Lista 1 RESMAT Soluções

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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˜� 
TCD 400N m˜� 
TDB 300N m˜� 
Solution:
Equations of equilibrium:
+ TA TC� 0=
TC TA� 
TC 250 N m˜ Ans
+ TA TCD� TD� 0=
TD TCD TA�� 
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˜� 
TBC 350N m˜� 
TAB 600N m˜� 
Solution:
Equations of equilibrium:6 Mx = 0; TB TBC TD�� 0=
TB TBC� TD�� 
TB 150 N m˜ Ans
6 Mx = 0; TC TD� 0=
TC TD� 
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� T 30deg� I 45deg� 
a 0.3m� b 0.1m� 
Solution:
Equations of equilibrium:
 + 6Fx'=0; NA P cos I T�� �˜� 0=
NA P cos I T�� �˜� 
NA 77.27 N Ans
+ 6Fy'=0; VA P sin I T�� �˜� 0=
VA P sin I T�� �˜� 
VA 20.71 N Ans
+ 60A=0; MA P cos I� �˜ a˜ cos T� �� P sin I� �˜ b a sin T� ��� �˜� 0=
MA P� cos I� �˜ a˜ cos T� � P sin I� �˜ b a sin T� ��� �˜�� 
MA 0.555� 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˜� 
a 0.05m� b 0.3m� 
Solution:
Segment AB: Support Reactions
+ 60A=0; �ME By 2 a˜ b�( )˜� 0=
By
ME�
2a b�� By 175� N 
+At B: Bx By
150
200
§¨© ·¹˜� Bx 131.25� N 
Segment DB: NB Bx�� VB By�� 
+ 6Fx=0; ND NB� 0=
ND NB�� ND 131.25� N Ans
+ 6Fy=0; VD VB� 0=
VD VB�� VD 175� N Ans
+ 60D=0; MD� ME� By a b�( )˜� 0=
MD ME� By a b�( )˜�� 
MD 8.75� N m˜ Ans
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
load weigh 1500 N, determine the resultant internal loadings in the crane on cross sections through
points A, B, and C.
Given: P 1500N� w 750 N
m
� 
a 2.1m� b 1.5m� 
c 0.6m� d 2.4m� e 0.9m� 
Solution:
Equations of Equilibrium: For point A
+ 6Fx=0; NA 0� Ans
+
VA w e˜� P� 0=6Fy=0;
VA w e˜ P�� VA 2.17 kN Ans
+ 60A=0; MA� w e˜( ) 0.5 e˜( )˜� P e( )˜� 0=
MA w e˜( )� 0.5 e˜( )˜ P e( )˜�� MA 1.654� 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� Ans
+
VB w d e�( )˜� P� 0=6Fy=0;
VB w d e�( )˜ P�� VB 3.98 kN Ans
+ 60B=0; �MB w d e�( )˜[ ] 0.5 d e�( )˜[ ]˜� P d e�( )˜� 0=
MB w d e�( )˜[ ]� 0.5 d e�( )˜[ ]˜ P d e�( )˜�� 
MB 9.034� 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� Ans
+
NC� w b c� d� e�( )˜� P� 0=6Fy=0;
NC w� b c� d� e�( )˜ P�� 
NC 5.55� kN Ans
+ 60B=0; MC� w c d� e�( )˜[ ] 0.5 c d� e�( )˜[ ]˜� P c d� e�( )˜� 0=
MC w c d� e�( )˜[ ]� 0.5 c d� e�( )˜[ ]˜ P c d� e�( )˜�� 
MC 11.554� kN m˜ Ans
Note: Negative sign indicates that NC and MC acts in the opposite direction to that shown
on FBD.
Problem 1-10
The beam supports the distributed load shown. Determine the resultant internal loadings on the cross
section through point C. Assume the reactions at the supports A and B are vertical.
Given: w1 4.5
kN
m
� w2 6.0 kNm� 
a 1.8m� b 1.8m� c 2.4m� 
d 1.35m� e 1.35m� 
Solution: L1 a b� c�� 
L2 d e�� 
Support Reactions: 
+ 60A=0; By L1˜ w1 L1˜� � 0.5 L1˜� �� 0.5w2 L2˜� � L1 L23�§¨© ·¹˜� 0=
By w1 L1˜� � 0.5( )˜ 0.5w2 L2˜� � 1 L23 L1˜�§¨© ·¹˜�� By 22.82 kN 
+ 6Fy=0; Ay By� w1 L1˜� 0.5w2 L2˜� 0=
Ay By� w1 L1˜� 0.5w2 L2˜�� Ay 12.29 kN 
Equations of Equilibrium: For point C
+ 6Fx=0; NC 0� Ans
+ 6Fy=0; Ay w1 a b�( )˜�ª¬ º¼ VC� 0=
VC Ay w1 a b�( )˜�ª¬ º¼�� VC 3.92 kN Ans
+ 60C=0; MC w1 a b�( )˜ª¬ º¼ 0.5˜ a b�( )˜� Ay a b�( )˜� 0=
MC w1 a b�( )˜ª¬ º¼� 0.5˜ a b�( )˜ Ay a b�( )˜�� 
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
The beam supports the distributed load shown. Determine the resultant internal loadings on the cross
sections through points D and E. Assume the reactions at the supports A and B are vertical.
Given: w1 4.5
kN
m
� w2 6.0 kNm� 
a 1.8m� b 1.8m� c 2.4m� 
d 1.35m� e 1.35m� 
Solution: L1 a b� c�� 
L2 d e�� 
Support Reactions: 
+ 60A=0; By L1˜ w1 L1˜� � 0.5 L1˜� �� 0.5w2 L2˜� � L1 L23�§¨© ·¹˜� 0=
By w1 L1˜� � 0.5( )˜ 0.5w2 L2˜� � 1 L23 L1˜�§¨© ·¹˜�� By 22.82 kN 
+ 6Fy=0; Ay By� w1 L1˜� 0.5w2 L2˜� 0=
Ay By� w1 L1˜� 0.5w2 L2˜�� Ay 12.29 kN 
Equations of Equilibrium: For point D
+ 6Fx=0; ND 0� Ans
+ 6Fy=0; Ay w1 a( )˜�ª¬ º¼ VD� 0=
VD Ay w1 a( )˜�� VD 4.18 kN Ans
+ 60D=0; MD w1 a( )˜ª¬ º¼ 0.5˜ a( )˜� Ay a( )˜� 0=
MD w1 a( )˜ª¬ º¼� 0.5˜ a( )˜ Ay a( )˜�� MD 14.823 kN m˜ Ans
Equations of Equilibrium: For point E
+ 6Fx=0; NE 0� Ans
+ 6Fy=0; VE 0.5w2 0.5 e˜( )˜� 0=
VE 0.5w2 0.5 e˜( )˜� VE 2.03 kN Ans
+ 60D=0; ME� 0.5w2 0.5 e˜( )˜ª¬ º¼ e3§¨© ·¹˜� 0=
ME 0.5� w2 0.5 e˜( )˜ª¬ º¼ e3§¨© ·¹˜� ME 0.911� kN m˜ Ans
Note: Negative sign indicates that ME acts in the opposite direction to that shown on FBD.
Problem 1-17
Determine the resultant internal loadings acting on the cross section at point B.
Given: w 900
kN
m
� 
a 1m� b 4m� 
Solution: L a b�� 
Equations of Equilibrium: For point B
+ 6Fx=0; NB 0� 
NB 0 kN Ans
+ 6Fy=0; VB 0.5 w bL§¨© ·¹˜ª«¬ º»¼˜ b( )˜� 0=
VB 0.5 w
b
L
§¨© ·¹˜ª«¬ º»¼˜ b( )˜� 
VB 1440 kN Ans
+ 60B=0; MB� 0.5 w bL§¨© ·¹˜ª«¬ º»¼˜ b( )˜ b3§¨© ·¹˜� 0=
MB 0.5� w bL§¨© ·¹˜ª«¬ º»¼˜ b( )˜ b3˜� 
MB 1920� kN m˜ Ans
Problem 1-18
The beam supports the distributed load shown. Determine the resultant internal loadings acting on the
cross section through point C. Assume the reactions at the supports A and B are vertical.
Given: w1 0.5
kN
m
� a 3m� 
w2 1.5
kN
m
� 
Solution: L 3 a˜� w w2 w1�� 
Support Reactions: 
+ 60A=0; By L˜ w1 L˜� � 0.5 L˜( )� 0.5 w( ) L˜[ ] 2L3§¨© ·¹˜� 0=
By w1 L˜� � 0.5( )˜ 0.5 w( ) L˜[ ] 23§¨© ·¹˜�� 
By 5.25 kN 
+ 6Fy=0; Ay By� w1 L˜� 0.5 w( ) L˜� 0=
Ay By� w1 L˜� 0.5 w( ) L˜�� 
Ay 3.75 kN 
Equations of Equilibrium: For point C
+ 6Fx=0; NC 0� NC 0 kN Ans
+ 6Fy=0; VC w1 a˜� 0.5 w aL§¨© ·¹˜ª«¬ º»¼˜ a( )˜� Ay� 0=
VC w1� a˜ 0.5 w aL§¨© ·¹˜ª«¬ º»¼˜ a( )˜� Ay�� 
VC 1.75 kN Ans
+ 60C=0; MC w1 a˜� � 0.5 a˜( )� 0.5 w aL§¨© ·¹˜ª«¬ º»¼˜ a( )˜ a3§¨© ·¹˜� Ay a˜� 0=
MC w1� a˜� � 0.5 a˜( ) 0.5 w aL§¨© ·¹˜ª«¬ º»¼˜ a( )˜ a3§¨© ·¹˜� Ay a˜�� 
MC 8.5 kN m˜ Ans
Problem 1-19
Determine the resultant internal loadings acting on the cross section through point D in Prob. 1-18.
Given: w1 0.5
kN
m
� a 3m� 
w2 1.5
kN
m
� 
Solution: L 3 a˜� w w2 w1�� 
Support Reactions: 
+ 60A=0; By L˜ w1 L˜� � 0.5 L˜( )� 0.5 w( ) L˜[ ] 2L3§¨© ·¹˜� 0=
By w1 L˜� � 0.5( )˜ 0.5 w( ) L˜[ ] 23§¨© ·¹˜�� 
By 5.25 kN 
+ 6Fy=0; Ay By� w1 L˜� 0.5 w( ) L˜� 0=
Ay By� w1 L˜� 0.5 w( ) L˜�� 
Ay 3.75 kN 
Equations of Equilibrium: For point D
+ 6Fx=0; ND 0� ND 0 kN Ans
+ 6Fy=0; VD w1 2a( )˜� 0.5 w 2aL§¨© ·¹˜ª«¬ º»¼˜ 2a( )˜� Ay� 0=
VD w1� 2a( )˜ 0.5 w 2 a˜L§¨© ·¹˜ª«¬ º»¼˜ 2a( )˜� Ay�� 
VD 1.25� kN Ans
+ 60D=0; MD w1 2a( )˜ª¬ º¼ a( )� 0.5 w 2 a˜L§¨© ·¹˜ª«¬ º»¼˜ 2a( )˜ 2a3§¨© ·¹˜� Ay 2a( )˜� 0=
MD w1� 2a( )˜ª¬ º¼ a( ) 0.5 w 2 a˜L§¨© ·¹˜ª«¬ º»¼˜ 2a( )˜ 2a3§¨© ·¹˜� Ay 2a( )˜�� 
MD 9.5 kN m˜ Ans
Problem 1-20
The wishbone construction of the power pole supports the three lines, each exerting a force of 4 kN
on the bracing struts. If the struts are pin connected at A, B, and C, determine the resultant internal
loadings at cross sections through points D, E, and F.
Given: P 4kN� a 1.2m� b 1.8m� 
Solution:
Support Reactions: FBD (a) and (b).
Given
+ 60A=0; By a( )˜ Bx0.5 b˜( )˜� P a( )˜� 0= [1]
+ 60C=0; Bx 0.5 b˜( )˜ P a( )˜� By a( )˜� P a( )˜� 0= [2]
Solving [1] and [2]: Initial guess: Bx 1kN� By 2kN� 
Bx
By
§¨¨© ·¹ Find Bx By�� �� BxBy§¨¨© ·¹ 2.672§¨© ·¹ kN 
From FBD (a):
+ 6Fx=0; Bx Ax� 0=
Ax Bx� Ax 2.67 kN 
+ 6Fy=0; Ay P� By� 0=
Ay P By�� Ay 6 kN 
From FBD (b):
+ 6Fx=0; Cx Bx� 0=
Cx Bx� Cx 2.67 kN 
+ 6Fy=0; Cy By� P� P� 0=
Cy 2P By�� Cy 6 kN 
Equations of Equilibrium: For point D [FBD (c)].
+ 6Fx=0; VD 0� VD 0 kN Ans
+ 6Fy=0; ND 0� ND 0 kN Ans
+ 60D=0; MD 0� MD 0 kN m˜ Ans
For point E [FBD (d)].
+ 6Fx=0; VF Ax Cx�� 0= VF Ax� Cx�� VF 0 kN Ans
+ 6Fy=0; NF Ay Cy�˜� 0= NF Ay Cy�� NF 12 kN Ans
+ 60F=0; MF Ax Cx�� � 0.5 b˜( )˜� 0= MF Ax Cx�� � 0.5 b˜( )˜� MF 4.8 kN m˜ Ans
+ 6Fx=0; Ax VE� 0= VE Ax� 
+ 6Fy=0; NE Ay� 0= NE Ay� 
+ 60E=0; ME Ax 0.5 b˜( )˜� 0= ME Ax 0.5 b˜( )˜� ME 2.4 kN m˜ Ans
For point F [FBD (e)].
VE 2.67 kN Ans
NE 6 kN Ans
Problem 1-23
The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings
acting on the cross section at B. Neglect the weight of the wrench CD.
Given: g 9.81
m
s
2
� U 12 kg
m
� P 60N� 
a 0.150m� b 0.400m� 
c 0.200m� d 0.300m� 
Solution: w U g˜� 
6Fx=0; NBx 0N� Ans
6Fy=0; VBy 0N� Ans
6Fz=0; VBz P� P� w b c�( )˜� 0=
VBz P P� w b c�( )˜�� 
VBz 70.6 N Ans
60x=0; TBx P b( )˜� P b( )˜� w b˜( ) 0.5 b˜( )˜� 0=
TBx P� b( )˜ P b( )˜� w b˜( ) 0.5 b˜( )˜�� TBx 9.42 N m˜ Ans
60y=0; MBy P 2a( )˜� w b˜( ) c( )˜� w c˜( ) 0.5 c˜( )˜� 0=
MBy P 2 a˜( )˜ w b˜( ) c( )˜� w c˜( ) 0.5 c˜( )˜�� MBy 6.23 N m˜ Ans
60z=0; MBz 0N m˜� Ans
Problem 1-24
The main beam AB supports the load on the wing of the airplane. The loads consist of the wheel
reaction of 175 kN at C, the 6-kN weight of fuel in the tank of the wing, having a center of gravity at
D, and the 2-kN weight of the wing, having a center of gravity at E. If it is fixed to the fuselage at A,
determine the resultant internal loadings on the beam at this point. Assume that the wing does not
transfer any of the loads to the fuselage, except through the beam.
Given: PC 175kN� PE 2kN� PD 6kN� 
a 1.8m� b 1.2m� e 0.3m� 
c 0.6m� d 0.45m� 
Solution:
6Fx=0; VAx 0kN� Ans
6Fy=0; NAy 0kN� Ans
6Fz=0; VAz PD� PE� PC� 0=
VAz PD PE PC��� 
VAz 167� kN Ans
60x=0;
MAx PD a( )˜ PE a b� c�( )˜� PC a b�( )˜�� MAx 507� kN m˜ Ans
60y=0; TAy PD d( )˜� PE e( )˜� 0=
TAy PD� d( )˜ PE e( )˜�� TAy 2.1� kN m˜ Ans
60z=0; MAz 0kN m˜� Ans
MAx PD a( )˜� PE a b� c�( )˜� PC a b�( )˜� 0=
Problem 1-25
Determine the resultant internal loadings acting on the cross section through point B of the signpost.
The post is fixed to the ground and a uniform pressure of 50 N/m2 acts perpendicular to the face of the
sign.
Given: a 4m� d 2m� 
p 50
N
m
2
� 
b 6m� e 3m� 
c 3m� 
Solution: P p c( )˜ d e�( )˜� 
6Fx=0; VBx P� 0=
VBx P� 
VBx 750 N Ans
6Fy=0; VBy 0N� Ans
6Fz=0; NBz 0N� Ans
60x=0; MBx 0N m˜� Ans
60y=0; MBy P b 0.5 c˜�( )˜� 0=
MBy P b 0.5 c˜�( )˜� 
MBy 5625 N m˜ Ans
60z=0; TBz P e 0.5 d e�( )˜�[ ]˜� 0=
TBz P e 0.5 d e�( )˜�[ ]˜� 
TBz 375 N m˜ Ans
Problem 1-26
The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the
pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section through
point D. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y
direction. The journal bearings at A and B exert only y and z components of force on the shaft.
Given: P1z 400N� P2y 200N� P3y 80N� 
a 0.3m� b 0.4m� c 0.3m� d 0.4m� 
Solution: L a b� c� d�� 
Support Reactions:60z=0; 2P3y d( )˜ 2P2y c d�( )˜� Ay L( )˜� 0=
Ay 2P3y
d
L
˜ 2P2y c d�L˜�� 
Ay 245.71 N 6Fy=0; Ay� By� 2 P2y˜� 2 P3y˜� 0=
By Ay� 2 P2y˜� 2 P3y˜�� By 314.29 N 
60y=0; 2P1z b c� d�( )˜ Az L( )˜� 0=
Az 2P1z
b c� d�
L
˜� Az 628.57 N 6Fz=0; Bz Az� 2 P1z˜� 0=
Bz Az� 2 P1z˜�� Bz 171.43 N 
Equations of Equilibrium: For point D.6Fx=0; NDx 0N� Ans6Fy=0; VDy By� 2 P3y˜� 0=
VDy By 2 P3y˜�� 
VDy 154.3 N Ans6Fz=0; VDz Bz� 0=
VDz Bz�� 
VDz 171.4� N Ans60x=0; TDx 0N m˜� Ans60y=0; MDy Bz d 0.5 c˜�( )˜� 0=
MDy Bz d 0.5 c˜�( )˜ª¬ º¼�� MDy 94.29� N m˜ Ans60z=0; MDz By d 0.5 c˜�( )˜� 2 P3y˜ 0.5 c˜( )˜� 0=
MDz By� d 0.5 c˜�( )˜ 2 P3y˜ 0.5 c˜( )˜�� MDz 148.86� N m˜ Ans
Problem 1-27
The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the
pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section through
point C. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y
direction. The journal bearings at A and B exert only y and z components of force on the shaft.
Given: P1z 400N� P2y 200N� P3y 80N� 
a 0.3m� b 0.4m� c 0.3m� d 0.4m� 
Solution: L a b� c� d�� 
Support Reactions:60z=0; 2P3y d( )˜ 2P2y c d�( )˜� Ay L( )˜� 0=
Ay 2P3y
d
L
˜ 2P2y c d�L˜�� Ay 245.71 N 6Fy=0; Ay� By� 2 P2y˜� 2 P3y˜� 0=
By Ay� 2 P2y˜� 2 P3y˜�� By 314.29 N 
60y=0; 2P1z b c� d�( )˜ Az L( )˜� 0=
Az 2P1z
b c� d�
L
˜� Az 628.57 N 6Fz=0; Bz Az� 2 P1z˜� 0=
Bz Az� 2 P1z˜�� Bz 171.43 N 
Equations of Equilibrium: For point C.
6Fx=0; NCx 0N� Ans6Fy=0; VCy Ay� 0=
VCy Ay� 
VCy 245.7 N Ans6Fz=0; VCz Az� 2P1z� 0=
VCz Az� 2P1z�� 
VCz 171.4 N Ans
60x=0; TCx 0N m˜� Ans60y=0; MCy Az a 0.5 b˜�( )˜� 2 P1z˜ 0.5 b˜( )˜� 0=
MCy Az a 0.5 b˜�( )˜ 2 P1z˜ 0.5 b˜( )˜�� MCy 154.29 N m˜ Ans60z=0; MCz Ay a 0.5 b˜�( )˜� 0=
MCz Ay� a 0.5 b˜�( )˜� MCz 122.86� N m˜ Ans
Problem 1-30
The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings
acting on the cross section through B.
Given: P 750N� MC 800N m˜� 
U 12 kg
m
� g 9.81 m
s
2
� 
a 1m� b 2m� c 2m� 
Solution:
Py
4
5
§¨© ·¹ P˜� Pz 35� P˜� 
Equations of Equilibrium: For point B.
6Fx=0; VBx 0kip� Ans
6Fy=0; NBy Py� 0=
NBy Py�� Ans
NBy 600� N Ans
6Fz=0; VBz Pz� U g˜ c˜� U g˜ b˜� 0=
VBz Pz� U g˜ c˜� U g˜ b˜�� 
VBz 920.9 N Ans60x=0; MBx Pz b( )˜� U g˜ c˜ b( )˜� U g˜ b˜ 0.5 b˜( )˜� 0=
MBx Pz� b( )˜ U g˜ c˜ b( )˜� U g˜ b˜ 0.5 b˜( )˜�� 
MBx 1606.3 N m˜ Ans60y=0; TBy 0N m˜� Ans
60z=0; MBy Mc� 0=
MBy MC�� 
MBy 800� N m˜ Ans
Problem 1-31
The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadings
acting on the cross section through A which is located at an angle T from the horizontal.
Solution: P kN� T deg� 
Equations of Equilibrium: For point A.
+ 6Fx=0; NA� P cos T� �˜� 0=
NA P cos T� �˜� Ans
 + 6Fy=0; VA P sin T� �˜� 0=
VA P sin T� �˜� Ans
+ 60A=0; MA P r˜ 1 cos T� ��� �˜� 0=
MA P r˜ 1 cos T� ��� �˜� r Ans