Soluções Resistência dos Materiais

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Problem 1-1
Determine the resultant internal normal force acting on the cross section through point A in each
column. In (a), segment BC weighs 300 kg/m and segment CD weighs 400kg/m. In (b), the column
has a mass of 200 kg/m.
a( ) Given: g 9.81
m
s2
wBC 300
kg
m
LBC 3m wCA 400
kg
m
FB 5kN LCA 1.2m
FC 3kN
Solution:
Fy = 0; FA wBC g LBC wCA g LCA FB 2FC 0=
FA wBC g LBC wCA g LCA FB 2FC
FA 24.5 kN Ans
b( ) Given: g 9.81
m
s2
w 200
kg
m
L 3m F1 6kN
FB 8kN F2 4.5kN
Solution:
Fy = 0; FA w L( ) g FB 2F1 2F2 0=
FA w L( ) g FB 2F1 2F2
FA 34.89 kN Ans
Problem 1-6
The beam AB is pin supported at A and supported by a cable BC. Determine the resultant internal
loadings acting on the cross section at point D.
Given: P 5000N
a 0.8m b 1.2m c 0.6m d 1.6m
e 0.6m
Solution:
atan
b
d
36.87 deg
atan
a b
d
14.47 deg
Member AB: 
+ A=0; FBC sin a b( ) P b( ) 0=
FBC
P b( )
sin a b( )
FBC 12.01 kN
Segment BD: 
 + Fx=0; ND FBC cos P cos 0=
ND FBC cos P cos
ND 15.63 kN Ans
+ Fy=0; VD FBC sin P sin 0=
VD FBC sin P sin
VD 0 kN Ans
+ D=0; FBC sin P sin
d c
sin
MD 0=
MD FBC sin P sin
d c
sin
MD 0 kN m Ans
Note: Member AB is the two-force member. Therefore the shear force and moment are zero.
Problem 1-7
Solve Prob. 1-6 for the resultant internal loadings acting at point E.
Given: P 5000N
a 0.8m b 1.2m c 0.6m d 1.6m
e 0.6m
Solution:
atan
b
d
36.87 deg
atan
a b
d
14.47 deg
Member AB: 
+ A=0; FBC sin a b( ) P b( ) 0=
FBC
P b( )
sin a b( )
FBC 12.01 kN
Segment BE: 
 + Fx=0; NE FBC cos P cos 0=
NE FBC cos P cos
NE 15.63 kN Ans
+ Fy=0; VE FBC sin P sin 0=
VE FBC sin P sin
VE 0 kN Ans
+ E=0; FBC sin P sin e ME 0=
ME FBC sin P sin e
ME 0 kN m Ans
Note: Member AB is the two-force member. Therefore the shear force and moment are zero.
Problem 1-42
The 250-N lamp is supported by three steel rods connected by a ring at A. Determine which rod is
subjected to the greater average normal stress and compute its value. Take = 30°. The diameter of
each rod is given in the figure.
Given: W 250N 30deg 45deg
dB 9mm dC 6mm dD 7.5mm
Solution: Initial guess: FAC 1N FAD 1N
Given
+ Fx=0; FAC cos FAD cos 0= [1]
+ Fy=0; FAC sin FAD sin W 0= [2]
Solving [1] and [2]:
FAC
FAD
Find FAC FAD
FAC
FAD
183.01
224.14
N
Rod AB:
AAB
dB
2
4
AAB 63.61725 mm
2
AB
W
AAB
AB 3.93 MPa
Rod AD :
AAD
dD
2
4
AAD 44.17865 mm
2
AD
FAD
AAD
AD 5.074 MPa
Rod AC:
AAC
dC
2
4
AAC 28.27433 mm
2
AC
FAC
AAC
AC 6.473 MPa Ans
Problem 1-43
Solve Prob. 1-42 for = 45°.
Given: W 250N 45deg 45deg
dB 9mm dC 6mm dD 7.5mm
Solution: Initial guess: FAC 1N FAD 1N
Given
+ Fx=0; FAC cos FAD cos 0= [1]
+ Fy=0; FAC sin FAD sin W 0= [2]
Solving [1] and [2]:
FAC
FAD
Find FAC FAD
FAC
FAD
176.78
176.78
N
Rod AB:
AAB
dB
2
4
AAB 63.61725 mm
2
AB
W
AAB
AB 3.93 MPa
Rod AD :
AAD
dD
2
4
AAD 44.17865 mm
2
AD
FAD
AAD
AD 4.001 MPa
Rod AC:
AAC
dC
2
4
AAC 28.27433 mm
2
AC
FAC
AAC
AC 6.252 MPa Ans
Problem 1-101
The hanger assembly is used to support a distributed loading of w = 12 kN/m. Determine the average
shear stress in the 10-mm-diameter bolt at A and the average tensile stress in rod AB, which has a
diameter of 12 mm. If the yield shear stress for the bolt is y = 175 MPa, and the yield tensile stress for
the rod is y = 266 MPa, determine the factor of safety with respect to yielding in each case.
Given: y 175MPa w 12
kN
m
y 266MPa
a 1.2m b 0.6m e 0.9m
do 10mm drod 12mm
Solution:
c a2 e2 h
a
c
v
e
c
Support Reactions: L a b
C=0; FAB v( ) a( ) w L( ) 0.5 L( ) 0=
FAB w
L
a v
0.5 L( )
FAB 27 kN
For bolt A: Bolt A is subjected to double shear, and V 0.5FAB V 13.5 kN
A
4
do
2 V
A
171.89 MPa Ans
FS
y
FS 1.02 Ans
For rod AB: N FAB N 27 kN
A
4
drod
2 N
A
238.73 MPa Ans
FS
y
FS 1.11 Ans
Problem 1-102
Determine the intensity w of the maximum distributed load that can be supported by the hanger
assembly so that an allowable shear stress of allow = 95 MPa is not exceeded in the 10-mm-diameter
bolts at A and B, and an allowable tensile stress of allow = 155 MPa is not exceeded in the
12-mm-diameter rod AB.
Given: allow 95MPa allow 155MPa
a 1.2m b 0.6m e 0.9m
do 10mm drod 12mm
Solution: c a2 e2 h
a
c
v
e
c
Support Reactions: L a b
C=0; FAB v( ) a( ) w L( ) 0.5 L( ) 0=
FAB w
L
a v
0.5 L( )=
Assume failure of pin A or B:
V 0.5FAB= V allow A= A 4
do
2
0.5 w
L
a v
0.5 L( ) allow 4
do
2=
w
a v
0.5L( )2
allow 4
do
2
w 6.632
kN
m
(controls!) Ans
Assuming failure of rod AB:
N FAB= N allow A= A 4
drod
2
w
L
a v
0.5 L( ) allow 4
drod
2=
w
a v
0.5L2
allow 4
drod
2
w 7.791
kN
m
Problem 2-2
A thin strip of rubber has an unstretched length of 375 mm. If it is stretched around a pipe having an
outer diameter of 125 mm, determine the average normal strain in the strip.
Given: L0 375mm
Solution:
L 125( ) mm
L L0
L0
0.0472
mm
mm
Ans
Problem 2-3
The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the
end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
Given: a 3m LCE 4m
b 4m LBD 4m
LCE 10mm
Solution:
LBD
a
a b
LCE LBD 4.2857 mm
CE
LCE
LCE
CE 0.00250
mm
mm
Ans
BD
LBD
LBD
BD 0.00107
mm
mm
Ans
Problem 2-5
The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam is displace
10 mm downward, determine the normal strain developed in wires CE and BD.
Given: a 3m b 2m c 2m
LCE 4m LBD 3m
tip 10mm
Solution:
LBD
a
LCE
a b
=
LCE
a b
tip
a b c
=
LCE
a b
a b c tip
LCE 7.1429 mm
LBD
a
a b c tip
LBD 4.2857 mm
Average Normal Strain:
CE
LCE
LCE
CE 0.00179
mm
mm
Ans
BD
LBD
LBD
BD 0.00143
mm
mm
Ans
Problem 2-6
The rigid beam is supported by a pin at A and wires BD and CE. If the maximum allowable normal
strain in each wire is max = 0.002 mm/mm, determine the maximum vertical displacement of the load
P.
Given: a 3m b 2m c 2m
LCE 4m LBD 3m
allow 0.002
mm
mm
Solution:
LBD
a
LCE
a b
=
LCE
a b
tip
a b c
=
Average Elongation/Vertical Displacement:
LBD LBD allow LBD 6.00 mm
tip
a b c
a
LBD
tip 14.00 mm
LCE LCE allow LCE 8.00 mm
tip
a b c
a b
LCE
tip 11.20 mm (Controls !) Ans
Problem 2-7
The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2
mm, determine the normal strain developed in each wire.
Given:
a 300mm 30deg
A 2mm
Solution:
Consider the triangle CAA':
A 180deg A 150 deg
LCA' a
2
A
2
2 a A cos A
LCA' 301.734 mm
CA
LCA' a
a
CA 0.00578
mm
mm
Ans
Problem 3-14
A specimen is originally 300 mm long, has a diameter of 12 mm, and is subjected to a force of 2.5 kN.
When the force is increased to 9 kN, the specimen elongates 22.5 mm. Determine the modulus of
elasticity for the material if it remains elastic.
Given: d 12mm L0 300mm
P1 2.5kN P2 9kN L 22.5mm
Solution:
A
4
d2
Normal Force: Applying equation = P / A .
1
P1
A 1
22.105 MPa
2
P2
A 2
79.577 MPa
Thus,
2 1 57.473 MPa
L
L0
E E 766.3 MPa Ans
Problem 3-15
A structural member in a nuclear reactor is made from a zirconium alloy. If an axial load of 20 kN is to
be supported by the member, determine its required cross-sectional area. Use a factor of safety of 3
with respect toyielding. What is the load on the member if it is 1-m long and its elongation is 0.5 mm?
Ezr = 100 GPa, Y = 400 MPa. The material has elastic behavior.
Given: P 20kN L0 1m L 0.5mm
Ezr 100 GPa Y 400 MPa FoS 3
Solution:
allow
Y
FoS allow
133.33 MPa
Areq
P
allow
Areq 150 mm
2 Ans
A Areq
L
L0
0.0005
mm
mm
Ezr 50 MPa
P A P 7.5 kN Ans
Problem 3-26
The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it,
determine the change in its length and the change in its diameter. Ep = 2.70 GPa, p = 0.4. 
Given: P 300N
L 200mm d 15mm
EP 2.70 GPa 0.4
Solution:
A
4
d2
P
A
1.698 MPa
long EP
long 0.0006288
mm
mm
long L 0.126 mm Ans
lat long lat 0.0002515
mm
mm
d lat d( ) d 0.003773 mm Ans
Problem 3-38
The rigid pipe is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5
mm., determine how much it stretches when a load of P = 1.5 kN acts on the pipe. The material
remains elastic.
Given: P 1.5kN Est 200 GPa
LBC 2.4m d 5mm
60deg
Solution:
Support Reactions: 
C=0; FAB cos LAB P LAB 0=
FAB
P
cos
FAB 3 kN
Normal Stress:
Area
4
d2
AB
FAB
Area AB
152.789 MPa
Normal Strain: LAB
LBC
sin
LAB 2.771 m
Applying Hook's law = E .
AB
AB
Est
AB 0.0007639
mm
mm
Thus, LAB AB LAB
LAB 2.1171 mm Ans
Problem 3-39
The rigid pipe is supported by a pin at C and an A-36 guy wire AB. If the wire has a diameter of 5
mm., determine the load P if the end B is displaced 2.5 mm. to the right. Est = 200 GPa.
Given: L 2.4m d 5mm
60deg Bx 2.5mm
Est 200 GPa
Solution:
Consider triangle BB'C: L sin C Bx=
C asin
Bx
L
Consider triangle AB'C: 'C 90deg C LAC L cot
L'AB LAC
2 L2 2 LAC L cos 'C
L'AB 2.7725 m
Normal Strain:
LAB
L
sin
AB
L'AB LAB
LAB
AB 0.0004510
mm
mm
Normal Stress:
Area
4
d2
Applying Hook's law = E .
AB AB Est AB 90.191 MPa
Thus, FAB AB Area( ) FAB 1.771 kN
Support Reactions:
C=0; FAB cos L( ) P L 0=
P FAB cos P 0.885 kN Ans