤⎦:=
Ui 2.211 J=
External Work: The external work done by the force P is
Ue
1
2
P⋅ ∆A.h( )⋅=
Conservation of Energy: Ui = Ue
Ui
1
2
P⋅ ∆A.h( )⋅=
∆A.h
2 Ui⋅
P
:= ∆A.h 0.442 mm= Ans
Problem 14-26
Determine the vertical displacement of joint D. AE is constant.
Given: L m:= a 0.6L:= b 0.8L:= c L:=
EA kN:= P kN:=
Solution:
Member Forces: 
By inspection of Joint D, FAD 0:=
FCD P:= (T)
Applying the method of joints to Joint C.
Given
+ ΣFy=0; FCA
b
c
⎛⎜⎝
⎞
⎠⋅ P− 0= (1)
+ ΣFx=0; FCB FCA
a
c
⎛⎜⎝
⎞
⎠⋅− 0= (2)
Solving Eqs. (1) and (2): Guess FCA 1kN:= FCB 1kN:=
(C)FCA
FCB
⎛⎜⎜⎝
⎞
⎠
Find FCA FCB,( ):= FCAFCB
⎛⎜⎜⎝
⎞
⎠
1.25
0.75
⎛⎜⎝
⎞
⎠ P= (T)
Applying the method of joints to Joint A.
+ ΣFy=0; FAB FCA
b
c
⋅− 0= FAB FCA
b
c
⋅:= FAB 1.00 P= (T)
Axial Strain Energy:
Ui
1
2 EA⋅ FCD
2 b( )⋅ FCA−( )2 c( )⋅+ FCB2 a( )⋅+ FAB2 b( )⋅+⎡⎣ ⎤⎦:=
Ui 1.750
P2 L⋅
EA
=
External Work: The external work done by the force P is
Ue
1
2
P⋅ ∆D.v( )⋅=
Conservation of Energy: Ui = Ue
Ui
1
2
P⋅ ∆D.v( )⋅=
∆D.v
2 Ui⋅
P
:=
∆D.v 3.50
P L⋅
EA
= Ans
Problem 14-27
Determine the slope at the end B of the A-36 steel beam. I = 80(106) mm4.
Given: L 8m:= I 80 106( )⋅ mm4:=
Ans
Mo 6kN m⋅:=
Solution:
Support Reactions :
+
 
ΣFy=0; A B+ 0= (1)
ΣΜB=0; A L⋅ Mo+ 0= (2)
Solving Eqs. (1) and (2): A
Mo−
L
:= B A−:=
A 0.75− kN= B 0.75 kN=
Internal Moment Function: 
M A x⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Ui
1
2E I⋅ 0
L
xA x⋅( )2⌠⎮⌡ d:=
Ui 3.000 J=
External Work: The external work done by the moment Mo is
Ue
1
2
Mo⋅ θB⋅=
Conservation of Energy: Ui = Ue
Ui
1
2
Mo⋅ θB⋅=
θB
2 Ui⋅
Mo
:=
θB 0.0010 rad=
E 200GPa:=
Problem 14-28
Determine the displacement of point B on the A-36 steel beam. I = 97.7(106) mm4.
Given: a 4.5m:= b 3m:= I 97.7 106( )⋅ mm4:=
E 200GPa:= P 40kN:=
Solution:
Internal Moment Function: 
M P x⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Ui
1
2E I⋅ 0
a
xP x⋅( )2⌠⎮⌡ d:=
Ui 1243.603 J=
External Work: The external work done by the force P is
Ue
1
2
P⋅ ∆B( )⋅=
Conservation of Energy: Ui = Ue
Ui
1
2
P⋅ ∆B( )⋅=
∆B
2 Ui⋅
P
:= ∆B 62.18 mm= Ans
Problem 14-29
The cantilevered beam has a rectangular cross-sectional area A, a moment of inertia I, and a modulus
of elasticity E. If a load P acts at point B as shown, determine the displacement at B in the direction of
P, accounting for bending, axial force, and shear.
Problem 14-30
Use the method of work and energy and determine the slope of the beam at point B. EI is constant.
Problem 14-31
Determine the slope at point A of the beam. EI is constant.
Problem 14-32
Determine the displacement of point B on the 2014-T6 aluminum beam.
Given: a 2m:= bf 150mm:= dw 150mm:=
b 6m:= df 25mm:= tw 25mm:=
P 20kN:= E 73.1GPa:=
Solution: L a b+:=
Section Property : h df dw+:=
yc
Σ yi
⎯
Ai⋅( )⋅
Σ Ai( )⋅=
yc
bf df⋅( ) 0.5df( )⋅ tw dw⋅( ) 0.5dw df+( )⋅+
bf df⋅( ) tw dw⋅( )+:= yc 56.25 mm=
I
1
12
bf⋅ df3⋅ bf df⋅( ) 0.5df yc−( )2⋅+ 112 tw⋅ dw3⋅+ tw dw⋅( ) 0.5dw df+ yc−( )2⋅+:=
I 21582031.25 mm4=Support Reactions :
+
 
ΣFy=0; A C+ P− 0= (1)
ΣΜC=0
;
A L⋅ P b⋅− 0= (2)
Solving Eqs. (1) and (2): A P
b
L
⋅:= A 15 kN=
C P A−:= C 5 kN=
Internal Moment Function: 
M1 A x1⋅= M2 C x2⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Ui
1
2E I⋅
0
a
x1M1
2⌠⎮⌡ d
⎛⎜⎜⎝
⎞
⎠ 0
b
x2M2
2⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
=
Ui
1
2E I⋅
0
a
x1A x1⋅( )2⌠⎮⌡ d 0
b
x2C x2⋅( )2⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ui 760.63 J=
External Work: The external work done by the force P is
Ue
1
2
P⋅ ∆B( )⋅=
Conservation of Energy: Ui = Ue
Ui
1
2
P⋅ ∆B( )⋅= ∆B 2 Ui⋅P:= ∆B 76.06 mm= Ans
Problem 14-33
The A-36 steel bars are pin connected at C. If they each have a diameter of 50 mm, determine the
displacement at E.
Given: a 3.6m:= b 3m:= c 2.4m:=
E 200GPa:= P 10kN:= do 50mm:=
Solution:
Section Properties : I
π do4⋅
64
:=
Support Reactions :
+
By symmetry A=B=R, and D=0
ΣFy=0; 2R P− 0= R 0.5P:= R 5 kN=
Internal Moment Function: 
M1 R x1⋅= M2 R x2⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Ui
1
2E I⋅
0
0.5a
x1M1
2⌠⎮⌡ d 0
0.5a
x2M2
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ui
1
2E I⋅
0
0.5a
x1R x1⋅( )2⌠⎮⌡ d 0
0.5a
x2R x2⋅( )2⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ui 792.057 J=
External Work: The external work done by the force P is
Ue
1
2
P⋅ ∆E( )⋅=
Conservation of Energy: Ui = Ue
Ui
1
2
P⋅ ∆E( )⋅= ∆E 2 Ui⋅P:= ∆E 158.41 mm= Ans
Problem 14-34
Determine the deflection of the beam at its center caused by shear. The shear modulus is G.
Problem 14-35
The A-36 steel bars are pin connected at B and C.If they each have a diameter of 30 mm, determine
the slope at E.
Given: a 3m:= b 2m:= do 30mm:=
E 200GPa:= Mo 0.3kN m⋅:=
Solution:
Section Properties : I
π do4⋅
64
:=
Equations of Equilibrium : For member BC,
+
 
ΣFy=0; B C+ 0= (1)
ΣΜB=0;
Ans
(2)
Solving Eqs. (1) and (2): B
Mo−
2 b⋅:= C B−:=
B 75− N= C 75 N=
Internal Moment Function: 
M1 B− x1⋅=
M2 B x2⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Using coocrdinates x1 and x2 and applying Eq. 14-17: 
Ubi
1
2E I⋅ 2
0
a
x1M1
2⌠⎮⌡ d 2 0
b
x2M2
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ubi
1
2E I⋅ 2
0
a
x1B− x1⋅( )2⌠⎮⌡ d 2 0
b
x2B x2⋅( )2⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ubi 8.252 J=
External Work: The external work done by the moment Mo is
Ue
1
2
Mo⋅ θE⋅=
Conservation of Energy: Ui = Ue
Ubi
1
2
Mo⋅ θE⋅=
θE
2 Ubi⋅
Mo
:= θE 0.05502 rad=
θE 3.152 deg=
B 2 b⋅( )⋅ Mo+ 0=
Problem 14-36
The A-36 steel bars are pin connected at B. If each has a square cross section, determine the vertical
displacement at B.
Given: a 2.4m:= b 1.2m:= c 3m:=
E 200GPa:= P 4kN:= do 75mm:=
Solution:
Section Properties : I
do
4
12
:=
+
Support Reactions : A 0:=
 
ΣFy=0; C D+ P− 0= (1)
ΣΜC=0
;
P− b⋅ D c⋅− 0= (2)
Solving Eqs. (1) and (2): D P− b
c
⋅:= D 1.6− kN=
C P D−:= C 5.6 kN=
Internal Moment Function: 
M1 P− x1⋅= M2 D− x2⋅=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Ui
1
2E I⋅
0
b
x1M1
2⌠⎮⌡ d 0
c
x2M2
2⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
Ui
1
2E I⋅
0
b
x1P− x1⋅( )2⌠⎮⌡ d 0
c
x2D x2⋅( )2⌠⎮⌡ d+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
Ui 30.583 J=
External Work: The external work done by the force P is
Ue
1
2
P⋅ ∆B( )⋅=
Conservation of Energy: Ui = Ue
Ui
1
2
P⋅ ∆B( )⋅= ∆B 2 Ui⋅P:= ∆B 15.29 mm= Ans
Problem 14-37
The rod has a circular cross section with a moment of inertia I. If a vertical force P is applied at A,
determine the vertical displacement at this point. Only consider the strain energy due to bending. The
modulus of elasticity is E.
Problem 14-38
The load P causes the open coils of the spring to make an angle θ with the horizontal when the spring
is stretched. Show that for this position this causes a torque T = PR cos θ and a bending moment M =
PR sin θ at the cross section. Use these results to determine the maximum normal stress in the
material.
Problem 14-39
The coiled spring has n coils and is made from a material having a shear modulus G. Determine the
stretch of the spring when it is subjected to the load P. Assume that the coils are close to each other so
that θ 0° and the deflection is caused entirely by the torsional stress in the coil.≈
Problem 14-40
A bar is 4 m long and has a diameter of 30 mm. If it is to be used to absorb energy in tension from an
impact
loading, determine the total amount of elastic energy that it can absorb if (a) it is made of steel
for which Est = 200 GPa, σY = 800 MPa, and (b) it is made from an aluminum alloy for which Eal = 70
GPa, σY = 405 MPa.
Unit used: kJ 1000J:=
Given: L 4m:= E1 200GPa:= E2 70GPa:=
do 30mm:= σY1 800MPa:= σY2 405MPa:=
Solution:
Geometric Property : V
π do2⋅
4
L⋅:= V 2827433.39 mm3=
a) Strain Energy: εY1
σY1
E1
:=
ur1
1
2
σY1⋅ εY1⋅:= ur1
1
2E1
σY12⋅:= ur1 1.600 106×
J
m3
=
Ui1 ur1 V⋅:=
Ui1 4.524 kJ= Ans
b) Strain Energy: εY2
σY2
E2
:=
ur2
1
2
σY2⋅ εY2⋅:= ur1
1
2E2
σY22⋅:= ur2 1.172 106×
J
m3
=
Ui2 ur2 V⋅:=
Ui2 3.313 kJ= Ans
Problem 14-41
Determine the diameter of a brass bar that is 2.4 m long if it is to be used to absorb 1200 J of energy in
tension from an impact loading. Take σY = 70 MPa, E = 100.GPa.
Given: L 2.4m:= Ui 1200J:=
E 100GPa:= σY 70MPa:=
Solution:
Geometric Property : V
π do2⋅
4
L⋅=
Strain Energy: εY
σY
E
:=
ur
1
2
σY⋅ εY⋅:= ur 24500.00
J
m3
=
Conservation of Energy: Ui = Ur
Ui ur V⋅=
Ui ur
π do2⋅
4
L⋅
⎛⎜⎝
⎞
⎠⋅=
do
4 Ui⋅
π L⋅ ur⋅
:=
do 161.20 mm= Ans
Problem 14-42
The collar has a weight of 250 N (~25 kg) and falls down the titanium bar. If the bar has a diameter of
12 mm determine the maximum stress developed in the bar if the weight is (a) dropped from a height
of h = 300 mm, (b) released from a height h 0, and (c) placed slowly on the flange at A. Eti = 120
GPa, σY = 420 MPa.
≈
Given: do 12mm:= L 1.2m:=
W 250N:= E 120GPa:= σY 420MPa:=
Solution:
Section Property : A
π do2⋅
4
:=
Static Displacement : ∆st
W L⋅
E A⋅:= ∆st 0.02210 mm=
a) Drop Height: h 300mm:=
Impact factor: η 1 1 2 h∆st
⋅++:= η 165.76=
Pmax W η⋅:=
σmax
Pmax
A
:=
σmax 366.4 MPa= Ans
b) Drop Height: h 0mm:=
Impact factor: η 1 1 2 h∆st
⋅++:= η 2.00=
Pmax W η⋅:=
σmax
Pmax
A
:=
σmax 4.421 MPa= Ans
c) No impact: η 1:=
Pmax W η⋅:=
σmax
Pmax
A
:=
σmax 2.210 MPa= Ans
Problem 14-43
The collar has a weight of 250 N (~25 kg) and falls down the titanium bar. If the bar has a diameter of
12 mm, determine the largest height h at which the weight can be released and not permanently damage
the bar after striking the flange at A. Eti = 120 GPa, σY = 420 MPa.
Given: do 12mm:= L 1.2m:=
W 250N:= E 120GPa:= σY 420MPa:=
Solution:
Section Property : A
π do2⋅
4
:=
Static Displacement : ∆st
W L⋅
E A⋅:= ∆st 0.02210 mm=
Maximum Drop Height: 
Impact factor: η 1 1 2 h∆st
⋅++=
σmax
W
A
⎛⎜⎝
⎞
⎠ 1 1 2
h
∆st
⋅++⎛⎜⎝
⎞
⎠
⋅= σmax σY=
h
∆st
2
σY A⋅
W
1−⎛⎜⎝
⎞
⎠
2
1−
⎡⎢⎢⎣
⎤⎥⎥⎦⋅:=
h 394.8 mm= Ans
Problem 14-44
The mass of 50 Mg is held just over the top of the steel post having a length of L = 2 m and a cross-
sectional area of 0.01 m2. If the mass is released, determine the maximum stress developed in the bar
and its maximum deflection. Est = 200 GPa, σY = 600 MPa.
Given: L 2m:= A 0.01m2:= mo 50 103( ) kg:=
E 200GPa:= σY 600MPa:=
Solution:
Weight : W mo g⋅:=
Static Displacement : ∆st
W L⋅
E A⋅:= ∆st 0.4903 mm=
Drop Height: h 0:=
Impact factor: η 1 1 2 h∆st
⋅++:= η 2.00=
Pmax W η⋅:=
σmax
Pmax
A
:= σmax 98.1 MPa= Ans
< σY = 600 MPa (O.K.!)
∆max ∆st η⋅:= ∆max 0.981 mm= Ans
Problem 14-45
Determine the speed v of the 50-Mg mass when it is just over the top of the steel post, if after impact,
the maximum stress developed in the post is 550 MPa. The post has a length of L = 1 m and a
cross-sectional area of 0.01 m2. Est = 200 GPa, σY = 600 MPa.
Given: L 1m:= A 0.01m2:= mo 50 103( ) kg:=
E 200GPa:= σY 600MPa:= σmax 550MPa:=
Solution:
Weight : W mo g⋅:=
Equivalent Spring Stiffness:
 
k
E A⋅
L
:= k 2.00 106× kN
m
=
Maximum Displacement :
Pmax A σmax⋅:= ∆max
Pmax
k
:=
∆max 2.75 mm=
Conservation of Energy: Ui = Ur
Given 1
2
mo⋅ v2⋅ W ∆max⋅+
1
2
k⋅ ∆max2⋅= (1)
Solving Eq. (1): Guess v 1
m
s
:=
v Find v( ):= v 0.499 m
s
= Ans
Problem 14-46
The composite aluminum bar is made from two segments having diameters of 5 mm and 10 mm.
Determine the maximum axial stress developed in the bar if the 5-kg collar is dropped from a height of
h = 100 mm. Eal = 70 GPa, σY = 410 MPa.
Given: d1 5mm:= d2 10mm:= E 70GPa:=
L1 0.2m:= L2 0.3m:= σY 410MPa:=
mo 5kg:= h 0.1m:=
Solution:
Weight : W mo g⋅:=
Section Property : A1
π d12⋅
4
:= A2
π d22⋅
4
:=
Static Displacement : ∆st
mo g⋅( ) L1⋅
E A1⋅
mo g⋅( ) L2⋅
E A2⋅
+:=
∆st 0.00981 mm=
Impact factor: η 1 1 2 h∆st
⋅++:= η 143.78=
Pmax W η⋅:=
σmax
Pmax
A1
:=
σmax 359.1 MPa= Ans
< σY = 410 MPa (O.K.!)
Problem 14-47
The composite aluminum bar is made from two segments having diameters of 5 mm and 10 mm.
Determine the maximum height h from which the 5-kg collar should be dropped so that it produces a
maximum axial stress in the bar of σmax = 300 MPa. Eal = 70 GPa, σY = 410 MPa.
Given: d1 5mm:= d2 10mm:= E 70GPa:=
L1 0.2m:= L2 0.3m:= σY 410MPa:=
mo 5kg:= σmax 300MPa:=
Solution:
Weight : W mo g⋅:=
Section Property : A1
π d12⋅
4
:= A2
π d22⋅
4
:=
Static Displacement : ∆st
W L1⋅
E A1⋅
W L2⋅
E A2⋅
+:=
∆st 0.00981 mm=
Maximum Drop Height: 
Impact factor: η 1 1 2 h∆st
⋅++=
σmax
W
A1
1 1 2
h
∆st
⋅++⎛⎜⎝
⎞
⎠
⋅=
h
∆st
2
σmax A1⋅
W
1−⎛⎜⎝
⎞
⎠
2
1−
⎡⎢⎢⎣
⎤⎥⎥⎦⋅:=
h 69.6 mm= Ans
Problem 14-48
A steel cable having a diameter of 16 mm wraps over a drum and is used to lower an elevator having a
weight of 4 kN (~400kg). The elevator is 45 m below the drum and is descending at the constant rate
of 0.6 m/s when the drum suddenly stops. Determine the maximum stress developed in the cable when
this occurs. Est = 200 GPa, σY = 350 MPa.
Given: do 16mm:= E 200GPa:= L 45m:=
W 4kN:= σY 350MPa:= v 0.6 m
s
:=
Solution:
Section Property : A
π do2⋅
4
:=
Eq. Spring Constant: k
E A⋅
L
:= k 893.61 kN
m
=
Conservation of Energy: Ui = Ur
Given 1
2
W
g
⎛⎜⎝
⎞
⎠⋅ v
2⋅ W ∆max⋅+
1
2
k⋅ ∆max2⋅= (1)
Solving Eq. (1): Guess ∆max 10mm:=
∆max Find ∆max( ):= ∆max 18.05 mm=
Maximum Stress:
Pmax k ∆max⋅:=
σmax
Pmax
A
:=
σmax 80.2 MPa= Ans
< σY = 350 MPa (O.K. !)
Problem 14-49
Solve Prob. 14-48 if the elevator is descending at the constant rate of 0.9 m/s.
Given: do 16mm:= E 200GPa:= L 45m:=
W 4kN:= σY 350MPa:= v 0.9 m
s
:=
Solution:
Section Property : A
π do2⋅
4
:=
Eq. Spring Constant: k
E A⋅
L
:= k 893.61 kN
m
=
Conservation of Energy: Ui = Ur
Given 1
2
W
g
⎛⎜⎝
⎞
⎠⋅ v
2⋅ W ∆max⋅+
1
2
k⋅ ∆max2⋅= (1)
Solving Eq. (1): Guess ∆max 10mm:=
∆max Find ∆max( ):= ∆max 24.22 mm=
Maximum Stress:
Pmax k ∆max⋅:=
σmax
Pmax
A
:=
σmax 107.6 MPa= Ans
< σY = 350 MPa (O.K. !)
Problem 14-50
The 250-N (~25-kg) weight is falling at 0.9 m/s at the instant it is 0.6 m above the spring and post
assembly. Determine the maximum stress in the post if the spring has a stiffness of k = 40 MN/m. The
post has a diameter of 75 mm and a modulus of elasticity of E = 48 GPa. Assume the material will not
yield.
Given: do 75mm:= h 0.6m:= W 250N:= L 0.6m:=
E 48GPa:= v 0.9 m
s
:= ksp
40 103( )⋅ kN
m
:=
Solution:
Section Property : A
π do2⋅
4
:=
Eq. Spring Constants: kp
E A⋅
L
:= kp 353429.17
kN
m
=
Equilibrium requires: Fsp Fp=
Hence, ksp ∆sp⋅ kp ∆p⋅= ∆sp
kp
ksp
∆p⋅= (1)
Conservation of Energy: Ui = Ur
1
2
W
g
⎛⎜⎝
⎞
⎠⋅ v
2⋅ W h ∆max+( )⋅+ 12 ksp⋅ ∆sp2⋅ 12 kp⋅ ∆p2⋅+= (2)
However, ∆max ∆p ∆sp+= (3)
Substituting Eqs.(1) and (3) into (2):
Given
1
2
W
g
⎛⎜⎝
⎞
⎠⋅ v
2⋅ W h ∆p+
kp
ksp
∆p⋅+
⎛⎜⎝
⎞
⎠
⋅+ 1
2
ksp⋅
kp
ksp
∆p⋅
⎛⎜⎝
⎞
⎠
2
⋅
1
2
kp⋅ ∆p2⋅+= (4)
Solving Eq. (4): Guess ∆p 10mm:=
∆p Find ∆p( ):= ∆p 0.3044 mm=
Maximum Stress:
Pmax kp ∆p⋅:=
σmax
Pmax
A
:=
σmax 24.4 MPa= Ans
Problem 14-51
The A-36 steel bolt is required to absorb the energy of a 2-kg mass that falls h = 30 mm. If the bolt has
a diameter of 4 mm, determine its required length L so the stress in the bolt does not exceed 150 MPa.
Given: do 4mm:= h 30mm:=
mo 2 kg⋅:= E 200GPa:= σallow 150MPa:=
Solution:
Weight : W mo g⋅:=
Section Property : A
π do2⋅
4
:=
Static Displacement : ∆st
W L⋅
E A⋅:=
Allowable Length:
Impact factor: η 1 1 2 h∆st
⋅++=
σallow
W
A
⎛⎜⎝
⎞
⎠ η⋅=
Given
σallow
W
A
⎛⎜⎝
⎞
⎠ 1 1 2 h⋅
E A⋅
W L⋅⋅++
⎛⎜⎝
⎞
⎠⋅= (1)
Solving Eq. (1): Guess L 10mm:=
L Find L( ):= L 850.1 mm= Ans
Problem 14-52
The A-36 steel bolt is required to absorb the energy of a 2-kg mass that falls h = 30 mm. If the bolt has
a diameter of 4 mm and a length of L = 200 mm, determine if the stress in the bolt will exceed 175
MPa.
Given: do 4mm:= L 200mm:= h 30mm:=
mo 2 kg⋅:= E 200GPa:= σallow 175MPa:=
Solution: W mo g⋅:=
Section Property : A
π do2⋅
4
:=
Static Displacement : ∆st
W L⋅
E A⋅:= ∆st 0.00156 mm=
Maximum Stress:
 
Impact factor: η 1 1 2 h∆st
⋅++:= η 197.07=
σmax
W
A
⎛⎜⎝
⎞
⎠ η⋅:=
σmax 307.6 MPa= > σallow = 175 MPa
(Exceeded !) Ans
Problem 14-53
The A-36 steel bolt is required to absorb the energy of a 2-kg mass that falls along the 4-mm-diameter
bolt shank that is 150 mm long. Determine the maximum height h of release so the stress in the bolt
does no exceed 150 MPa. 
Given: do 4mm:= L 150mm:=
mo 2 kg⋅:= E 200GPa:= σallow 150MPa:=
Solution:
Weight : W mo g⋅:=
Section Property : A
π do2⋅
4
:=
Static Displacement : ∆st
W L⋅
E A⋅:=
Allowable Length:
Impact factor: η 1 1 2 h∆st
⋅++=
σmax
W
A
⎛⎜⎝
⎞
⎠ 1 1 2
h
∆st
⋅++⎛⎜⎝
⎞
⎠
⋅=
h
∆st
2
σallow A⋅
W
1−⎛⎜⎝
⎞
⎠
2
1−
⎡⎢⎢⎣
⎤⎥⎥⎦⋅:=
h 5.293 mm= Ans
Problem 14-54
The collar has a mass of 5 kg and falls down the titanium Ti-6A1-4V bar. If the bar has a diameter of
20 mm, determine the maximum stress developed in the bar if the weight is (a) dropped from a height
of h = 1 m, (b) released from a height h 0, and (c) placed slowly on the flange at A.≈
Given: do 20mm:= L 1.5m:=
mo 5kg:= E 120GPa:= σY 924MPa:=
Solution:
Weight : W mo g⋅:=
Section Property : A
π do2⋅
4
:=
Static Displacement : ∆st
W L⋅
E A⋅:= ∆st 0.00195 mm=
a) Drop Height: h 1m:=
Impact factor: η 1 1 2 h∆st
⋅++:= η 1013.49=
Pmax W η⋅:=
σmax
Pmax
A
:=
σmax 158.2 MPa= Ans
b) Drop Height: h 0mm:=
Impact factor: η 1 1 2 h∆st
⋅++:= η 2.00=
Pmax W η⋅:=
σmax
Pmax
A
:=
σmax 0.312 MPa= Ans
c) No impact: η 1:=
Pmax W η⋅:=
σmax
Pmax
A
:=
σmax 0.156 MPa= Ans
Note: since all of the σmax < σY = 924 MPa, the above analysis is valid.
Problem 14-55
The collar has a mass of 5 kg and falls down the titanium Ti-6A1-4V bar. If the bar has a diameter of
20 mm, determine if the weight can be released from rest at any point along the bar and not
permanently damage the bar after striking the flange at A.
Given: do 20mm:= L 1.5m:=
mo 5kg:= E 120GPa:= σY 924MPa:=
Solution:
Weight : W mo g⋅:=
Section Property : A
π do2⋅
4
:=
Static Displacement : ∆st
W L⋅
E A⋅:= ∆st 0.00195 mm=
Drop Height: hmax 1.5m:=
Impact factor: ηmax 1 1 2
hmax
∆st
⋅++:= ηmax 1241.04=
Pmax W ηmax⋅:=
σmax
Pmax
A
:=
σmax 193.7 MPa=
Note: since σmax < σY = 924 MPa, the weight can be released from rest
at any point along the bar without causing permanently damage to the bar. Ans
Problem 14-56
A cylinder having the dimensions shown is made from magnesium Am 1004-T61. If it is struck by a
rigid block having a weight of 4 kN and traveling at 0.6 m/s, determine the maximum stress in the
cylinder. Neglect the mass of the cylinder.
Given: do 150mm:= E 44.7GPa:=
W 4kN:= σY 152MPa:=
L 0.5m:= v 0.6 m
s
:=
Solution:
Section Property : A
π do2⋅
4
:=
Eq. Spring Constant: k
E A⋅
L
:= k 1579828.41 kN
m
=
Conservation of Energy: Ui = Ur
1
2
W
g
⎛⎜⎝
⎞
⎠⋅ v
2⋅ 1
2
k⋅ ∆max2⋅=
∆max
W
g k⋅ v⋅:= ∆max 0.3049 mm=
Maximum Stress:
Pmax k ∆max⋅:=
σmax
Pmax
A
:=
σmax 27.3 MPa= Ans
< σY = 152 MPa (O.K. !)
Problem 14-57
The wide-flange beam has a length of 2L, a depth 2c, and a constant EI. Determine the maximum
height h at which a weight W can be dropped on its end without exceeding a maximum elastic stress
σmax in the beam.
Problem 14-58
The sack of cement has a weight of 450 N (~45 kg). If it is dropped from rest at a height of h = 1.2 m
onto the center of the W250 X 58 structural steel A-36 beam, determine the maximum bending stress
developed in the beam due to the impact. Also, what is the impact factor?
Given: L 7.2m:= h 1.2m:= W 450N:=
E 200GPa:= σY 250MPa:=
Solution:
Use W 250x58 : I 87.3 106( )⋅ mm4:=
d 252mm:=
Static Displacement : ∆st
W L3⋅
48E I⋅:= ∆st 0.20041 mm=(From Appendix C)
Maximum Stress: 
Impact factor: η 1 1 2 h∆st
⋅++:= η 110.44= Ans
Maximum moment occurs at mid-span, where Mmax
W L⋅
4
:= Mmax 0.810 kN m⋅=
c 0.5d:= σmax
Mmax c⋅
I
⎛⎜⎝
⎞
⎠ η⋅:=
σmax 129.1 MPa= Ans
Since σmax < σY = 250 MPa, the above analysis is valid.
Problem 14-59
The sack of cement has a weight of 450 N (~45 kg). Determine the maximum height h from which it
can be dropped from rest onto the center of the W250 X 58 structural steel A-36 beam so that the
maximum bending stress due to impact does not exceed 210 MPa.
Given: L 7.2m:= W 450N:=
E 200GPa:= σallow 210MPa:=
Solution:
Use W 250x58 : I 87.3 106( )⋅ mm4:=
d 252mm:=
Static Displacement : ∆st
W L3⋅
48E I⋅:= ∆st 0.20041 mm=(From Appendix C)
Maximum Stress: 
Impact factor: η 1 1 2 h∆st
⋅++=
Maximum moment occurs at mid-span, where Mmax
W L⋅
4
:=
c 0.5d:= σmax
Mmax c⋅
I
⎛⎜⎝
⎞
⎠ η⋅= σmax σallow=
σallow
W L⋅ d⋅
8I
⎛⎜⎝
⎞
⎠ 1 1 2
h
∆st
⋅++⎛⎜⎝
⎞
⎠
⋅=
h
∆st
2
8 I⋅ σallow
W
⎯
L d⋅
1−⎛⎜⎝
⎞
⎠
2
1−
⎡⎢⎢⎣
⎤⎥⎥⎦⋅:=
h 3.197 m= Ans
Problem 14-60
A 200-N (~20-kg) weight is dropped from a height of h = 0.6 m onto the center of the cantilevered
A-36 steel beam. If the beam is a W250 X 22, determine the maximum bending stress developed in the
beam.
Given: L 1.5m:= h 0.6m:= W 200N:=
E 200GPa:= σY 250MPa:=
Solution:
Use W 250x22 : I 28.8 106( )⋅ mm4:=
d 254mm:=
Static Displacement : ∆st
W L3⋅
3E I⋅:= ∆st 0.03906 mm=(From Appendix C)
Maximum Stress: 
Impact factor: η 1 1 2 h∆st
⋅++:= η 176.27=
Maximum moment occurs at support A, where Mmax W L⋅:= Mmax 0.300 kN m⋅=
c 0.5d:= σmax
Mmax c⋅
I
⎛⎜⎝
⎞
⎠ η⋅:=
σmax 233.2 MPa= Ans
Since σmax < σY = 250 MPa, the above analysis is valid.
Problem 14-61
If the maximum allowable bending stress for the W250 X 22 structural A-36 steel beam is σallow = 140
MPa, determine the maximum height h from which a 250-N (~25-kg) weight can be released from rest
and strike the center of the beam.
Given: L 1.5m:= W 250N:=
E 200GPa:= σallow 140MPa:=
Solution:
Use W 250x22 : I 28.8 106( )⋅ mm4:=
d 254mm:=
Static Displacement : ∆st
W L3⋅
3E I⋅:= ∆st 0.04883 mm=(From Appendix C)
Maximum Stress: 
Impact factor: η 1 1 2 h∆st
⋅++=
Maximum moment occurs at mid-span, where Mmax W L⋅:=
c 0.5d:= σmax
Mmax c⋅
I
⎛⎜⎝
⎞
⎠ η⋅= σmax σallow=
σallow
W L⋅ d⋅
2I
⎛⎜⎝
⎞
⎠ 1 1 2
h
∆st
⋅++⎛⎜⎝
⎞
⎠
⋅=
h
∆st
2
2 I⋅ σallow
W
⎯
L d⋅
1−⎛⎜⎝
⎞
⎠
2
1−
⎡⎢⎢⎣
⎤⎥⎥⎦⋅:=
h 0.171 m= Ans
Problem 14-62
A 200-N (~20-kg) weight is dropped from a height of h = 0.6 m onto the center of the cantilevered
A-36 steel beam. If the beam is a W250 X 22,
determine the vertical displacement of its end B due to
the impact.
Given: L 1.5m:= h 0.6m:= W 200N:=
E 200GPa:= σY 250MPa:=
Solution:
Use W 250x22 : I 28.8 106( )⋅ mm4:=
d 254mm:=
Static Displacement : 
(From Appendix C)
At mid-span,
∆st
W L3⋅
3E I⋅:= ∆st 0.03906 mm=
θst
W L2⋅
2E I⋅:= θst 0.00003906 rad=
At end B, ∆'st ∆st θst L⋅+:= ∆'st 0.09766 mm=
Maximum Displacement : 
Impact factor: η 1 1 2 h∆st
⋅++:= η 176.27=
∆'B.max ∆'st( ) η⋅:=
∆'B.max 17.21 mm= Ans
v 0.5
m
s
:=
Problem 14-63
The steel beam AB acts to stop the oncoming railroad car, which has a mass of 10 Mg and is coasting
towards it at v = 0.5 m/s. Determine the maximum stress developed in the beam if it is struck at its
center by the car. The beam is simply supported and only horizontal forces occur at A and B. Assume
that the railroad car and the supporting framework for the beam remains rigid. Also, compute the
maximum deflection of the beam. Est = 200 GPa, σY = 250 MPa.
Given: L 2m:= mo 10 103( ) kg:=
E 200GPa:= do 200mm:=
σY 250MPa:=
d 153mm:=Solution:
Weight : W mo g⋅:=
Section Properties : I
do
4
12
:=
Static Displacement : At mid-span,
(From Appendix C)
∆st
W L3⋅
48E I⋅:= ∆st 0.61292 mm=
Equivalent Spring Stiffness: k
W
∆st
:= k 160000 kN
m
=
Maximum Displacement : ∆max
∆st v2⋅
g
:= ∆max 3.953 mm=
Maximum Force : Pmax k ∆max⋅:= Pmax 632.46 kN=
Maximum Stress: c 0.5do:=
Maximum moment occurs at mid-span: Mmax 0.25Pmax L⋅:=
σmax
Mmax c⋅
I
:=
σmax 237.2 MPa= Ans
< σY = 250 MPa (O.K.!)
Problem 14-64
The tugboat has a weight of 600 kN (~60-tonne) and is traveling forward at 0.6 m/s when it strikes the
300-mm-diameter fender post AB used to protect a bridge pier. If the post is made from treated white
spruce and is assumed fixed at the river bed, determine the maximum horizontal distance the top of the
post will move due to the impact. Assume the tugboat is rigid and neglect the effect of the water.
Given: LBC 3.6m:= LCA 0.9m:=
∆A.max 410.8 mm=
E 9.65GPa:= v 0.6 m
s
:=
Ans
Solution:
Section Properties :
I
π do4⋅
64
:=
Maximum Displacement : 
(From Appendix C)
At point C, Pmax
3 E⋅ I⋅
LBC
3
∆C.max⋅= (1)
Conservation of Energy :
1
2
m⋅ v2⋅ 1
2
Pmax⋅ ∆C.max⋅=
1
2
W
g
⎛⎜⎝
⎞
⎠⋅ v
2⋅ 1
2
3 E⋅ I⋅
LBC
3
∆C.max⋅⎛⎜⎝
⎞
⎠
⋅ ∆C.max⋅=
∆C.max
W v2⋅ LBC3⋅
3g E I⋅:= ∆C.max 298.79 mm=
From Eq.(1): Pmax
3 E⋅ I⋅
LBC
3
∆C.max⋅:= Pmax 73.72 kN=
(From Appendix C)
At point C, θC.max
Pmax LBC
2⋅
2E I⋅:= θC.max 0.12450 rad=
At end A, ∆A.max ∆C.max θC.max LCA⋅+:=
do 300mm:=
W 600kN:=
Problem 14-65
The W250 X18 beam is made from A-36 steel and is cantilevered from the wall at B. The spring
mounted on the beam has a stiffness of k = 200 kN/m. If a weight of 40 N (~4 kg) is dropped onto the
spring from a height of 1 m, determine the maximum bending stress developed in the beam.
Given: L 2.5m:= h 1m:= W 40N:=
E 200GPa:=
ksp
200kN
m
:=σY 250MPa:=
Solution:
Use W 250x18 : I 22.5 106( )⋅ mm4:=
d 251mm:=
Eq. Spring Constants: kb
3E I⋅
L3
:=
(From Appendix C)
kb 864
kN
m
=
Equilibrium requires: Fsp Fb=
Hence, ksp ∆sp⋅ kb ∆b⋅= ∆sp
kb
ksp
∆b⋅= (1)
Conservation of Energy: Ui = Ur
W h ∆max+( )⋅ 12 ksp⋅ ∆sp2⋅ 12 kb⋅ ∆b2⋅+= (2)
However, ∆max ∆b ∆sp+= (3)
Substituting Eqs.(1) and (3) into (2):
Given
W h ∆b+
kb
ksp
∆b⋅+
⎛⎜⎝
⎞
⎠
⋅ 1
2
ksp⋅
kb
ksp
∆b⋅
⎛⎜⎝
⎞
⎠
2
⋅ 1
2
kb⋅ ∆b2⋅+= (4)
Solving Eq. (4): Guess ∆b 10mm:=
∆b Find ∆b( ):= ∆b 4.2184 mm=
Maximum Stress: c 0.5d:=
Pmax kb ∆b⋅:= Pmax 3.64 kN=
Mmax Pmax L⋅:= Mmax 9.11 kN m⋅=
σmax
Mmax c⋅
I
:=
σmax 50.82 MPa= Ans
< σY = 250 MPa (O.K. !)
Problem 14-66
The 375-N (~37.5-kg) block has a downward velocity of 0.6 m/s when it is 0.9 m from the top of the
wooden beam. Determine the maximum bending stress in the beam due to the impact, and compute the
maximum deflection of its end D. Ew = 14 GPa. Assume the material will not yield.
Given: d 300mm:= L 3.0m:= h 0.9m:= LCD 1.5m:=
E 14GPa:= W 375N:= v 0.6 m
s
:=
Solution:
Section Property : I
d4
12
:=
Eq. Spring Constants:
kb
48E I⋅
L3
:=(From Appendix C)
kb 16800
kN
m
=
Conservation of Energy: Ui = Ur
1
2
W
g
⎛⎜⎝
⎞
⎠⋅ v
2⋅ W h ∆max+( )⋅+ 12 kb⋅ ∆b2⋅= (1)
However, ∆b ∆max= (2)
Substituting Eq.(2) into (1):
Given
1
2
W
g
⎛⎜⎝
⎞
⎠⋅ v
2⋅ W h ∆max+( )⋅+ 12 kb⋅ ∆max2⋅= (3)
Solving Eq. (3): Guess ∆max 10mm:=
∆max Find ∆max( ):= ∆max 6.425 mm=
Maximum Stress: c 0.5d:= ∆b ∆max:=
Pmax kb ∆b⋅:= Pmax 107.95 kN=
Maximum moment occurs at mid-span:
Mmax Pmax
L
4
⎛⎜⎝
⎞
⎠⋅:= Mmax 80.96 kN m⋅=
σmax
Mmax c⋅
I
:= σmax 17.99 MPa= Ans
Maximum Displacement : 
(From Appendix C)
At point C, θC.max
Pmax L
2⋅
16E I⋅:= θC.max 0.006425 rad=
At end D, ∆D.max θC.max LCD⋅:= ∆D.max 9.638 mm= Ans
Problem 14-67
The 375-N (~37.5-kg) block has a downward velocity of 0.6 m/s when it is 0.9 m from the top of the
wood beam. Determine the maximum bending stress in the beam due to the impact, and compute the
maximum deflection of point B. Ew = 14 GPa.
Given: d 300mm:= L 3.0m:= h 0.9m:= LCD 1.5m:=
E 14GPa:= W 375N:= v 0.6 m
s
:=
Solution:
Section Property : I
d4
12
:=
Eq. Spring Constants:
kb
48E I⋅
L3
:=(From Appendix C)
kb 16800
kN
m
=
Conservation of Energy: Ui = Ur
1
2
W
g
⎛⎜⎝
⎞
⎠⋅ v
2⋅ W h ∆max+( )⋅+ 12 kb⋅ ∆b2⋅= (1)
However, ∆b ∆max= (2)
Substituting Eq.(2) into (1):
Given
1
2
W
g
⎛⎜⎝
⎞
⎠⋅ v
2⋅ W h ∆max+( )⋅+ 12 kb⋅ ∆max2⋅= (3)
Solving Eq. (3): Guess ∆max 10mm:=
∆max Find ∆max( ):= ∆max 6.425 mm=
Maximum Stress: c 0.5d:= ∆b ∆max:=
Pmax kb ∆b⋅:= Pmax 107.95 kN=
Maximum moment occurs at mid-span:
Mmax Pmax
L
4
⎛⎜⎝
⎞
⎠⋅:= Mmax 80.96 kN m⋅=
σmax
Mmax c⋅
I
:= σmax 17.99 MPa= Ans
Problem 14-68
Determine the maximum height h from which an 400 N (~40 kg) weight can be dropped onto the end
of the A-36 steel W150 X 18 beam without exceeding the maximum elastic stress.
Given: L 3m:= W 0.4kN:=
E 200GPa:= σY 250MPa:=
Solution:
Use W 150x18: I 9.19 106( )⋅ mm4:= d 153mm:=
Static Displacement at B : 
+
Support Reactions :
 
ΣFy=0; A C+ W− 0= (1)
ΣΜC=0
;
A L⋅ W L⋅+ 0= (2)
Solving Eqs. (1) and (2): A W−:= A 0.4− kN=
C W A−:= C 0.8 kN=
Internal Moment Function: 
M1 A x1⋅= M2 M1=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Ui
1
2E I⋅ 2( )
0
L
x1M1
2⌠⎮⌡ d= Ui
2
2E I⋅
0
L
x1A x1⋅( )2⌠⎮⌡ d:= Ui 0.783 J=
External Work: The external work done by the weight W is Ue
1
2
W⋅ ∆B.st( )⋅=
Conservation of Energy: Ui = Ue
Ui
1
2
W⋅ ∆B.st( )⋅= ∆B.st 2 Ui⋅W:= ∆B.st 3.917 mm=
Eq. Spring Constant: kB
W
∆B.st
:= kB 102.11
kN
m
=
Maximum Force : c 0.5d:=
Maximum moment occurs at mid-support: Mmax Pmax L⋅=
σmax
Mmax c⋅
I
= σY
Pmax L⋅ d⋅
2I
= Pmax
2IσY
L d⋅:= Pmax 10.01 kN=
Maximum Displacement at B : 
∆max
Pmax
kB
:= ∆max 98.04 mm=
Conservation of Energy: Ui = Ur
W h ∆max+( )⋅ 12 kB⋅ ∆max2⋅= h 12W kB⋅ ∆max2⋅ ∆max−:= h 1.129 m= Ans
Problem 14-69
The 400 N (~40 kg) weight is dropped from rest at a height of h = 1.125 m onto the end of the A-36
steel W150 X 18 beam. Determine the maximum bending stress developed in the beam.
Given: L 3m:= W 0.4kN:= h 1.125m:=
E 200GPa:= σY 250MPa:=
Solution:
Use W 150x18: I 9.19 106( )⋅ mm4:= d 153mm:=
Static Displacement at B : 
+
Support Reactions :
 
ΣFy=0; A C+ W− 0= (1)
ΣΜC=0
;
A L⋅ W L⋅+ 0= (2)
Solving Eqs. (1) and (2): A W−:= A 0.4− kN=
C W A−:= C 0.8 kN=
Internal Moment Function: 
M1 A x1⋅= M2 M1=
∫= L0 2i dx/(2EI)MUBending Strain Energy: 
Ui
1
2E I⋅ 2( )
0
L
x1M1
2⌠⎮⌡ d= Ui
2
2E I⋅
0
L
x1A
x1⋅( )2⌠⎮⌡ d:= Ui 0.783 J=
External Work: The external work done by the weight W is Ue
1
2
W⋅ ∆B.st( )⋅=
Conservation of Energy: Ui = Ue
Ui
1
2
W⋅ ∆B.st( )⋅= ∆B.st 2 Ui⋅W:= ∆B.st 3.917 mm=
Eq. Spring Constant: kB
W
∆B.st
:= kB 102.11
kN
m
=
Maximum Displacement at B : ∆B ∆max=
Conservation of Energy: Ui = Ur
Given W h ∆max+( )⋅ 12 kB⋅ ∆max2⋅= (1)
Solving Eq. (1): Guess ∆max 10mm:=
∆max Find ∆max( ):= ∆max 97.88 mm=
Maximum Stress: c 0.5d:= ∆B ∆max:=
Pmax kB ∆B⋅:= Pmax 9.99 kN=
Maximum moment occurs at mid-support:
Mmax Pmax L⋅:= σmax
Mmax c⋅
I
:= σmax 249.60 MPa= Ans
< σY = 250 MPa (O.K. !)
1
2
mo⋅ v2⋅
1
2
kb⋅ ∆beam2⋅
1
4
kb
2
ksp
⋅ ∆beam2⋅+=
Given: L 1.8m:= c 75mm:=
mo 1.8 10
3( ) kg:= I 300 106( ) mm4:=
E 2GPa:= σY 30MPa:=
ksp 1500
kN
m
:= v 0.75 m
s
:=
Solution:
Equilibrium : This requires, Fsp
1
2
Pbeam⋅=
Then, ksp ∆sp⋅
1
2
kb⋅ ∆beam⋅=
∆sp
1
2
kb
ksp
⋅ ∆beam⋅= (1)
Weight : W mo g⋅:=
Static Displacement : At mid-span,
(From Appendix C)
∆st
W L3⋅
48E I⋅:= ∆st 3.575 mm=
Equivalent Spring Stiffness: kb
W
∆st
:= kb 4938.27
kN
m
=
Conservation of Energy: Ui = Ur
1
2
mo⋅ v2⋅
1
2
kb⋅ ∆beam2⋅ 2
1
2
ksp⋅ ∆sp2⋅⎛⎜⎝
⎞
⎠+= (2)
Substituting Eq.(1) into (2): Given
Problem 14-70
The car bumper is made of polycarbonate-polybutylene terephthalate. If E = 2.0 GPa, determine the
maximum deflection and maximum stress in the bumper if it strikes the rigid post when the car is
coasting at v = 0.75 m/s. The car has a mass of 1.80 Mg, and the bumper can be considered simply
supported on two spring supports connected to the rigid frame of the car. For the bumper take I =
300(106) mm4, c = 75 mm, σY = 30 MPa, and k = 1.5 MN/m.
(3)
Solving Eq. (3): Guess ∆beam 10mm:=
∆beam Find ∆beam( ):= ∆beam 8.803 mm=
Maximum Displacement : 
(From Eq.(1) ∆sp
1
2
kb
ksp
⋅ ∆beam⋅:= ∆sp 14.490 mm=
Hence, ∆max ∆sp ∆beam+:= ∆max 23.29 mm= Ans
Maximum Stress:
Pmax kb ∆beam⋅:= Pmax 43.47 kN=
Maximum moment occurs at mid-span:
Mmax Pmax
L
4
⎛⎜⎝
⎞
⎠⋅:= Mmax 19.56 kN m⋅=
σmax
Mmax c⋅
I
:= σmax 4.89 MPa= Ans
< σY = 30 MPa (O.K. !)
θ 30deg:=Given: LBC 1.5m:= P 4kN:=
LAB
LBC
sin θ( ):=
E 200GPa:= A 1250mm2:=
Solution:
Member Lengths:
Ans
LAB 3.000 m=
Member Real Forces: 
NAB P:=
NBC 0:=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nAB
F'
cos θ( ):= nAB 1.1547 kN=
nBC nAB− sin θ( )⋅:= nBC 0.5774− kN=
Member Virtual
n
Real
N
Length
L
Product
p
AB n1 nAB:= N1 NAB:= L1 LAB:= p1 n1 N1⋅ L1⋅:=
BC n2 nBC:= N2 NBC:= L2 LBC:= p2 n2 N2⋅ L2⋅:=
Sum p1 p2+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆B.h⋅
Sum
E A⋅=
∆B.h
Sum
E A⋅
1
F'
:=
∆B.h 0.0554 mm=
Problem 14-71
Determine the horizontal displacement of joint B on the two-member frame. Each A-36 steel member
has a cross-sectional area of 1250 mm2.
Problem 14-72
Determine the horizontal displacement of joint B. Each A-36 steel member has a cross-sectional area of
1250 mm2.
Given: LBC 1m:= LAC 1.5m:=
P 3kN:= E 200GPa:= A 1250mm2:=
Solution: θ atan
LAC
LBC
⎛⎜⎝
⎞
⎠
:= θ 56.310 deg=
Member Lengths:
LAB LBC
2 LAC
2+:= LAB 1.803 m=
Member Real Forces: 
NAB
P
sin θ( ):= NAB 3.6056 kN=
NBC NAB− cos θ( )⋅:= NBC 2.0000− kN=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nAB 0:= nAB 0.0000 kN=
nBC F':= nBC 1.0000 kN=
Member Virtual
n
Real
N
Length
L
Product
p
AB n1 nAB:= N1 NAB:= L1 LAB:= p1 n1 N1⋅ L1⋅:=
BC n2 nBC:= N2 NBC:= L2 LBC:= p2 n2 N2⋅ L2⋅:=
Sum p1 p2+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆B.h⋅
Sum
E A⋅=
∆B.h
Sum
E A⋅
1
F'
:=
∆B.h 8.00− 10 3−× mm= Ans
Problem 14-73
Determine the vertical displacement of joint B. Each A-36 steel member has a cross-sectional area of
1250 mm2.
Given: LBC 1m:= LAC 1.5m:=
P 3kN:= E 200GPa:= A 1250mm2:=
Solution: θ atan
LAC
LBC
⎛⎜⎝
⎞
⎠
:= θ 56.310 deg=
Member Lengths:
LAB LBC
2 LAC
2+:= LAB 1.803 m=
Member Real Forces: 
NAB
P
sin θ( ):= NAB 3.6056 kN=
NBC NAB− cos θ( )⋅:= NBC 2.0000− kN=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nAB
F'
sin θ( ):= nAB 1.2019 kN=
nBC nAB− cos θ( )⋅:= nBC 0.6667− kN=
Member Virtual
n
Real
N
Length
L
Product
p
AB n1 nAB:= N1 NAB:= L1 LAB:= p1 n1 N1⋅ L1⋅:=
BC n2 nBC:= N2 NBC:= L2 LBC:= p2 n2 N2⋅ L2⋅:=
Sum p1 p2+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆B.v⋅
Sum
E A⋅=
∆B.v
Sum
E A⋅
1
F'
:=
∆B.v 36.58 10 3−× mm= Ans
Problem 14-74
Determine the horizontal displacement of point B. Each A-36 steel member has a cross-sectional area
of 1250 mm2.
Given: b 1.8m:= h 2.4m:=
P 1kN:= E 200GPa:= A 1250mm2:=
Solution: θ atan h
b
⎛⎜⎝
⎞
⎠:= θ 53.130 deg=
Member Lengths:
LAB b
2 h2+:= LAB 3.000 m=
LBC b
2 h2+:= LBC 3.000 m=
LAC 2 b⋅:= LAC 3.600 m=
Member Real Forces: 
NAB
P−
2 cos θ( ):= NAB 0.8333− kN=
NBC NAB−:= NBC 0.8333 kN=
NAC NAB− cos θ( )⋅:= NAC 0.5000 kN=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nAB
F'−
2 cos θ( ):= nAB 0.8333− kN=
nBC nAB−:= nBC 0.8333 kN=
nAC nAB− cos θ( )⋅:= nAC 0.5000 kN=
Member Virtual
n
Real
N
Length
L
Product
p
AB n1 nAB:= N1 NAB:= L1 LAB:= p1 n1 N1⋅ L1⋅:=
BC n2 nBC:= N2 NBC:= L2 LBC:= p2 n2 N2⋅ L2⋅:=
AC n3 nAC:= N3 NAC:= L3 LAC:= p3 n3 N3⋅ L3⋅:=
Sum p1 p2+ p3+:=∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆B.h⋅
Sum
E A⋅=
∆B.h
Sum
E A⋅
1
F'
:=
∆B.h 20.27 10 3−× mm= Ans
Problem 14-75
Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of
1250 mm2.
Given: b 1.8m:= h 2.4m:=
P 1kN:= E 200GPa:= A 1250mm2:=
Solution: θ atan h
b
⎛⎜⎝
⎞
⎠:= θ 53.130 deg=
Member Lengths:
LAB b
2 h2+:= LAB 3.000 m=
LBC b
2 h2+:= LBC 3.000 m=
LAC 2 b⋅:= LAC 3.600 m=
Member Real Forces: 
NAB
P−
2 cos θ( ):= NAB 0.8333− kN=
NBC NAB−:= NBC 0.8333 kN=
NAC NAB− cos θ( )⋅:= NAC 0.5000 kN=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nAB
F'−
2 sin θ( ):= nAB 0.6250− kN=
nBC nAB:= nBC 0.6250− kN=
nAC nAB− cos θ( )⋅:= nAC 0.3750 kN=
Member Virtual
n
Real
N
Length
L
Product
p
AB n1 nAB:= N1 NAB:= L1 LAB:= p1 n1 N1⋅ L1⋅:=
BC n2 nBC:= N2 NBC:= L2 LBC:= p2 n2 N2⋅ L2⋅:=
AC n3 nAC:= N3 NAC:= L3 LAC:= p3 n3 N3⋅ L3⋅:=
Sum p1 p2+ p3+:=∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆B.v⋅
Sum
E A⋅=
∆B.v
Sum
E A⋅
1
F'
:=
∆B.v 2.70 10 3−× mm= Ans
Problem 14-76
Determine the horizontal displacement of point C. Each A-36 steel member has a cross-sectional area
of 400 mm2.
Given: b 1.5m:= h 2m:= P1 5kN:=
E 200GPa:= A 400mm2:= P2 10kN:=
Solution: θ atan h
b
⎛⎜⎝
⎞
⎠:= θ 53.130 deg=
Member Lengths:
LCD h:= LAB h:= LAC b2 h2+:= LAC 2.5 m=
LDA b:= LBC b:=
Member Real Forces: NDA 0:=
NCD P2:= NCD 10 kN=
NAC
P2−
sin θ( ):= NAC 12.5− kN=
NBC P1 NAC cos θ( )⋅−:= NBC 12.5 kN=
NAB NAC− sin θ( )⋅:= NAB 10 kN=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nDA 0:=
nCD 0:=
nAC 0:=
nBC F':=
nAB 0:=
Member Virtual
n
Real
N
Length
L
Product
p
DA n1 nDA:= N1 NDA:= L1 LDA:= p1 n1 N1⋅ L1⋅:=
CD n2 nCD:= N2 NCD:= L2 LCD:= p2 n2 N2⋅ L2⋅:=
AC n3 nAC:= N3 NAC:= L3 LAC:= p3 n3 N3⋅ L3⋅:=
BC n4 nBC:= N4 NBC:= L4 LBC:= p4 n4 N4⋅ L4⋅:=
AB n5 nAB:= N5 NAB:= L5 LAB:= p5 n5 N5⋅ L5⋅:=
Sum p1 p2+ p3+ p4+ p5+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆C.h⋅
Sum
E A⋅=
∆C.h
Sum
E A⋅
1
F'
:=
∆C.h 0.234 mm= Ans
Problem 14-77
Determine the vertical displacement of point D. Each A-36 steel member has a cross-sectional area of
400 mm2.
Given: b 1.5m:= h 2m:= P1 5kN:=
E 200GPa:= A 400mm2:= P2 10kN:=
Solution: θ atan h
b
⎛⎜⎝
⎞
⎠:= θ 53.130 deg=
Member Lengths:
LCD h:=
LAB h:= LAC b2 h2+:= LAC 2.5 m=
LDA b:= LBC b:=
Member Real Forces: NDA 0:=
NCD P2:= NCD 10 kN=
NAC
P2−
sin θ( ):= NAC 12.5− kN=
NBC P1 NAC cos θ( )⋅−:= NBC 12.5 kN=
NAB NAC− sin θ( )⋅:= NAB 10 kN=
Member Virtual Forces: 
Apply Virual Force F' 1kN:=
nDA 0:= nDA 0:=
nCD F':= nCD 1 kN=
nAC
F'−
sin θ( ):= nAC 1.25− kN=
nBC nAC− cos θ( )⋅:= nBC 0.75 kN=
nAB nAC− sin θ( )⋅:= nAB 1 kN=
Member Virtual
n
Real
N
Length
L
Product
p
DA n1 nDA:= N1 NDA:= L1 LDA:= p1 n1 N1⋅ L1⋅:=
CD n2 nCD:= N2 NCD:= L2 LCD:= p2 n2 N2⋅ L2⋅:=
AC n3 nAC:= N3 NAC:= L3 LAC:= p3 n3 N3⋅ L3⋅:=
BC n4 nBC:= N4 NBC:= L4 LBC:= p4 n4 N4⋅ L4⋅:=
AB n5 nAB:= N5 NAB:= L5 LAB:= p5 n5 N5⋅ L5⋅:=
Sum p1 p2+ p3+ p4+ p5+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆D.v⋅
Sum
E A⋅=
∆D.v
Sum
E A⋅
1
F'
:=
∆D.v 1.164 mm= Ans
A 2800mm2:=
Given: b 2.4m:= h 1.8m:= P 25kN:=
E 200GPa:=
LAE 3.000 m=
Solution: θ atan h
b
⎛⎜⎝
⎞
⎠:= θ 36.870 deg=
Member Lengths:
LBE h:=
LAE b
2 h2+:= LCE b2 h2+:= LCE 3.000 m=
LAB b:= LEF b:= LAF h:= LBC b:= LDE b:= LCD h:=
Member Real Forces: 
NBE P:=
NAE
P−
2 sin θ( ):= NAE 20.8333− kN= NCE
P−
2 sin θ( ):= NCE 20.8333− kN=
NAB NAE− cos θ( )⋅:= NAB 16.6667 kN= NBC NCE− cos θ( )⋅:= NBC 16.6667 kN=
NEF 0:= NDE 0:=
NAF 0:= NCD 0:=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nBE F':=
nAE
F'−
2 sin θ( ):= nAE 0.8333− kN= nCE
F'−
2 sin θ( ):= nCE 0.8333− kN=
nAB nAE− cos θ( )⋅:= nAB 0.6667 kN= nBC nCE− cos θ( )⋅:= nBC 0.6667 kN=
nEF 0:= nDE 0:=
nAF 0:= nCD 0:=
Problem 14-78
Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of
2800 mm2. Est = 200 GPa.
Member Virtual
n
Real
N
Length
L
Product
p
AB n1 nAB:= N1 NAB:= L1 LAB:= p1 n1 N1⋅ L1⋅:=
BC n2 nBC:= N2 NBC:= L2 LBC:= p2 n2 N2⋅ L2⋅:=
EF n3 nEF:= N3 NEF:= L3 LEF:= p3 n3 N3⋅ L3⋅:=
DE n4 nDE:= N4 NDE:= L4 LDE:= p4 n4 N4⋅ L4⋅:=
AF n5 nAF:= N5 NAF:= L5 LAF:= p5 n5 N5⋅ L5⋅:=
CD n6 nCD:= N6 NCD:= L6 LCD:= p6 n6 N6⋅ L6⋅:=
AE n7 nAE:= N7 NAE:= L7 LAE:= p7 n7 N7⋅ L7⋅:=
CE n8 nCE:= N8 NCE:= L8 LCE:= p8 n8 N8⋅ L8⋅:=
BE n9 nBE:= N9 NBE:= L9 LBE:= p9 n9 N9⋅ L9⋅:=
Sum p1 p2+ p3+ p4+ p5+ p6+ p7+ p8+ p9+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆B.v⋅
Sum
E A⋅=
∆B.v
Sum
E A⋅
1
F'
:=
∆B.v 0.3616 mm= Ans
A 2800mm2:=
Given: b 2.4m:= h 1.8m:= P 25kN:=
E 200GPa:=
LAE 3.000 m=
Solution: θ atan h
b
⎛⎜⎝
⎞
⎠:= θ 36.870 deg=
Member Lengths:
LBE h:=
LAE b
2 h2+:=
nCD 0:=
LCE b
2 h2+:= LCE 3.000 m=
LAB b:= LEF b:= LAF h:= LBC b:= LDE b:= LCD h:=
Member Real Forces: 
NBE P:=
NAE
P−
2 sin θ( ):= NAE 20.8333− kN= NCE
P−
2 sin θ( ):= NCE 20.8333− kN=
NAB NAE− cos θ( )⋅:= NAB 16.6667 kN= NBC NCE− cos θ( )⋅:= NBC 16.6667 kN=
NEF 0:= NDE 0:=
NAF 0:= NCD 0:=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nBE 0:=
nAE
F'−
2 sin θ( ):= nAE 0.8333− kN= nCE
F'−
2 sin θ( ):= nCE 0.8333− kN=
nAB nAE− cos θ( )⋅:= nAB 0.6667 kN= nBC nCE− cos θ( )⋅:= nBC 0.6667 kN=
nEF 0:= nDE 0:=
nAF 0:=
Problem 14-79
Determine the vertical displacement of point E. Each A-36 steel member has a cross-sectional area of
2800 mm2.
Member Virtual
n
Real
N
Length
L
Product
p
AB n1 nAB:= N1 NAB:= L1 LAB:= p1 n1 N1⋅ L1⋅:=
BC n2 nBC:= N2 NBC:= L2 LBC:= p2 n2 N2⋅ L2⋅:=
EF n3 nEF:= N3 NEF:= L3 LEF:= p3 n3 N3⋅ L3⋅:=
DE n4 nDE:= N4 NDE:= L4 LDE:= p4 n4 N4⋅ L4⋅:=
AF n5 nAF:= N5 NAF:= L5 LAF:= p5 n5 N5⋅ L5⋅:=
CD n6 nCD:= N6 NCD:= L6 LCD:= p6 n6 N6⋅ L6⋅:=
AE n7 nAE:= N7 NAE:= L7 LAE:= p7 n7 N7⋅ L7⋅:=
CE n8 nCE:= N8 NCE:= L8 LCE:= p8 n8 N8⋅ L8⋅:=
BE n9 nBE:= N9 NBE:= L9 LBE:= p9 n9 N9⋅ L9⋅:=
Sum p1 p2+ p3+ p4+ p5+ p6+ p7+ p8+ p9+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆B.v⋅
Sum
E A⋅=
∆B.v
Sum
E A⋅
1
F'
:=
∆B.v 0.2813 mm= Ans
Problem 14-80
Determine the horizontal displacement of point D. Each A-36 steel member has a cross-sectional area
of 300 mm2.
Given: h 3m:= b 3m:= P1 4kN:=
E 200GPa:= A 300mm2:= P2 2kN:=
Solution: θ atan 2 h⋅
b
⎛⎜⎝
⎞
⎠:= θ 63.435 deg=
Member Lengths: c 0.5 b⋅( )2 h2+:=
LAE h:= LED h:= LCD c:=
LEC 0.5 b⋅:= LBC c:= LAC c:=
Member Real Forces: 
NCD
P1−
cos θ( ):= NCD 8.9443− kN=
NED NCD− sin θ( )⋅:= NED 8 kN=
NAE NED:= NAE 8 kN=
NEC P2−:= NEC 2− kN=
NAC
0.5− NEC
cos θ( ):= NAC 2.2361 kN=
NBC NCD
0.5− NEC
cos θ( )−:= NBC 11.1803− kN=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nCD
F'−
cos θ( ):= nCD 2.2361− kN=
nED nCD− sin θ( )⋅:= nED 2 kN=
nAE nED:= nAE 2 kN=
nEC 0:= nEC 0 kN=
nAC 0:= nAC 0 kN=
nBC nCD:= nBC 2.2361− kN=
Member Virtual
n
Real
N
Length
L
Product
p
CD n1 nCD:= N1 NCD:= L1 LCD:= p1 n1 N1⋅ L1⋅:=
ED n2 nED:= N2 NED:= L2 LED:= p2 n2 N2⋅ L2⋅:=
AE n3 nAE:= N3 NAE:= L3 LAE:= p3 n3 N3⋅ L3⋅:=
EC n4 nEC:= N4 NEC:= L4 LEC:= p4 n4 N4⋅ L4⋅:=
AC n5 nAC:= N5 NAC:= L5 LAC:= p5 n5 N5⋅ L5⋅:=
BC n6 nBC:= N6 NBC:= L6 LBC:= p6 n6 N6⋅ L6⋅:=
Sum p1 p2+ p3+ p4+ p5+ p6+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆D.h⋅
Sum
E A⋅=
∆D.h
Sum
E A⋅
1
F'
:=
∆D.h 4.116 mm= Ans
Problem 14-81
Determine the horizontal displacement of point E. Each A-36 steel member has a cross-sectional area
of 300 mm2.
Given: h 3m:= b 3m:= P1 4kN:=
E 200GPa:= A 300mm2:= P2 2kN:=
Solution: θ atan 2 h⋅
b
⎛⎜⎝
⎞
⎠:= θ 63.435 deg=
Member Lengths: c 0.5 b⋅( )2 h2+:=
LAE h:= LED h:= LCD c:=
LEC 0.5 b⋅:= LBC c:= LAC c:=
Member Real Forces: 
NCD
P1−
cos θ( ):= NCD 8.9443− kN=
NED NCD− sin θ( )⋅:= NED 8 kN=
NAE NED:= NAE 8 kN=
NEC P2−:= NEC 2− kN=
NAC
0.5− NEC
cos θ( ):= NAC 2.2361 kN=
NBC NCD
0.5− NEC
cos θ( )−:= NBC 11.1803− kN=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nCD 0:= nCD 0 kN=
nED 0:= nED 0 kN=
nAE 0:= nAE 0 kN=
nEC F'−:= nEC 1− kN=
nAC
0.5− nEC
cos θ( ):= nAC 1.118 kN=
nBC
0.5− nEC
cos θ( )−:= nBC 1.118− kN=
Member Virtual
n
Real
N
Length
L
Product
p
CD n1 nCD:= N1 NCD:= L1 LCD:= p1 n1 N1⋅ L1⋅:=
ED n2 nED:= N2 NED:= L2 LED:= p2 n2 N2⋅ L2⋅:=
AE n3 nAE:= N3 NAE:= L3 LAE:= p3 n3 N3⋅ L3⋅:=
EC n4 nEC:= N4 NEC:= L4 LEC:= p4 n4 N4⋅ L4⋅:=
AC n5 nAC:= N5 NAC:= L5 LAC:= p5 n5 N5⋅ L5⋅:=
BC n6 nBC:= N6 NBC:= L6 LBC:= p6 n6 N6⋅ L6⋅:=
Sum p1 p2+ p3+ p4+ p5+ p6+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆D.h⋅
Sum
E A⋅=
∆D.h
Sum
E A⋅
1
F'
:=
∆D.h 0.8885 mm= Ans
Problem 14-82
Determine the horizontal displacement of joint B of the truss. Each A-36 steel member has a
cross-sectional area of 400 mm2.
Given: h 1.5m:= b 2m:= P2 5kN:=
E 200GPa:= A 400mm2:= P1 4kN:=
Solution: θ atan h
b
⎛⎜⎝
⎞
⎠:= θ 36.870 deg=
Member Lengths:
LCD h:= LAB h:= LAC b2 h2+:= LAC 2.5 m=
LDA b:= LBC b:=
Member Real Forces: NAB 0:=
NBC P1:= NBC 4 kN=
NAC
P1−
cos θ( ):= NAC 5− kN=
NCD P2− NAC sin θ( )⋅−:= NCD 2− kN=
NDA NAC− cos θ( )⋅:= NDA 4 kN=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nAB 0:=
nBC F':= nBC 1 kN=
nAC
F'−
cos θ( ):= nAC 1.25− kN=
nCD nAC− sin θ( )⋅:= nCD 0.75 kN=
nDA nAC− cos θ( )⋅:= nDA 1 kN=
Member Virtual
n
Real
N
Length
L
Product
p
DA n1 nDA:= N1 NDA:= L1 LDA:= p1 n1 N1⋅ L1⋅:=
CD n2 nCD:= N2 NCD:= L2 LCD:= p2 n2 N2⋅ L2⋅:=
AC n3 nAC:= N3 NAC:= L3 LAC:= p3 n3 N3⋅ L3⋅:=
BC n4 nBC:= N4 NBC:= L4 LBC:= p4 n4 N4⋅ L4⋅:=
AB n5 nAB:= N5 NAB:= L5 LAB:= p5 n5 N5⋅ L5⋅:=
Sum p1 p2+ p3+ p4+ p5+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆B.h⋅
Sum
E A⋅=
∆B.h
Sum
E A⋅
1
F'
:=
∆B.h 0.367 mm= Ans
Problem 14-83
Determine the vertical displacement of joint C of the truss. Each A-36 steel member has a
cross-sectional area of 400 mm2.
Given: h 1.5m:= b 2m:= P2 5kN:=
E 200GPa:= A 400mm2:= P1 4kN:=
Solution: θ atan h
b
⎛⎜⎝
⎞
⎠:= θ 36.870 deg=
Member Lengths:
LCD h:= LAB h:= LAC b2 h2+:= LAC
2.5 m=
LDA b:= LBC b:=
Member Real Forces: NAB 0:=
NBC P1:= NBC 4 kN=
NAC
P1−
cos θ( ):= NAC 5− kN=
NCD P2− NAC sin θ( )⋅−:= NCD 2− kN=
NDA NAC− cos θ( )⋅:= NDA 4 kN=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
nAB 0:=
nBC 0:=
nAC 0:=
nCD F'−:=
nDA 0:=
Member Virtual
n
Real
N
Length
L
Product
p
DA n1 nDA:= N1 NDA:= L1 LDA:= p1 n1 N1⋅ L1⋅:=
CD n2 nCD:= N2 NCD:= L2 LCD:= p2 n2 N2⋅ L2⋅:=
AC n3 nAC:= N3 NAC:= L3 LAC:= p3 n3 N3⋅ L3⋅:=
BC n4 nBC:= N4 NBC:= L4 LBC:= p4 n4 N4⋅ L4⋅:=
AB n5 nAB:= N5 NAB:= L5 LAB:= p5 n5 N5⋅ L5⋅:=
Sum p1 p2+ p3+ p4+ p5+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆C.v⋅
Sum
E A⋅=
∆C.v
Sum
E A⋅
1
F'
:=
∆C.v 0.0375 mm= Ans
Problem 14-84
Determine the vertical displacement of joint A. Each A-36 steel member has a cross-sectional area of
400 mm2.
Given: h 2m:= b1 1.5m:= b2 3m:=
E 200GPa:= A 400mm2:=
P1 40kN:= P2 60kN:=
Solution: θ1 atan
h
b1
⎛⎜⎝
⎞
⎠
:= θ1 53.130 deg=
θ2 atan
h
b2
⎛⎜⎝
⎞
⎠
:= θ2 33.690 deg=
Support Reactions :
+
Cv 0:=
+
ΣFy=0; Dv P1− P2− 0= (1)
 
ΣFx=0; Ch Dh+ 0= (2)
ΣΜD=0
;
P1− b1 b2+( )⋅ P2 b2⋅− Ch h⋅+ 0= (3)
Solving Eqs. (2) and (3): Ch
P1 b1 b2+( )⋅ P2 b2⋅+
h
:= Ch 180 kN=
Dh Ch−:= Dh 180− kN=
Solving Eq. (1): Dv P1 P2+:= Dv 100 kN=
Member Lengths: c1 b1
2 h2+:= c2 b22 h2+:=
LAE b1:= LED b2:= LBE h:=
LAB c1:= LBC b2:= LBD c2:=
Member Virtual Forces: 
Member Real Forces: Apply Virtual Force F' 1kN:=
NAB
P1
sin θ1( ):= NAB 50 kN= nAB
F'
sin θ1( ):= nAB 1.25 kN=
NAE NAB− cos θ1( )⋅:= NAE 30− kN= nAE nAB− cos θ1( )⋅:= nAE 0.75− kN=
NBE P2:= NBE 60 kN= nBE 0:= nBE 0 kN=
NED NAE:= NED 30− kN= nED nAE:= nED 0.75− kN=
NBC Ch:= NBC 180 kN= nBC
F' b1 b2+( )⋅
h
:= nBC 2.25 kN=
NBD
Dv
sin θ2( )−:= NBD 180.3− kN= nBD
F'
sin θ2( )−:= nBD 1.80− kN=
Member Virtual
n
Real
N
Length
L
Product
p
AB n1 nAB:= N1 NAB:= L1 LAB:= p1 n1 N1⋅ L1⋅:=
AE n2 nAE:= N2 NAE:= L2 LAE:= p2 n2 N2⋅ L2⋅:=
BE n3 nBE:= N3 NBE:= L3 LBE:= p3 n3 N3⋅ L3⋅:=
ED n4 nED:= N4 NED:= L4 LED:= p4 n4 N4⋅ L4⋅:=
BC n5 nBC:= N5 NBC:= L5 LBC:= p5 n5 N5⋅ L5⋅:=
BD n6 nBD:= N6 NBD:= L6 LBD:= p6 n6 N6⋅ L6⋅:=
Sum p1 p2+ p3+ p4+ p5+ p6+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆A.v⋅
Sum
E A⋅=
∆A.v
Sum
E A⋅
1
F'
:=
∆A.v 33.05 mm= Ans
Problem 14-85
DetermiNe the vertical displacemeNt of joiNt C. Each A-36 steel member has a cross-sectioNal area of
2800 mm2.
Given: b 4m:= h 3m:=
E 200GPa:= A 2800mm2:=
P1 30kN:= P2 40kN:=
Solution: θ atan h
b
⎛⎜⎝
⎞
⎠:= θ 36.870 deg=
Support Reactions :
+
By symmetry, A=E=R
ΣFy=0; 2R 2P1 P2+( )− 0= R 0.5 2P1 P2+( )⋅:= R 50 kN=
Member Lengths: c b2 h2+:= c 5 m=
LCH h:=
LAI c:= LBH c:= LDH LBH:= LEG LAI:=
LAB b:= LBC b:= LCD LBC:= LDE LAB:=
LAJ h:= LBI h:= LDG LBI:= LEF LAJ:=
LJI b:= LIH b:= LHG LIH:= LGF LJI:=
Member Real Forces: 
NCH P2:=
NAI
R−
sin θ( ):= NBH
P2−
2 sin θ( ):= NDH NBH:= NEG NAI:=
NAB NAI cos θ( )⋅:= NBC NAB NBH cos θ( )⋅+:= NCD NBC:= NDE NAB:=
NAJ 0:= NBI NAI sin θ( )⋅:= NDG NBI:= NEF NAJ:=
NJI 0:= NIH NAI cos θ( )⋅:= NHG NIH:= NGF NJI:=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
Support Reactions :
+
By symmetry, A'=E'=R'
ΣFy=0; 2R' F'− 0= R' 0.5 F'⋅:= R' 0.5 kN=
nCH F':=
nAI
R'−
sin θ( ):= nBH
F'−
2 sin θ( ):= nDH nBH:= nEG nAI:=
nAB nAI cos θ( )⋅:= nBC nAB nBH cos θ( )⋅+:= nCD nBC:= nDE nAB:=
nAJ 0:= nBI nAI sin θ( )⋅:= nDG nBI:= nEF nAJ:=
nJI 0:= nIH nAI cos θ( )⋅:= nHG nIH:= nGF nJI:=
Member Virtual
N
Real
N
Length
L
Product
p
CH n1 nCH:= N1 NCH:= L1 LCH:= p1 n1 N1⋅ L1⋅:=
AI n2 nAI:= N2 NAI:= L2 LAI:= p2 n2 N2⋅ L2⋅:=
AB n3 nAB:= N3 NAB:= L3 LAB:= p3 n3 N3⋅ L3⋅:=
AJ n4 nAJ:= N4 NAJ:= L4 LAJ:= p4 n4 N4⋅ L4⋅:=
JI n5 nJI:= N5 NJI:= L5 LJI:= p5 n5 N5⋅ L5⋅:=
BH n6 nBH:= N6 NBH:= L6 LBH:= p6 n6 N6⋅ L6⋅:=
BC n7 nBC:= N7 NBC:= L7 LBC:= p7 n7 N7⋅ L7⋅:=
BI n8 nBI:= N8 NBI:= L8 LBI:= p8 n8 N8⋅ L8⋅:=
IH n9 nIH:= N9 NIH:= L9 LIH:= p9 n9 N9⋅ L9⋅:=
DH n10 nDH:= N10 NDH:= L10 LDH:= p10 n10 N10⋅ L10⋅:=
CD n11 nCD:= N11 NCD:= L11 LCD:= p11 n11 N11⋅ L11⋅:=
DG n12 nDG:= N12 NDG:= L12 LDG:= p12 n12 N12⋅ L12⋅:=
HG n13 nHG:= N13 NHG:= L13 LHG:= p13 n13 N13⋅ L13⋅:=
EG n14 nEG:= N14 NEG:= L14 LEG:= p14 n14 N14⋅ L14⋅:=
DE n15 nDE:= N15 NDE:= L15 LDE:= p15 n15 N15⋅ L15⋅:=
EF n16 nEF:= N16 NEF:= L16 LEF:= p16 n16 N16⋅ L16⋅:=
GF n17 nGF:= N17 NGF:= L17 LGF:= p17 n17 N17⋅ L17⋅:=
Sum1 p1 p2+ p3+ p4+ p5+ p6+ p7+ p8+ p9+:=
Sum2 p10 p11+ p12+ p13+ p14+ p15+ p16+ p17+:=
Sum Sum1 Sum2+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆C.v⋅
Sum
E A⋅=
∆C.v
Sum
E A⋅
1
F'
:=
∆C.v 5.266 mm= Ans
Problem 14-86
Determine the vertical displacement of joint H. Each A-36 steel member has a cross-sectional area of
2800 mm2.
Given: b 4m:= h 3m:=
E 200GPa:= A 2800mm2:=
P1 30kN:= P2 40kN:=
Solution: θ atan h
b
⎛⎜⎝
⎞
⎠:= θ 36.870 deg=
Support Reactions :
+
By symmetry, A=E=R
ΣFy=0; 2R 2P1 P2+( )− 0= R 0.5 2P1 P2+( )⋅:= R 50 kN=
Member Lengths: c b2 h2+:= c 5 m=
LCH h:=
LAI c:= LBH c:= LDH LBH:= LEG LAI:=
LAB b:= LBC b:= LCD LBC:= LDE LAB:=
LAJ h:= LBI h:= LDG LBI:= LEF LAJ:=
LJI b:= LIH b:= LHG LIH:= LGF LJI:=
Member Real Forces: 
NCH P2:=
NAI
R−
sin θ( ):= NBH
P2−
2 sin θ( ):= NDH NBH:= NEG NAI:=
NAB NAI cos θ( )⋅:= NBC NAB NBH cos θ( )⋅+:= NCD NBC:= NDE NAB:=
NAJ 0:= NBI NAI sin θ( )⋅:= NDG NBI:= NEF NAJ:=
NJI 0:= NIH NAI cos θ( )⋅:= NHG NIH:= NGF NJI:=
Member Virtual Forces: 
Apply Virtual Force F' 1kN:=
Support Reactions :
+
By symmetry, A'=E'=R'
ΣFy=0; 2R' F'− 0= R' 0.5 F'⋅:= R' 0.5 kN=
nCH 0:=
nAI
R'−
sin θ( ):= nBH
F'−
2 sin θ( ):= nDH nBH:= nEG nAI:=
nAB nAI cos θ( )⋅:= nBC nAB nBH cos θ( )⋅+:= nCD nBC:= nDE nAB:=
nAJ 0:= nBI nAI sin θ( )⋅:= nDG nBI:= nEF nAJ:=
nJI 0:= nIH nAI cos θ( )⋅:= nHG nIH:= nGF nJI:=
Member Virtual
N
Real
N
Length
L
Product
p
CH n1 nCH:= N1 NCH:= L1 LCH:= p1 n1 N1⋅ L1⋅:=
AI n2 nAI:= N2 NAI:= L2 LAI:= p2 n2 N2⋅ L2⋅:=
AB n3 nAB:= N3 NAB:= L3 LAB:= p3 n3 N3⋅ L3⋅:=
AJ n4 nAJ:= N4 NAJ:= L4 LAJ:= p4 n4 N4⋅ L4⋅:=
JI n5 nJI:= N5 NJI:= L5 LJI:= p5 n5 N5⋅ L5⋅:=
BH n6 nBH:= N6 NBH:= L6 LBH:= p6 n6 N6⋅ L6⋅:=
BC n7 nBC:= N7 NBC:= L7 LBC:= p7 n7 N7⋅ L7⋅:=
BI n8 nBI:= N8 NBI:= L8 LBI:= p8 n8 N8⋅ L8⋅:=
IH n9 nIH:= N9 NIH:= L9 LIH:= p9 n9 N9⋅ L9⋅:=
DH n10 nDH:= N10 NDH:= L10 LDH:= p10 n10 N10⋅ L10⋅:=
CD n11 nCD:= N11 NCD:= L11 LCD:= p11 n11 N11⋅ L11⋅:=
DG n12 nDG:= N12 NDG:= L12 LDG:= p12 n12 N12⋅ L12⋅:=
HG n13 nHG:= N13 NHG:= L13 LHG:= p13 n13 N13⋅ L13⋅:=
EG n14 nEG:= N14 NEG:= L14 LEG:= p14 n14 N14⋅ L14⋅:=
DE n15 nDE:= N15 NDE:= L15 LDE:= p15 n15 N15⋅ L15⋅:=
EF n16 nEF:= N16 NEF:= L16 LEF:= p16 n16 N16⋅ L16⋅:=
GF n17 nGF:= N17 NGF:= L17 LGF:= p17 n17 N17⋅ L17⋅:=
Sum1 p1 p2+ p3+ p4+ p5+ p6+ p7+ p8+ p9+:=
Sum2 p10 p11+ p12+ p13+ p14+ p15+ p16+ p17+:=
Sum Sum1 Sum2+:=
∑=⋅ nNL/(EA)∆1Virtual Work Equation: 
F' ∆C.v⋅
Sum
E A⋅=
∆C.v
Sum
E A⋅
1
F'
:=
∆C.v 5.052 mm= Ans
Problem 14-87
Determine the displacement of point C and the slope at point B. EI is constant.
Problem 14-88
Determine the displacement at point C. EI is constant.
Problem 14-89
Determine the slope at point C. EI is constant.
Problem 14-90
Determine the slope at point A. EI is constant.
Problem 14-91
Determine the displacement of point C of the beam made from A-36 steel and having a moment of
inertia of I = 21(106) mm4.
Given: a 1.5m:= b 3m:= P 40kN:=
E 200GPa:= I 21 106( ) mm4:=
Solution: Virtual Force F' 1kN:=
Real Support Reactions : 
+
Support Reactions due to Virtual Force at C:
+
 
ΣFy=0; A B+ P− 0= (1) ΣFy=0; A' B'+ F'− 0= (1')
ΣΜB=0; A b⋅ P a b+( )⋅− 0= (2) ΣΜB=0; A' b⋅ F' a⋅+ 0= (2')
Solving Eqs. (1) and (2): Solving Eqs. (1') and (2'):
A
P a b+( )⋅
b
:= B P A−:= A' F'− a⋅
b
:= B' F' A'−:=
A 60 kN= B 20− kN= A' 0.5− kN= B' 1.5 kN=
Internal Moment Functions: 
O->A B->A C->B
Real M1 P x1⋅= M2 B x2⋅= M3 0=
Virtual m1 0= m2 A' b x2−( )⋅= m3 F'− x3⋅=
∫=⋅ L0 dxmM/(EI)∆1Virtual Work Equation: 
F' ∆C⋅
1
E I⋅
0
a
x1m1 M1⋅
⌠⎮⌡ d 0
b
x2m2 M2⋅
⌠⎮⌡ d+ 0
a
x3m3 M3⋅
⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
∆C
1
E I⋅
1
F'
0
0
b
x2A' b x2−( )⋅ B x2⋅( )⋅⌠⎮⌡ d+ 0+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
∆C 10.71 mm= Ans
I 21 106( ) mm4:=Given: a 1.5m:= b 3m:= P 40kN:=E 200GPa:=
Support Reactions due to Virtual Moment at B:
 Solution: Virtual Moment m' 1kN m⋅:=
Real Support Reactions : 
+
Ans
+
 
ΣFy=0; A B+ P− 0= (1) ΣFy=0; A' B'+ 0= (1')
ΣΜB=0; A b⋅ P a b+( )⋅− 0= (2) ΣΜB=0; A' b⋅ m'+ 0= (2')
Solving Eqs. (1) and (2): Solving Eqs. (1') and (2'):
A
P a b+( )⋅
b
:= B P A−:= A' m'−
b
:= B' A'−:=
A 60 kN= B 20− kN= A' 0.33− kN= B' 0.33 kN=
Internal Moment Functions: 
O->A B->A C->B
Real M1 P− x1⋅= M2 B x2⋅= M3 0=
Virtual m1 0= m2 A' b x2−( )⋅= m3 0=
∫=⋅ L0 dxmM/(EI)θ1Virtual Work Equation: 
m' θB⋅
1
E I⋅
0
a
x1m1 M1⋅
⌠⎮⌡ d 0
b
x2m2 M2⋅
⌠⎮⌡ d+ 0
a
x3m3 M3⋅
⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
θB
1
E I⋅
1
m'
0
0
b
x2A' b x2−( )⋅ B x2⋅( )⋅⌠⎮⌡ d+ 0+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
θB 7.143 10 3−× rad=
θB 0.409 deg= (clockwise)
Problem 14-92
Determine the slope at B of the beam made from A-36 steel and having a moment of inertia of I =
21(106) mm4.
Problem 14-93
Determine the displacement of point C of the W360 X 39 beam made from A-36 steel.
Given: a 1.5m:= b 3m:= P 40kN:=
E 200GPa:=
Solution: Virtual Force F' 1kN:=
Use W 360x39: I 102 106( )⋅ mm4:=
Real Support Reactions : 
+
Support Reactions due to Virtual Force at C:
+ΣFy=0; A B+ 2P− 0= (1) ΣFy=0; A' B'+ F'− 0= (1')
Due to symmetry: A B= (2) Due to symmetry: A' B'= (2')
Solving Eqs. (1) and (2): Solving Eqs. (1') and (2'):
A P:= B P:= A' 0.5F':= B' 0.5F':=
A 40 kN= B 40 kN= A' 0.5 kN= B' 0.5 kN=
Internal Moment Functions: 
O->A B->A D->B
Real M1 P− x1⋅= M2 P− a⋅= M3 P− x3⋅=
Virtual m1 0= m2 B' x2⋅= m3 0=
(for x2 <0.5b)
∫=⋅ L0 dxmM/(EI)∆1Virtual Work Equation: 
F' ∆C⋅
1
E I⋅
0
a
x1m1 M1⋅
⌠⎮⌡ d 2 0
0.5 b⋅
x2m2 M2⋅
⌠⎮⌡ d+ 0
a
x3m3 M3⋅
⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
∆C
1
E I⋅
1
F'
0 2
0
0.5 b⋅
x2B' x2⋅ P− a⋅( )⋅
⌠⎮⌡ d+ 0+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
∆C 3.31− mm= Ans
Problem 14-94
Determine the slope at A of the W360 X 39 beam made from A36 steel.
Given: a 1.5m:= b 3m:= P 40kN:=
E 200GPa:=
 Solution: Virtual Moment m' 1kN m⋅:=
Use W 360x39: I 102 106( )⋅ mm4:=
Real Support Reactions : 
+
Support Reactions due to Virtual Moment at A: 
+ΣFy=0; A B+ 2P− 0= (1) ΣFy=0; A' B'+ 0= (1')
Due to symmetry: A B= (2) ΣΜB=0; A' b⋅ m'+ 0= (2')
Solving Eqs. (1) and (2): Solving Eqs. (1') and (2'):
A P:= B P:= A' m'−
b
:= B' A'−:=
A 40 kN= B 40 kN=
A' 0.33− kN= B' 0.33 kN=
Internal Moment Functions: 
O->A B->A D->B
Real M1 P− x1⋅= M2 P− a⋅= M3 P− x3⋅=
Virtual m1 0= m2 B' x2⋅= m3 0=
∫=⋅ L0 dxmM/(EI)θ1Virtual Work Equation: 
m' θA⋅
1
E I⋅
0
a
x1m1 M1⋅
⌠⎮⌡ d 0
b
x2m2 M2⋅
⌠⎮⌡ d+ 0
a
x3m3 M3⋅
⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
θA
1
E I⋅
1
m'
0
0
b
x2P− a⋅( ) B' x2⋅( )⋅⌠⎮⌡ d+ 0+
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
θA 4.412− 10 3−× rad=
θA 0.253− deg= (anticlockwise) Ans
do 38mm:=
Given: a 0.6m:= b 1.6m:= c 0.45m:=
PD 1.4kN:=
F' 1kN:=
PE 3.2kN:= E 200GPa:=
L a b+ c+:=
Solution: Virtual Force
m3 C' x3⋅=
Section Property : 
I
π do4⋅
64
:= I 102.35 103× mm4=
Real Support Reactions : 
+
 
ΣFy=0; A C+ PD PE+( )− 0= (1)
C PD PE+( ) A−:=
ΣΜC=0
;
A L⋅ PD b c+( )⋅− PE c⋅− 0= (2)
Solving Eqs. (1) and (2):
A
1
L
PD b c+( )⋅ PE c⋅+⎡⎣ ⎤⎦:= A 1.626 kN=
A' C'+ F'− 0=
C 2.974 kN=
Support Reactions due to Virtual Force at B:
+ ΣFy=0; (1')
ΣΜC=0
;
A' L⋅ F' b a− c+( )⋅− 0= (2')
Solving Eqs. (1') and (2'):
A'
1
L
F' b a− c+( )⋅[ ]:= C' F' A'−:=
A' 0.547 kN= C' 0.453 kN=
Internal Moment Functions: 
A->D D->B E->B C->E
Real M1 A x1⋅= M2 A a x2+( )⋅ PD x2⋅−= M4 C c x4+( )⋅ PE x4⋅−= M3 C x3⋅=
Virtual m1 A' x1⋅= m2 A' a x2+( )⋅= m4 C' c x4+( )⋅=
Problem 14-95
Determine the displacement at B of the 38-mm-diameter A-36 steel shaft.
∫=⋅ L0 dxmM/(EI)∆1Virtual Work Equation: 
F' ∆B⋅
1
E I⋅
0
a
x1m1 M1⋅
⌠⎮⌡ d 0
a
x2m2 M2⋅
⌠⎮⌡ d+ 0
b a−
x4m4 M4⋅
⌠⎮⌡ d+ 0
c
x3m3 M3⋅
⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
I1
0
a
x1m1 M1⋅
⌠⎮⌡ d= I1 0
a
x1A' x1⋅ A x1⋅( )⋅⌠⎮⌡ d
⎡⎢⎢⎣
⎤⎥⎥⎦
:= I1 0.06407 kN2 m3⋅=
I2
0
a
x2m2 M2⋅
⌠⎮⌡ d= I2 0
a
x2A' a x2+( )⋅ A a x2+( )⋅ PD x2⋅−⎡⎣ ⎤⎦⋅⌠⎮⌡ d
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
I2 0.31064 kN
2 m3⋅=
I4
0
b a−
x4m4 M4⋅
⌠⎮⌡ d= I4 0
b a−
x4C' c x4+( )⋅ C c x4+( )⋅ PE x4⋅−⎡⎣ ⎤⎦⋅⌠⎮⌡ d
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
I4 0.51840 kN
2 m3⋅=
I3
0
c
x3m3 M3⋅
⌠⎮⌡ d= I3 0
c
x3C' x3⋅ C x3⋅( )⋅⌠⎮⌡ d
⎡⎢⎢⎣
⎤⎥⎥⎦
:= I3 0.04090 kN2 m3⋅=
∆B
1
E I⋅
1
F'
I1 I2+ I4+ I3+( ):=
∆B 45.63 mm= Ans
Problem 14-96
Determine the slope of the 30-mm-diameter A-36 steel shaft at the bearing support A.
Given: a 0.6m:= b 1.6m:= c 0.45m:=
PD 1.4kN:= do 38mm:=
PE 3.2kN:= E 200GPa:=
L a b+ c+:=
 Solution: Virtual Moment m' 1kN m⋅:=
Section Property : 
I
π do4⋅
64
:= I 102.35 103× mm4=
Real Support Reactions : 
+
 
ΣFy=0; A C+ PD PE+( )− 0= (1)
ΣΜC=0
;
A L⋅ PD b c+( )⋅− PE c⋅− 0= (2)
Solving Eqs. (1) and (2):
A
1
L
PD b c+( )⋅ PE c⋅+⎡⎣ ⎤⎦:= A 1.626 kN=
C PD PE+( ) A−:= C 2.974 kN=
+
Support Reactions due to Virtual Moment at A: 
ΣFy=0; A' C'+ 0= (1')
ΣΜC=0
;
A' L⋅ m'+ 0= (2')
Solving Eqs. (1') and (2'):
A'
m'−
L
:= C' A'−:=
A' 0.377− kN= C' 0.377 kN=
Internal Moment Functions: 
A->D E->D C->E
Real M1 A x1⋅= M4 C c x4+( )⋅ PE x4⋅−= M3 C x3⋅=
Virtual m1 m' A' x1⋅+= m4 C' c x4+( )⋅= m3 C' x3⋅=
∫=⋅ L0 dxmM/(EI)θ1Virtual Work Equation: 
m' θA⋅
1
E I⋅
0
a
x1m1 M1⋅
⌠⎮⌡ d 0
b
x4m4 M4⋅
⌠⎮⌡ d+ 0
c
x3m3 M3⋅
⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
I1
0
a
x1m1 M1⋅
⌠⎮⌡ d= I1 0
a
x1m' A' x1⋅+( ) A x1⋅( )⋅⌠⎮⌡ d
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
I1 0.24857 kN
2 m3⋅=
I4
0
b
x4m4 M4⋅
⌠⎮⌡ d= I4 0
b
x4C' c x4+( )⋅ C c x4+( )⋅ PE x4⋅−⎡⎣ ⎤⎦⋅⌠⎮⌡ d
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
I4 0.84403 kN
2 m3⋅=
I3
0
c
x3m3 M3⋅
⌠⎮⌡ d= I3 0
c
x3C' x3⋅ C x3⋅( )⋅⌠⎮⌡ d
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
I3 0.03408 kN
2 m3⋅=
θA
1
E I⋅
1
m'
I1 I4+ I3+( ):=
θA 55.038 10 3−× rad=
θA 3.153 deg= (clockwise) Ans
Problem 14-97
Determine the displacement at pulley B. The A-36 steel shaft has a diameter of 30 mm.
Given: a 0.4m:= b 0.3m:=
PD 4kN:= do 30mm:=
PE 3kN:= E 200GPa:=
PC 2kN:= L 2a 2 b⋅+:=
Solution: Virtual Force F' 1kN:=
(1')Section Property : 
(2')
I
π do4⋅
64
:= I 39.76 103× mm4=
Real Support Reactions : 
+
 
ΣFy=0; A C+ PD PE+ PC+( )− 0= (1)
ΣΜC=0
;
A L⋅ PD L a−( )⋅− PE b⋅− PC 2 b⋅( )⋅− 0= (2)
Solving Eqs. (1) and (2):
A
1
L
PD L a−( )⋅ PE b⋅+ PC 2 b⋅( )⋅+⎡⎣ ⎤⎦:= A 4.357 kN=
C PD PE+ PC+( ) A−:= C 4.643 kN=
+
Support Reactions due to Virtual Force at B: 
ΣFy=0; A' C'+ F'= (1')
ΣΜC=0
;
A' L⋅ F' 2 b⋅( )− 0= (2')
Solving Eqs. (1') and (2'):
A'
2 b⋅ F'⋅
L
:= C' F' A'−:=
A' 0.4286 kN= C' 0.5714 kN=
Internal Moment Functions: 
A->D D->B E->D C->E
Real M1 A x1⋅= M2 A a x2+( )⋅ PD x2⋅−= M4 C b x4+( )⋅ PE x4⋅−= M3 C x3⋅=
Virtual m1 A' x1⋅= m2 A' a x2+( )⋅= m4 C' b x4+( )⋅= m3 C' x3⋅=
∫=⋅ L0 dxmM/(EI)∆1Virtual Work Equation: 
F' ∆B⋅
1
E I⋅
0
a
x1m1 M1⋅
⌠⎮⌡ d 0
a
x1m2 M2⋅
⌠⎮⌡ d+ 0
b
x4m4 M4⋅
⌠⎮⌡ d+ 0
b
x3m3 M3⋅
⌠⎮⌡ d+
⎛⎜⎜⎝
⎞
⎠
=
I1
0
a
x1m1 M1⋅
⌠⎮⌡ d= I1 0
a
x1A' x1⋅ A x1⋅( )⋅⌠⎮⌡ d
⎡⎢⎢⎣
⎤⎥⎥⎦
:=
I1 0.03984 kN
2 m3⋅=
I2
0
a
x2m2 M2⋅
⌠⎮⌡ d= I2 0
a
x2A' a x2+( )⋅ A a x2+( )⋅ PD x2⋅−⎡⎣ ⎤⎦⋅⌠⎮⌡