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# 1. Hibbeler Resistências dos Materiais 7ªed Soluções

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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
s
2
\ufffd wBC 300
kg
m
\ufffd
LBC 3m\ufffd wCA 400
kg
m
\ufffd
FB 5kN\ufffd LCA 1.2m\ufffd
FC 3kN\ufffd
Solution:
\ufffdn6 Fy = 0; FA wBC g\ufffd \ufffd LBC\ufffd wCA g\ufffd \ufffd LCA\ufffd FB\ufffd 2FC\ufffd 0=
FA wBC g\ufffd \ufffd LBC wCA g\ufffd \ufffd LCA\ufffd FB\ufffd 2FC\ufffd\ufffd
FA 24.5 kN Ans
b( ) Given: g 9.81
m
s
2
\ufffd w 200 kg
m
\ufffd
L 3m\ufffd F1 6kN\ufffd
FB 8kN\ufffd F2 4.5kN\ufffd
Solution:
\ufffdn6 Fy = 0; FA w L( ) g\ufffd FB\ufffd 2F1\ufffd 2F2\ufffd 0=
FA w L( ) g FB\ufffd 2F1\ufffd 2F2\ufffd\ufffd
FA 34.89 kN Ans
Problem 1-2
Determine the resultant internal torque acting on the cross sections through points C and D of the
shaft. The shaft is fixed at B.
Given: TA 250N m\ufffd
TCD 400N m\ufffd
TDB 300N m\ufffd
Solution:
Equations of equilibrium:
+ TA TC\ufffd 0=
TC TA\ufffd
TC 250 N m Ans
+ TA TCD\ufffd TD\ufffd 0=
TD TCD TA\ufffd\ufffd
TD 150 N m Ans
Problem 1-3
Determine the resultant internal torque acting on the cross sections through points B and C.
Given: TD 500N m\ufffd
TBC 350N m\ufffd
TAB 600N m\ufffd
Solution:
Equations of equilibrium:
6 Mx = 0; TB TBC TD\ufffd\ufffd 0=
TB TBC\ufffd TD\ufffd\ufffd
TB 150 N m Ans
6 Mx = 0; TC TD\ufffd 0=
TC TD\ufffd
TC 500 N m Ans
Problem 1-4
A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting
on the section through point A.
Given: P 80N\ufffd
T 30deg\ufffd I 45deg\ufffd
a 0.3m\ufffd b 0.1m\ufffd
Solution:
Equations of equilibrium:
+ 6Fx'=0; NA P cos I T\ufffd\ufffd \ufffd\ufffd 0=
NA P cos I T\ufffd\ufffd \ufffd\ufffd
NA 77.27 N Ans
+ 6Fy'=0; VA P sin I T\ufffd\ufffd \ufffd\ufffd 0=
VA P sin I T\ufffd\ufffd \ufffd\ufffd
VA 20.71 N Ans
+ 60A=0; MA P cos I\ufffd \ufffd a cos T\ufffd \ufffd\ufffd P sin I\ufffd \ufffd b a sin T\ufffd \ufffd\ufffd\ufffd \ufffd\ufffd 0=
MA P\ufffd cos I\ufffd \ufffd a cos T\ufffd \ufffd P sin I\ufffd \ufffd b a sin T\ufffd \ufffd\ufffd\ufffd \ufffd\ufffd\ufffd
MA 0.555\ufffd N m Ans
Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Problem 1-5
Determine the resultant internal loadings acting on the cross section through point D of member AB.
Given: ME 70N m\ufffd
a 0.05m\ufffd b 0.3m\ufffd
Solution:
Segment AB: Support Reactions
+ 60A=0; \ufffdME By 2 a b\ufffd( )\ufffd 0=
By
ME\ufffd
2a b\ufffd\ufffd By 175\ufffd N
+At B: Bx By
150
200
§¨
·
¹\ufffd Bx 131.25\ufffd N
Segment DB: NB Bx\ufffd\ufffd VB By\ufffd\ufffd
+ 6Fx=0; ND NB\ufffd 0=
ND NB\ufffd\ufffd ND 131.25\ufffd N Ans
+ 6Fy=0; VD VB\ufffd 0=
VD VB\ufffd\ufffd VD 175\ufffd N Ans
+ 60D=0; MD\ufffd ME\ufffd By a b\ufffd( )\ufffd 0=
MD ME\ufffd By a b\ufffd( )\ufffd\ufffd
MD 8.75\ufffd N m Ans
Problem 1-6
The beam AB is pin supported at A and supported by a cable BC. Determine the resultant internal
Given: P 5000N\ufffd
a 0.8m\ufffd b 1.2m\ufffd c 0.6m\ufffd d 1.6m\ufffd
e 0.6m\ufffd
Solution:
T atan b
d
§¨
·
¹\ufffd T 36.87 deg
I atan a b\ufffd
d
§¨
·
¹ T\ufffd\ufffd I 14.47 deg
Member AB:
+ 60A=0; FBC sin I\ufffd \ufffd a b\ufffd( ) P b( )\ufffd 0=
FBC
P b( )
sin I\ufffd \ufffd a b\ufffd( )\ufffd
FBC 12.01 kN
Segment BD:
+ 6Fx=0; ND\ufffd FBC cos I\ufffd \ufffd\ufffd P cos T\ufffd \ufffd\ufffd 0=
ND FBC\ufffd cos I\ufffd \ufffd P cos T\ufffd \ufffd\ufffd\ufffd
ND 15.63\ufffd kN Ans
+ 6Fy=0; VD FBC sin I\ufffd \ufffd\ufffd P sin T\ufffd \ufffd\ufffd 0=
VD FBC\ufffd sin I\ufffd \ufffd P sin T\ufffd \ufffd\ufffd\ufffd
VD 0 kN Ans
+ 60D=0; FBC sin I\ufffd \ufffd P sin T\ufffd \ufffd\ufffd\ufffd \ufffd d c\ufffd
sin T\ufffd \ufffd MD\ufffd 0=
MD FBC sin I\ufffd \ufffd P sin T\ufffd \ufffd\ufffd\ufffd \ufffd d c\ufffd
sin T\ufffd \ufffd\ufffd
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\ufffd
a 0.8m\ufffd b 1.2m\ufffd c 0.6m\ufffd d 1.6m\ufffd
e 0.6m\ufffd
Solution:
T atan b
d
§¨
·
¹\ufffd T 36.87 deg
I atan a b\ufffd
d
§¨
·
¹ T\ufffd\ufffd I 14.47 deg
Member AB:
+ 60A=0; FBC sin I\ufffd \ufffd a b\ufffd( ) P b( )\ufffd 0=
FBC
P b( )
sin I\ufffd \ufffd a b\ufffd( )\ufffd
FBC 12.01 kN
Segment BE:
+ 6Fx=0; NE\ufffd FBC cos I\ufffd \ufffd\ufffd P cos T\ufffd \ufffd\ufffd 0=
NE FBC\ufffd cos I\ufffd \ufffd P cos T\ufffd \ufffd\ufffd\ufffd
NE 15.63\ufffd kN Ans
+ 6Fy=0; VE FBC sin I\ufffd \ufffd\ufffd P sin T\ufffd \ufffd\ufffd 0=
VE FBC\ufffd sin I\ufffd \ufffd P sin T\ufffd \ufffd\ufffd\ufffd
VE 0 kN Ans
+ 60E=0; FBC sin I\ufffd \ufffd P sin T\ufffd \ufffd\ufffd\ufffd \ufffd e ME\ufffd 0=
ME FBC sin I\ufffd \ufffd P sin T\ufffd \ufffd\ufffd\ufffd \ufffd e\ufffd
ME 0 kN m Ans
Note: Member AB is the two-force member. Therefore the shear force and moment are zero.
Problem 1-8
The boom DF of the jib crane and the column DE have a uniform weight of 750 N/m. If the hoist and
points A, B, and C.
Given: P 1500N\ufffd w 750 N
m
\ufffd
a 2.1m\ufffd b 1.5m\ufffd
c 0.6m\ufffd d 2.4m\ufffd e 0.9m\ufffd
Solution:
Equations of Equilibrium: For point A
+ 6Fx=0; NA 0\ufffd Ans
+
VA w e\ufffd P\ufffd 0=6Fy=0;
VA w e P\ufffd\ufffd VA 2.17 kN Ans
+ 60A=0; MA\ufffd w e( ) 0.5 e( )\ufffd P e( )\ufffd 0=
MA w e( )\ufffd 0.5 e( ) P e( )\ufffd\ufffd MA 1.654\ufffd kN m Ans
Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point B
+ 6Fx=0; NB 0\ufffd Ans
+
VB w d e\ufffd( )\ufffd P\ufffd 0=6Fy=0;
VB w d e\ufffd( ) P\ufffd\ufffd VB 3.98 kN Ans
+ 60B=0; \ufffdMB w d e\ufffd( )[ ] 0.5 d e\ufffd( )[ ]\ufffd P d e\ufffd( )\ufffd 0=
MB w d e\ufffd( )[ ]\ufffd 0.5 d e\ufffd( )[ ] P d e\ufffd( )\ufffd\ufffd
MB 9.034\ufffd kN m Ans
Note: Negative sign indicates that MB acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point C
+ 6Fx=0; VC 0\ufffd Ans
+
NC\ufffd w b c\ufffd d\ufffd e\ufffd( )\ufffd P\ufffd 0=6Fy=0;
NC w\ufffd b c\ufffd d\ufffd e\ufffd( ) P\ufffd\ufffd
NC 5.55\ufffd kN Ans
+ 60B=0; MC\ufffd w c d\ufffd e\ufffd( )[ ] 0.5 c d\ufffd e\ufffd( )[ ]\ufffd P c d\ufffd e\ufffd( )\ufffd 0=
MC w c d\ufffd e\ufffd( )[ ]\ufffd 0.5 c d\ufffd e\ufffd( )[ ] P c d\ufffd e\ufffd( )\ufffd\ufffd
MC 11.554\ufffd kN m Ans
Note: Negative sign indicates that NC and MC acts in the opposite direction to that shown
on FBD.
Problem 1-9
The force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of the
tooth, i.e., at the centroid point A of section a-a.
Given: P 400N\ufffd
T 30deg\ufffd I 45deg\ufffd
a 4mm\ufffd b 5.75mm\ufffd
Solution: D I T\ufffd\ufffd
Equations of equilibrium: For section a -a
+ 6Fx'=0; VA P cos D\ufffd \ufffd\ufffd 0=
VA P cos D\ufffd \ufffd\ufffd
VA 386.37 N Ans
+ 6Fy'=0; NA sin D\ufffd \ufffd\ufffd 0=
NA P sin D\ufffd \ufffd\ufffd
NA 103.53 N Ans
+ 60A=0; MA\ufffd P sin D\ufffd \ufffd a\ufffd P cos D\ufffd \ufffd b\ufffd 0=
MA P\ufffd sin D\ufffd \ufffd a P cos D\ufffd \ufffd b\ufffd\ufffd
MA 1.808 N m Ans
Problem 1-10
section through point C. Assume the reactions at the supports A and B are vertical.
Given: w1 4.5
kN
m
\ufffd w2 6.0
kN
m
\ufffd
a 1.8m\ufffd b 1.8m\ufffd c 2.4m\ufffd
d 1.35m\ufffd e 1.35m\ufffd
Solution: L1 a b\ufffd c\ufffd\ufffd
L2 d e\ufffd\ufffd
Support Reactions:
+ 60A=0; By L1 w1 L1\ufffd \ufffd 0.5 L1\ufffd \ufffd\ufffd 0.5w2 L2\ufffd \ufffd L1 L23\ufffd§¨©
·
¹\ufffd 0=
By w1 L1\ufffd \ufffd 0.5( ) 0.5w2 L2\ufffd \ufffd 1 L23 L1\ufffd
§¨
·
¹
\ufffd\ufffd By 22.82 kN
+ 6Fy=0; Ay By\ufffd w1 L1\ufffd 0.5w2 L2\ufffd 0=
Ay By\ufffd w1 L1\ufffd 0.5w2 L2\ufffd\ufffd Ay 12.29 kN
Equations of Equilibrium: For point C
+ 6Fx=0; NC 0\ufffd Ans
+ 6Fy=0; Ay w1 a b\ufffd( )\ufffdª¬ º¼ VC\ufffd 0=
VC Ay w1 a b\ufffd( )\ufffdª¬ º¼\ufffd\ufffd VC 3.92 kN Ans
+ 60C=0; MC w1 a b\ufffd( )ª¬ º¼ 0.5 a b\ufffd( )\ufffd Ay a b\ufffd( )\ufffd 0=
MC w1 a b\ufffd( )ª¬ º¼\ufffd 0.5 a b\ufffd( ) Ay a b\ufffd( )\ufffd\ufffd
MC 15.07 kN m Ans
Note: Negative sign indicates that VC acts in the opposite direction to that shown on FBD.
Problem 1-11
sections through points D and E. Assume the reactions at the supports A and B are vertical.
Given: w1 4.5
kN
m
\ufffd w2 6.0
kN
m
\ufffd
a 1.8m\ufffd b 1.8m\ufffd c 2.4m\ufffd
d 1.35m\ufffd e 1.35m\ufffd
Solution: L1 a b\ufffd c\ufffd\ufffd
L2 d e\ufffd\ufffd
Support Reactions:
+ 60A=0; By L1 w1 L1\ufffd \ufffd 0.5 L1\ufffd \ufffd\ufffd 0.5w2 L2\ufffd \ufffd L1 L23\ufffd§¨©
·
¹\ufffd 0=
By w1 L1\ufffd \ufffd 0.5( ) 0.5w2 L2\ufffd \ufffd 1 L23 L1\ufffd
§¨
·
¹
\ufffd\ufffd By 22.82 kN
+ 6Fy=0; Ay By\ufffd w1 L1\ufffd 0.5w2 L2\ufffd 0=
Ay By\ufffd w1 L1\ufffd 0.5w2 L2\ufffd\ufffd Ay 12.29 kN
Equations of Equilibrium: For point D
+ 6Fx=0; ND 0\ufffd Ans
+ 6Fy=0; Ay w1 a( )\ufffdª¬ º¼ VD\ufffd 0=
VD Ay w1 a( )\ufffd\ufffd VD 4.18 kN Ans
+ 60D=0; MD w1 a( )ª¬ º¼ 0.5 a( )\ufffd Ay a( )\ufffd 0=
MD w1 a( )ª¬ º¼\ufffd 0.5 a(```
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