SOLUTION CAP 14

Prévia do material em texto

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 and the stress components x , y , and xy .
Problem 14-02
The strain-energy density must be the same whether the state of stress is represented by x , y , and
xy , or by the principal stresses 1 and 2 . This being the case, equate the strain-energy expressions
for each of these two cases and show that G = E/[2(1 + )].
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 10kN
Solution:
Section Properties :
Ast
dst
2
4
Ast 176.71 mm
2
Abr
dbr
2
4
Abr 314.16 mm
2
Aal
dal
2
4
Aal 490.87 mm
2
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 6kN m G 75GPa
LCD 0.5m TC 12kN m
Solution:
Section Properties :
J
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 4kN m
Solution:
Section Properties :
J
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
Fy=0; 2R P 0=
R 0.5P=
Internal Moment Function:
M R x=
L
0
2
i 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 30kNGiven: a 4m b 2m
I 22.5 106 mm4
E 200GPa
Solution: L a b
Use W 250x18 :
Ans
Support Reactions :
+
 
Fy=0; A B P 0= (1)
B=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=
L
0
2
i 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 :
+
 
Fy=0; Av Bv w L 0= (1)
B=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
+ Fx=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 x
2=
N' Ah=
L
0
2
i 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
L
0
2
i 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 x
2 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
30deg P2 15kN
E 200GPa
Solution: L 2a
Use W 200x86 : I 94.7 106 mm4
A 11000mm2
Support Reactions :
+
 
Fy=0; Av Bv P1 P2 sin 0= (1)
B=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
+ Fx=0; Ah P2 cos 0=
Ah P2 cos Ah 12.99 kN
Internal Moment and Force Function:
M1 Av x1= M2 Bv P2 sin x2= M2 Av x2=
N1 Ah= N2 Ah=
L
0
2
i 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
L
0
2
i 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
MB=0; R L Mo Mo 0=
R 0=
Internal Moment Function:
M Mo=
L
0
2
i 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 :
+
 
Fy=0; A B w b 0= (1)
B=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 0.5w x4
2= M3 0.5w b x3
2=
L
0
2
i 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
x40.5w x4
2 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
x30.5w 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
dst
2
4
Ast 2945.24 mm
2
Aconc rc
2 Ast Aconc 279798.10 mm
2
Given
Equilibrium : 
+ Fy=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
Pconc
Pst
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
dst
2
4
Ial
dal
4
12
Support Reactions :
+
By symmetry, A = B = Fst
Fy=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
L
0
2
i 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
L
0
2
i 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
=
L
0
2
i 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
do
2
4
I
d t3
12
Support Reactions :
Ans
Fx=0; P A B 0= (1)
B=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
L
0
2
i 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
+
L
0
2
i 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
mb 2m
Solution:
Support Reactions :
+
By symmetry, A = B = R
Fy=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 kSpring Strain Energy:
The spring deforms
R
k
0.500 mm
Usp.i
1
2
k 2
Usp.i 1.000 J Ans
L
0
2
i 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
do
4
64
Internal Moment Function:
M1 P x1= M2 P a=
L
0
2
i 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, avg = 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
+ Fy=0; FAB FAD
a
c
0= (1)
+ Fx=0; FAD
b
c
P 0= (2)
Solving Eqs. (1) and (2): Guess FAB 1kN FAD 1kN
(C)FAB
FAD
Find FAB FAD
FAB
FAD
7.50
12.50
kN
(T)
Applying the method of joints to Joint D.
Given
+ Fy=0; FCD FBD
a
c
FAD
a
c
0= (3)
+ Fx=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
FBD
FCD
12.50
15.00
kN
(T)
L/(2EA)NU 2iAxial Strain Energy:
Ui
1
2 E A
FAB
2 2a( ) FAD
2 c( ) FBD
2 c( ) FCD
2 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
+ Fy=0; FCA
b
c
P 0= (1)
+ Fx=0; FCB FCA
a
c
0= (2)
Solving Eqs. (1) and (2): Guess FCA 1kN FCB 1kN
(C)FCA
FCB
Find FCA FCB
FCA
FCB
1.25
0.75
P (T)
Applying the method of joints to Joint A.
+ Fy=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( ) FCB
2 a( ) FAB
2 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 :
+
 
Fy=0; A B 0= (1)
B=0; A LMo 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=
L
0
2
i 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 B=
Conservation of Energy: Ui = Ue
Ui
1
2
Mo B=
B
2 Ui
Mo
B 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=
L
0
2
i 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
yi Ai
Ai
=
yc
bf df 0.5df tw dw 0.5dw df
bf df tw dw yc 56.25 mm
I
1
12
bf df
3 bf df 0.5df yc
2 1
12
tw dw
3 tw dw 0.5dw df yc
2
I 21582031.25 mm4Support Reactions :
+
 
Fy=0; A C P 0= (1)
C=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=
L
0
2
i 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
do
4
64
Support Reactions :
+
By symmetry A=B=R, and D=0
Fy=0; 2R P 0= R 0.5P R 5 kN
Internal Moment Function:
M1 R x1= M2 R x2=
L
0
2
i 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
do
4
64
Equations of Equilibrium : For member BC,
+
 
Fy=0; B C 0= (1)
B=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=
L
0
2
i 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 E=
Conservation of Energy: Ui = Ue
Ubi
1
2
Mo E=
E
2 Ubi
Mo E 0.05502 rad
E 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
 
Fy=0; C D P 0= (1)
C=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=
L
0
2
i 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 with the horizontal when the spring
is stretched. Show that for this position this causes a torque T = PR cos and a bending moment M =
PR sin 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 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, Y = 800 MPa, and (b) it is made from an aluminum alloy for which Eal = 70
GPa, Y = 405 MPa.
Unit used: kJ 1000J
Given: L 4m E1 200GPa E2 70GPa
do 30mm Y1 800MPa Y2 405MPa
Solution:
Geometric Property : V
do
2
4
L V 2827433.39 mm3
a) Strain Energy: Y1
Y1
E1
ur1
1
2 Y1 Y1
ur1
1
2E1
Y1
2 ur1 1.600 10
6 J
m3
Ui1 ur1 V
Ui1 4.524 kJ Ans
b) Strain Energy: Y2
Y2
E2
ur2
1
2 Y2 Y2
ur1
1
2E2
Y2
2 ur2 1.172 10
6 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 Y = 70 MPa, E = 100.GPa.
Given: L 2.4m Ui 1200J
E 100GPa Y 70MPa
Solution:
Geometric Property : V
do
2
4
L=
Strain Energy: Y
Y
E
ur
1
2 Y Y
ur 24500.00
J
m3
Conservation of Energy: Ui = Ur
Ui ur V=
Ui ur
do
2
4
L=
do
4 Ui
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, Y = 420 MPa.
Given: do 12mm L 1.2m
W 250N E 120GPa Y 420MPa
Solution:
Section Property : A
do
2
4
Static Displacement : st
W L
E A st
0.02210 mm
a) Drop Height: h 300mm
Impact factor: 1 1 2
h
st
165.76
Pmax W
max
Pmax
A
max 366.4 MPa Ans
b) Drop Height: h 0mm
Impact factor: 1 1 2
h
st
2.00
Pmax W
max
Pmax
A
max 4.421 MPa Ans
c) No impact: 1
Pmax W
max
Pmax
A
max 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 canbe released and not permanently damage
the bar after striking the flange at A. Eti = 120 GPa, Y = 420 MPa.
Given: do 12mm L 1.2m
W 250N E 120GPa Y 420MPa
Solution:
Section Property : A
do
2
4
Static Displacement : st
W L
E A st
0.02210 mm
Maximum Drop Height:
Impact factor: 1 1 2
h
st
=
max
W
A
1 1 2
h
st
= max Y=
h
st
2
Y 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, Y = 600 MPa.
Given: L 2m A 0.01m2 mo 50 10
3 kg
E 200GPa Y 600MPa
Solution:
Weight : W mo g
Static Displacement : st
W L
E A st
0.4903 mm
Drop Height: h 0
Impact factor: 1 1 2
h
st
2.00
Pmax W
max
Pmax
A max
98.1 MPa Ans
< Y = 600 MPa (O.K.!)
max st 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, Y = 600 MPa.
Given: L 1m A 0.01m2 mo 50 10
3 kg
E 200GPa Y 600MPa max 550MPa
Solution:
Weight : W mo g
Equivalent Spring Stiffness:
 
k
E A
L
k 2.00 106
kN
m
Maximum Displacement :
Pmax A max max
Pmax
k
max 2.75 mm
Conservation of Energy: Ui = Ur
Given 1
2
mo v
2 W max
1
2
k max
2= (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, Y = 410 MPa.
Given: d1 5mm d2 10mm E 70GPa
L1 0.2m L2 0.3m Y 410MPa
mo 5kg h 0.1m
Solution:
Weight : W mo g
Section Property : A1
d1
2
4
A2
d2
2
4
Static Displacement : st
mo g L1
E A1
mo g L2
E A2
st 0.00981 mm
Impact factor: 1 1 2
h
st
143.78
Pmax W
max
Pmax
A1
max 359.1 MPa Ans
< Y = 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 max = 300 MPa. Eal = 70 GPa, Y = 410 MPa.
Given: d1 5mm d2 10mm E 70GPa
L1 0.2m L2 0.3m Y 410MPa
mo 5kg max 300MPa
Solution:
Weight : W mo g
Section Property : A1
d1
2
4
A2
d2
2
4
Static Displacement : st
W L1
E A1
W L2
E A2
st 0.00981 mm
Maximum Drop Height: 
Impact factor: 1 1 2
h
st
=
max
W
A1
1 1 2
h
st
=
h
st
2
max 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, Y = 350 MPa.
Given: do 16mm E 200GPa L 45m
W 4kN Y 350MPa v 0.6
m
s
Solution:
Section Property : A
do
2
4
Eq. Spring Constant: k
E A
L
k 893.61
kN
m
Conservation of Energy: Ui = Ur
Given 1
2
W
g
v2 W max
1
2
k max
2= (1)
Solving Eq. (1): Guess max 10mm
max Find max max 18.05 mm
Maximum Stress:
Pmax k max
max
Pmax
A
max 80.2 MPa Ans
< Y = 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 Y 350MPa v 0.9
m
s
Solution:
Section Property : A
do
2
4
Eq. Spring Constant: k
E A
L
k 893.61
kN
m
Conservation of Energy: Ui = Ur
Given 1
2
W
g
v2 W max
1
2
k max
2= (1)
Solving Eq. (1): Guess max 10mm
max Find max max 24.22 mm
Maximum Stress:
Pmax k max
max
Pmax
A
max 107.6 MPa Ans
< Y = 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
do
2
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
v2 W h max
1
2
ksp sp
2 1
2
kp p
2= (2)
However, max p sp= (3)
Substituting Eqs.(1) and (3) into (2):
Given
1
2
W
g
v2 W h p
kp
ksp
p
1
2
ksp
kp
ksp
p
2
1
2
kp p
2= (4)
Solving Eq. (4): Guess p 10mm
p Find p p 0.3044 mm
Maximum Stress:
Pmax kp p
max
Pmax
A
max 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 allow 150MPa
Solution:
Weight : W mo g
Section Property : A
do
2
4
Static Displacement : st
W L
E A
Allowable Length:
Impact factor: 1 1 2
h
st
=
allow
W
A
=
Given
allow
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 allow 175MPa
Solution: W mo g
Section Property : A
do
2
4
Static Displacement : st
W L
E A st
0.00156 mm
Maximum Stress:
 
Impact factor: 1 1 2
h
st
197.07
max
W
A
max 307.6 MPa > allow = 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 allow 150MPa
Solution:
Weight : W mo g
Section Property : A
do
2
4
Static Displacement : st
W L
E A
Allowable Length:
Impact factor: 1 1 2
h
st
=
max
W
A
1 1 2
h
st
=
h
st
2
allow 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 Y 924MPa
Solution:
Weight : W mo g
Section Property : A
do
2
4
Static Displacement : st
W L
E A st
0.00195 mm
a) Drop Height: h 1m
Impact factor: 1 1 2
h
st
1013.49
Pmax W
max
Pmax
A
max 158.2 MPa Ans
b) Drop Height: h 0mm
Impact factor: 1 1 2
h
st
2.00
Pmax W
max
Pmax
A
max 0.312 MPa Ans
c) No impact: 1
Pmax W
max
Pmax
A
max 0.156 MPa Ans
Note: since all of the max < Y = 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 canbe 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 Y 924MPa
Solution:
Weight : W mo g
Section Property : A
do
2
4
Static Displacement : st
W L
E A st
0.00195 mm
Drop Height: hmax 1.5m
Impact factor: max 1 1 2
hmax
st
max 1241.04
Pmax W max
max
Pmax
A
max 193.7 MPa
Note: since max < Y = 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 Y 152MPa
L 0.5m v 0.6
m
s
Solution:
Section Property : A
do
2
4
Eq. Spring Constant: k
E A
L
k 1579828.41
kN
m
Conservation of Energy: Ui = Ur
1
2
W
g
v2
1
2
k max
2=
max
W
g k
v max 0.3049 mm
Maximum Stress:
Pmax k max
max
Pmax
A
max 27.3 MPa Ans
< Y = 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
max 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 Y 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: 1 1 2
h
st
110.44 Ans
Maximum moment occurs at mid-span, where Mmax
W L
4
Mmax 0.810 kN m
c 0.5d max
Mmax c
I
max 129.1 MPa Ans
Since max < Y = 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 allow 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: 1 1 2
h
st
=
Maximum moment occurs at mid-span, where Mmax
W L
4
c 0.5d max
Mmax c
I
= max allow=
allow
W L d
8I
1 1 2
h
st
=
h
st
2
8 I allow
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 Y 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: 1 1 2
h
st
176.27
Maximum moment occurs at support A, where Mmax W L Mmax 0.300 kN m
c 0.5d max
Mmax c
I
max 233.2 MPa Ans
Since max < Y = 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 allow = 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 allow 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: 1 1 2
h
st
=
Maximum moment occurs at mid-span, where Mmax W L
c 0.5d max
Mmax c
I
= max allow=
allow
W L d
2I
1 1 2
h
st
=
h
st
2
2 I allow
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 Y 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
st
W L2
2E I st
0.00003906 rad
At end B, 'st st st L 'st 0.09766 mm
Maximum Displacement :
Impact factor: 1 1 2
h
st
176.27
'B.max 'st
'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, Y = 250 MPa.
Given: L 2m mo 10 10
3 kg
E 200GPa do 200mm
Y 250MPa
d 153mmSolution:
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 v
2
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
max
Mmax c
I
max 237.2 MPa Ans
< Y = 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
do
4
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
v2
1
2
3 E I
LBC
3 C.max C.max
=
C.max
W v2 LBC
3
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, C.max
Pmax LBC
2
2E I C.max
0.12450 rad
At end A, A.max C.max C.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
mY 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
1
2
ksp sp
2 1
2
kb b
2= (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 b
2= (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
max
Mmax c
I
max 50.82 MPa Ans
< Y = 250 MPa (O.K. !)
Problem14-66
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
sSolution:
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
v2 W h max
1
2
kb b
2= (1)
However, b max= (2)
Substituting Eq.(2) into (1):
Given
1
2
W
g
v2 W h max
1
2
kb max
2= (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
max
Mmax c
I max
17.99 MPa Ans
Maximum Displacement : 
(From Appendix C)
At point C, C.max
Pmax L
2
16E I C.max
0.006425 rad
At end D, D.max C.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
sSolution:
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
v2 W h max
1
2
kb b
2= (1)
However, b max= (2)
Substituting Eq.(2) into (1):
Given
1
2
W
g
v2 W h max
1
2
kb max
2= (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
max
Mmax c
I max
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 Y 250MPa
Solution:
Use W 150x18: I 9.19 106 mm4 d 153mm
Static Displacement at B : 
+
Support Reactions :
 
Fy=0; A C W 0= (1)
C=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=
L
0
2
i 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=
max
Mmax c
I
= Y
Pmax L d
2I
= Pmax
2I Y
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
1
2
kB max
2= h
1
2W
kB max
2
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 Y 250MPa
Solution:
Use W 150x18: I 9.19 106 mm4 d 153mm
Static Displacement at B : 
+
Support Reactions :
 
Fy=0; A C W 0= (1)
C=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=
L
0
2
i 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
1
2
kB max
2= (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 max
Mmax c
I max
249.60 MPa Ans
< Y = 250 MPa (O.K. !)
1
2
mo v
2 1
2
kb beam
2 1
4
kb
2
ksp
beam
2=
Given: L 1.8m c 75mm
mo 1.8 10
3 kg I 300 106 mm4
E 2GPa Y 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 v
2 1
2
kb beam
2 2
1
2
ksp sp
2= (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, Y = 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
max
Mmax c
I max
4.89 MPa Ans
< Y = 30 MPa (O.K. !)
30degGiven: LBC 1.5m P 4kN
LAB
LBC
sin
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
nAB 1.1547 kN
nBC nAB sin 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: atan
LAC
LBC
56.310 deg
Member Lengths:
LAB LBC
2 LAC
2 LAB 1.803 m
Member Real Forces:
NAB
P
sin
NAB 3.6056 kN
NBC NAB cos 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: atan
LAC
LBC
56.310 deg
Member Lengths:
LAB LBC
2 LAC
2 LAB 1.803 m
Member Real Forces:
NAB
P
sin
NAB 3.6056 kN
NBC NAB cos NBC 2.0000 kN
Member Virtual Forces:
Apply Virtual Force F' 1kN
nAB
F'
sin
nAB 1.2019 kN
nBC nAB cos 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: atan
h
b
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
NAB 0.8333 kN
NBC NAB NBC 0.8333 kN
NAC NAB cos NAC 0.5000 kN
Member Virtual Forces:
Apply Virtual Force F' 1kN
nAB
F'
2 cos
nAB 0.8333 kN
nBC nAB nBC 0.8333 kN
nAC nAB cos 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: atan
h
b
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
NAB 0.8333 kN
NBC NAB NBC 0.8333 kN
NAC NAB cos NAC 0.5000 kN
Member Virtual Forces:
Apply Virtual Force F' 1kN
nAB
F'
2 sin
nAB 0.6250 kN
nBC nAB nBC 0.6250 kN
nAC nAB cos 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: atan
h
b
53.130 deg
Member Lengths:
LCD h LAB h LAC b
2 h2 LAC 2.5 m
LDA b LBC b
Member Real Forces: NDA 0
NCD P2 NCD 10 kN
NAC
P2
sin
NAC 12.5 kN
NBC P1 NAC cos NBC 12.5 kN
NAB NAC sin 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: atan
h
b
53.130 deg
Member Lengths:
LCD h LAB h LAC b
2 h2 LAC 2.5 m
LDA b LBC b
Member Real Forces: NDA 0
NCD P2 NCD 10 kN
NAC
P2
sin
NAC 12.5 kN
NBC P1 NAC cos NBC 12.5 kN
NAB NAC sin NAB 10 kN
Member Virtual Forces:
Apply Virual Force F' 1kN
nDA 0 nDA 0
nCD F' nCD 1 kN
nAC
F'
sin
nAC 1.25 kN
nBC nAC cos nBC 0.75 kN
nAB nAC sin 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: atan
h
b
36.870 deg
Member Lengths:
LBE h
LAE b
2 h2 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
NAE 20.8333 kN NCE
P
2 sin
NCE 20.8333 kN
NAB NAE cos NAB 16.6667 kN NBC NCE cos 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
nAE 0.8333 kN nCE
F'
2 sin
nCE 0.8333 kN
nAB nAE cos nAB 0.6667 kN nBC nCE cos 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: atan
h
b
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
NAE 20.8333 kN NCE
P
2 sin
NCE 20.8333 kN
NAB NAE cos NAB 16.6667 kN NBC NCE cos 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
nAE 0.8333 kN nCE
F'
2 sin
nCE 0.8333 kN
nAB nAE cos nAB 0.6667 kN nBC nCE cos 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: atan
2 h
b
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
NCD 8.9443 kN
NED NCD sin NED 8 kN
NAE NED NAE 8 kN
NEC P2 NEC 2 kN
NAC
0.5NEC
cos
NAC 2.2361 kN
NBC NCD
0.5NEC
cos
NBC 11.1803 kN
Member Virtual Forces:
Apply Virtual Force F' 1kN
nCD
F'
cos
nCD 2.2361 kN
nED nCD sin 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: atan
2 h
b
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
NCD 8.9443 kN
NED NCD sin NED 8 kN
NAE NED NAE 8 kN
NEC P2 NEC 2 kN
NAC
0.5NEC
cos
NAC 2.2361 kN
NBC NCD
0.5NEC
cos
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 kNnEC F' nEC 1 kN
nAC
0.5nEC
cos
nAC 1.118 kN
nBC
0.5nEC
cos
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: atan
h
b
36.870 deg
Member Lengths:
LCD h LAB h LAC b
2 h2 LAC 2.5 m
LDA b LBC b
Member Real Forces: NAB 0
NBC P1 NBC 4 kN
NAC
P1
cos
NAC 5 kN
NCD P2 NAC sin NCD 2 kN
NDA NAC cos NDA 4 kN
Member Virtual Forces:
Apply Virtual Force F' 1kN
nAB 0
nBC F' nBC 1 kN
nAC
F'
cos
nAC 1.25 kN
nCD nAC sin nCD 0.75 kN
nDA nAC cos 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: atan
h
b
36.870 deg
Member Lengths:
LCD h LAB h LAC b
2 h2 LAC 2.5 m
LDA b LBC b
Member Real Forces: NAB 0
NBC P1 NBC 4 kN
NAC
P1
cos
NAC 5 kN
NCD P2 NAC sin NCD 2 kN
NDA NAC cos 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: 1 atan
h
b1
1 53.130 deg
2 atan
h
b2
2 33.690 deg
Support Reactions :
+
Cv 0
+
Fy=0; Dv P1 P2 0= (1)
Fx=0; Ch Dh 0= (2)
D=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 b2
2 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 1
NAB 50 kN nAB
F'
sin 1
nAB 1.25 kN
NAE NAB cos 1 NAE 30 kN nAE nAB cos 1 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 2
NBD 180.3 kN nBD
F'
sin 2
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: atan
h
b
36.870 deg
Support Reactions :
+
By symmetry, A=E=R
Fy=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
NBH
P2
2 sin NDH NBH NEG NAI
NAB NAI cos NBC NAB NBH cos NCD NBC NDE NAB
NAJ 0 NBI NAI sin NDG NBI NEF NAJ
NJI 0 NIH NAI cos NHG NIH NGF NJI
Member Virtual Forces:
Apply Virtual Force F' 1kN
Support Reactions :
+
By symmetry, A'=E'=R'
Fy=0; 2R' F' 0= R' 0.5 F' R' 0.5 kN
nCH F'
nAI
R'
sin
nBH
F'
2 sin nDH nBH nEG nAI
nAB nAI cos nBC nAB nBH cos nCD nBC nDE nAB
nAJ 0 nBI nAI sin nDG nBI nEF nAJ
nJI 0 nIH nAI cos 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: atan
h
b
36.870 deg
Support Reactions :
+
By symmetry, A=E=R
Fy=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
NBH
P2
2 sin NDH NBH NEG NAI
NAB NAI cos NBC NAB NBH cos NCD NBC NDE NAB
NAJ 0 NBI NAI sin NDG NBI NEF NAJ
NJI 0 NIH NAI cos NHG NIH NGF NJI
Member Virtual Forces:
Apply Virtual Force F' 1kN
Support Reactions :
+
By symmetry, A'=E'=R'
Fy=0; 2R' F' 0= R' 0.5 F' R' 0.5 kN
nCH 0
nAI
R'
sin
nBH
F'
2 sin nDH nBH nEG nAI
nAB nAI cos nBC nAB nBH cos nCD nBC nDE nAB
nAJ 0 nBI nAI sin nDG nBI nEF nAJ
nJI 0 nIH nAI cos 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:
+
 
Fy=0; A B P 0= (1) Fy=0; A' B' F' 0= (1')
B=0; A b P a b( ) 0= (2) B=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=
L
0
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
+
 
Fy=0; A B P 0= (1) Fy=0; A' B' 0= (1')
B=0; A b P a b( ) 0= (2) B=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=
L
0
dxmM/(EI)1Virtual Work Equation:
m' B
1
E I
0
a
x1m1 M1 d
0
b
x2m2 M2 d
0
a
x3m3 M3 d=
B
1
E I
1
m'
0
0
b
x2A' b x2 B x2 d 0
B 7.143 10
3 rad
B 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:
+Fy=0; A B 2P 0= (1) Fy=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)
L
0
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: 
+Fy=0; A B 2P 0= (1) Fy=0; A' B' 0= (1')
Due to symmetry: A B= (2) B=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=
L
0
dxmM/(EI)1Virtual Work Equation:
m' A
1
E I
0
a
x1m1 M1 d
0
b
x2m2 M2 d
0
a
x3m3 M3 d=
A
1
E I
1
m'
0
0
b
x2P a( ) B' x2 d 0
A 4.412 10
3 rad
A 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
do
4
64
I 102.35 103 mm4
Real Support Reactions : 
+ Fy=0; A C PD PE 0= (1)
C PD PE A
C=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:
+ Fy=0; (1')
C=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.
L
0
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 kN
2 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 kN
2 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
do
4
64
I 102.35 103 mm4
Real Support Reactions : 
+ Fy=0; A C PD PE 0= (1)
C=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: 
Fy=0; A' C' 0= (1')
C=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=
L
0
dxmM/(EI)1Virtual Work Equation:
m' A
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
A
1
E I
1
m'
I1 I4 I3
A 55.038 10
3 rad
A 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
do
4
64
I 39.76 103 mm4
Real Support Reactions : 
+ Fy=0; A C PD PE PC 0= (1)
C=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: 
Fy=0; A' C' F'= (1')
C=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=
L
0
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 d
I2 0.18743 kN
2 m3
I4
0
b
x4m4 M4 d= I4
0
b
x4C' b x4 C b x4 PE x4 d
I4 0.12857 kN
2 m3
I3
0
b
x3m3 M3 d= I3
0
b
x3C' x3 C x3 d
I3 0.02388 kN
2 m3
B
1
E I
1
F'
I1 I2 I4 I3
B 47.75 mm Ans
Problem 14-98
The simply supported beam having a square cross section is subjected to a uniform load w. Determine
the maximum deflection of the beam caused only by bending, and caused by bending and shear. Take
E = 3G.
Problem 14-99
Determine the displacement at point C. EI is constant.
Problem 14-100
Determine theslope at point B. EI is constant.
Problem 14-101
The A-36 steel beam has a moment of inertia of I = 125(106) mm4. Determine the displacement at D.
Given: a 4mMo 18kN m
b 3m E 200GPa
I 125 106 mm4
Solution: Virtual Force F' 1kN
Real Support Reactions : 
+ Fy=0; B C 0= (1)
Due to symmetry: B C= (2)
Solving Eqs. (1) and (2):
B 0 C 0
Support Reactions due to Virtual Force at D:
+ Fy=0; B' C' F' 0= (1')
Due to symmetry: B' C'= (2')
Solving Eqs. (1') and (2'):
B' 0.5F' C' 0.5F'
B' 0.5 kN C' 0.5 kN
Internal Moment Functions:
A->B B->D C->D E->C
Real M1 Mo= M2 Mo= M3 Mo= M4 Mo=
Virtual m1 0= m2 B' x2= m3 C' x3= m4 0=
(for x2 < b)
L
0
dxmM/(EI)1Virtual Work Equation:
F' D
1
E I
0
a
x1m1 M1 d
0
b
x2m2 M2 d
0
b
x3m3 M3 d
0
a
x4m4 M4 d=
D
1
E I
1
F'
0
0
b
x2B' x2 Mo d
0
b
x3C' x3 Mo d 0
D 3.24 mm Ans
Problem 14-102
The A-36 steel beam has a moment of inertia of I = 125(106) mm4. Determine the slope at A.
Given: a 4m Mo 18kN m
b 3m E 200GPa
I 125 106 mm4
 Solution: Virtual Moment m' 1kN m
Real Support Reactions : 
+ Fy=0; B C 0= (1)
Due to symmetry: B C= (2)
Solving Eqs. (1) and (2):
B 0 C 0
Support Reactions due to Virtual Moment at A:
+ Fy=0; B' C' 0= (1')
C=0; B' 2 b( ) m' 0= (2')
Solving Eqs. (1') and (2'):
B'
m'
2 b
C' B'
B' 0.1667 kN C' 0.1667 kN
Internal Moment Functions:
A->B C->B E->C
Real M1 Mo= M3 Mo= M4 Mo=
Virtual m1 m'= m3 C' x3= m4 0=
(for x3 <2b)
L
0
dxmM/(EI)1Virtual Work Equation:
m' A
1
E I
0
a
x1m1 M1 d
0
2 b
x3m3 M3 d
0
a
x4m4 M4 d=
A
1
E I
1
m'
0
a
x1m' Mo d
0
2 b
x3C' x3 Mo d 0
A 5.040 10
3 rad
A 0.289 deg (anticlockwise) Ans
Problem 14-103
The A-36 structural steel beam has a moment of inertia of I = 125(106) mm4. Determine the slope of
the beam at B.
Given: a 4m Mo 18kN m
b 3m E 200GPa
I 125 106 mm4
 Solution: Virtual Moment m' 1kN m
Real Support Reactions : 
+ Fy=0; B C 0= (1)
Due to symmetry: B C= (2)
Solving Eqs. (1) and (2):
B 0 C 0
Support Reactions due to Virtual Moment at B:
+ Fy=0; B' C' 0= (1')
C=0; B' 2 b( ) m' 0= (2')
Solving Eqs. (1') and (2'):
B'
m'
2 b
C' B'
B' 0.1667 kN C' 0.1667 kN
Internal Moment Functions:
A->B B->C E->C
Real M1 Mo= M2 Mo= M4 Mo=
Virtual m1 0= m2 m' B' x2= m4 0=
(for x2 <2b)
L
0
dxmM/(EI)1Virtual Work Equation:
m' A
1
E I
0
a
x1m1 M1 d
0
2 b
x2m2 M2 d
0
a
x4m4 M4 d=
B
1
E I
1
m'
0
0
2 b
x2m' B' x2 Mo d 0
B 2.160 10
3 rad
B 0.124 deg (anticlockwise) Ans
Problem 14-104
Determine the slope at A. EI is constant.
Problem 14-105
Determine the displacement at C. EI is constant.
Problem 14-106
Determine the slope at B. EI is constant.
Problem 14-107
The beam is made of oak, for which Eo = 11 GPa. Determine the slope and displacement at A.
Given: L 3m E 11GPa
h 400mm wo
4kN
mt 200mm
Solution:
Section Property : I
t h3
12
a) Applying Virtual Force at A: F' 1kN
Internal Moment Functions:
(clockwise) Ans
Real M1
x1
2
wo
x1
L
x1
3
= M2
wo
2
L
L
3
x2
wo
2
x2
2=
Virtual m1 F' x1= m2 F' L x2=
L
0
dxmM/(EI)1Virtual Work Equation:
F' A
1
E I
0
L
x1m1 M1 d
0
L
x2m2 M2 d=
A
1
E I
0
L
x1x1
x1
2
wo
x1
L
x1
3
d
0
L
x2L x2
wo
2
L
L
3
x2
wo
2
x2
2 d
A 27.38 mm Ans
b) Applying Virtual Moment at A: m' 1kN m
Internal Moment Functions:
Real M1
x1
2
wo
x1
L
x1
3
= M2
wo
2
L
L
3
x2
wo
2
x2
2=
Virtual m1 m'= m2 m'=
L
0
dxmM/(EI)1Virtual Work Equation:
m' A
1
E I
0
L
x1m1 M1 d
0
L
x2m2 M2 d=
A
1
E I
0
L
x1
x1
2
wo
x1
L
x1
3
d
0
L
x2
wo
2
L
L
3
x2
wo
2
x2
2 d
A 5.753 10
3 rad
C->BA->C
C->BA->C
Problem 14-108
Determine the displacement at B. EI is constant.
Problem 14-109
Determine the slope and displacement at point C. EI is constant.
Problem 14-110
Bar ABC has a rectangular cross section of 300mm by 100 mm. Attached rod DB has a diameter of 20
mm. If both members are made of A-36 steel, determine the vertical displacement of point C due to the
loading. Consider only the effect of bending in ABC and axial force in DB.
Given: a 4m h 300mm P 20kN E 200GPa
b 3m t 100mm do 20mm
Solution: Virtual Force F' 1kN
Section Properties : 
Lo a
2 b2 Ao
do
2
4
I
t h3
12
Virtual Support Reactions : 
+
 
Fy=0; A'v B'v F' 0= (1)
B=0; A'v b F' b 0= (2)
Solving Eqs. (1) and (2): A'v F' A'v 1 F'
B'v F' A'v B'v 2 F'
Axial Force in the Rod BD : n'BD B'v
Lo
a
n'BD 2.5 F'
Real Support Reactions : Similarly, Av 1 P
Bv 2 P
NBD 2.5 P
Internal Moment and Force Functions:
A->B B->C B->D
Real M1 Av x1= M2 P x2= N3 NBD=
Virtual m1 A'v x1= m2 F' x2= n3 n'BD=
nNL/(EA)dxmM/(EI)1
L
0Virtual Work Equation:
F' C
1
E I
0
b
x1m1 M1 d
0
b
x2m2 M2 d
n3 N3 Lo
E A
=
C
1
E I
1
F'
0
b
x1A'v x1 Av x1 d
0
b
x2F' x2 P x2 d
2.5 F' 2.5 P( ) Lo
E Ao F'
C 17.95 mm Ans
(2')
Given: a 4m h 300mm P 20kN E 200GPa
b 3m t 100mm do 20mm
 Solution: Virtual Moment m' 1kN m
Section Properties : 
Lo a
2 b2 Ao
do
2
4
I
t h3
12
+
Real Support Reactions : 
Fy=0; Av Bv P 0= (1)
B=0; Av b P b 0= (2)
Solving Eqs. (1) and (2): Av P Av 1 P
Bv P Av Bv 2 P
Axial Force in the Rod BD : NBD Bv
Lo
a
NBD 2.5 P
Virtual Support Reactions : 
+ Fy=0; A'v B'v 0= (1')
B=0; A'v b m' 0=
Problem 14-111
Bar ABC has a rectangular cross section of 300mm by 100 mm. Attached rod DB has a diameter of 20
mm. If both members are made of A-36 steel, determine the slope at A due to the loading. Consider
only the effect of bending in ABC and axial force in DB.
Solving Eqs. (1) and (2): A'v
m'
b
A'v 0.3333 kN
B'v A'v B'v 0.3333 kN
Axial Force in the Rod BD :
n'BD B'v
Lo
a
n'BD 0.417 kN
Internal Moment and Force Functions:
A->B B->C B->D
Real M1 Av x1= M2 P x2= N3 NBD=
Virtual m1 m' A'v x1= m2 0= n3 n'BD=
nNL/(EA)dxmM/(EI)1
L
0
Virtual Work Equation:
m' A
1
E I
0
b
x1m1 M1 d
0
b
x2m2 M2 d
n3 N3 Lo
E A
=
A
1
E I
1
m'
0
b
x1m' A'v x1 Av x1 d 0
n'BD 2.5 P( ) Lo
E Ao m'
A 991.20 10
6 rad (anticlockwise) Ans
Problem 14-112
Determine the vertical displacement of point A on the angle bracket due to the concentrated force P.
The bracket is fixed connected to its support. EI is constant. Consider only the effect of bending.
Problem 14-113
The L-shaped frame is made from two segments, each of length L and flexural stiffness EI. If it is
subjected to the uniform distributed load, determine the horizontal displacement of the end C.
Problem 14-114
The L-shaped frame is made from two segments, each of length L and flexural stiffness EI. If it is
subjected to the uniform distributed load, determine the vertical displacement of the point B.
Problem 14-115
Determine the horizontal displacement of point C. EI is constant. There is a fixed support at A.
Consider only the effect of bending.
Given: a 1.6m b 1.5m c 1.5m
Pv 3kN Ph 4kN
Mo 1.8kN m
Solution: Virtual Force F' 1kN
Internal Moment Functions:
C->B B->E E->A
Real M1 Mo Pv x1= M2 Mo Pv a= M3 Mo Pv a Ph x3=
Virtual m1 0= m2 F' x2= m3 F' b x3=
L
0
dxmM/(EI)1Virtual Work Equation:
Set EI 1kN m2
F' C
1
E I
0
a
x1m1 M1 d
0
b
x2m2 M2 d
0
c
x3m3 M3 d=
C
1
EI
1
F'
0
0
b
x2F' x2 Mo Pv a d
0
c
x3F' b x3 Mo Pv a Ph x3 d
C 40.95 m
C
40.95
EI
m= Ans
Problem 14-116
The ring rests on the rigid surface and is subjected to the vertical load P. Determine the vertical
displacement at B. EI is constant.
Problem 14-117
Solve Prob. 14-71 using Castigliano's theorem.
Given: LBC 1.5m P 4kN