Soluções - Resistência dos Materiais - HIBBELER 7ª Ed

Soluções - Resistência dos Materiais - HIBBELER 7ª Ed

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Stamp
CD1
4/13 US
 ISM for
\ufffd
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:
+\u2191\u3a3 Fy = 0; FA wBC g\u22c5( ) LBC\u22c5\u2212 wCA g\u22c5( ) LCA\u22c5\u2212 FB\u2212 2FC\u2212 0=
FA wBC g\u22c5( ) LBC\u22c5 wCA g\u22c5( ) LCA\u22c5+ 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:
+\u2191\u3a3 Fy = 0; FA w L\u22c5( ) g\u22c5\u2212 FB\u2212 2F1\u2212 2F2\u2212 0=
FA w L\u22c5( ) g\u22c5 FB+ 2F1+ 2F2+:=
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\u22c5:=
TCD 400N m\u22c5:=
TDB 300N m\u22c5:=
Solution:
Equations of equilibrium:
+ TA TC\u2212 0=
TC TA:=
TC 250 N m\u22c5= Ans
+ TA TCD\u2212 TD+ 0=
TD TCD TA\u2212:=
TD 150 N m\u22c5= Ans
Problem 1-3
Determine the resultant internal torque acting on the cross sections through points B and C.
Given: TD 500N m\u22c5:=
TBC 350N m\u22c5:=
TAB 600N m\u22c5:=
Solution:
Equations of equilibrium:
\u3a3 Mx = 0; TB TBC TD\u2212+ 0=
TB TBC\u2212 TD+:=
TB 150 N m\u22c5= Ans
\u3a3 Mx = 0; TC TD\u2212 0=
TC TD:=
TC 500 N m\u22c5= 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:=
\u3b8 30deg:= \u3c6 45deg:=
a 0.3m:= b 0.1m:=
Solution:
Equations of equilibrium:
 + \u3a3Fx'=0; NA P cos \u3c6 \u3b8\u2212( )\u22c5\u2212 0=
NA P cos \u3c6 \u3b8\u2212( )\u22c5:=
NA 77.27 N= Ans
+ \u3a3Fy'=0; VA P sin \u3c6 \u3b8\u2212( )\u22c5\u2212 0=
VA P sin \u3c6 \u3b8\u2212( )\u22c5:=
VA 20.71 N= Ans
+ \u3a3\u39cA=0; MA P cos \u3c6( )\u22c5 a\u22c5 cos \u3b8( )+ P sin \u3c6( )\u22c5 b a sin \u3b8( )+( )\u22c5\u2212 0=
MA P\u2212 cos \u3c6( )\u22c5 a\u22c5 cos \u3b8( ) P sin \u3c6( )\u22c5 b a sin \u3b8( )+( )\u22c5+:=
MA 0.555\u2212 N m\u22c5= 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\u22c5:=
a 0.05m:= b 0.3m:=
Solution:
Segment AB: Support Reactions
+ \u3a3\u39cA=0; \u2212ME By 2 a\u22c5 b+( )\u22c5\u2212 0=
By
ME\u2212
2a b+:= By 175\u2212 N=
+At B: Bx By
150
200
\u239b\u239c\u239d
\u239e
\u23a0\u22c5:= Bx 131.25\u2212 N=
Segment DB: NB Bx\u2212:= VB By\u2212:=
+ \u3a3Fx=0; ND NB+ 0=
ND NB\u2212:= ND 131.25\u2212 N= Ans
+ \u3a3Fy=0; VD VB+ 0=
VD VB\u2212:= VD 175\u2212 N= Ans
+ \u3a3\u39cD=0; MD\u2212 ME\u2212 By a b+( )\u22c5\u2212 0=
MD ME\u2212 By a b+( )\u22c5\u2212:=
MD 8.75\u2212 N m\u22c5= 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:
\u3b8 atan b
d
\u239b\u239c\u239d
\u239e
\u23a0:= \u3b8 36.87 deg=
\u3c6 atan a b+
d
\u239b\u239c\u239d
\u239e
\u23a0 \u3b8\u2212:= \u3c6 14.47 deg=
Member AB: 
+ \u3a3\u39cA=0; FBC sin \u3c6( )\u22c5 a b+( )\u22c5 P b( )\u22c5\u2212 0=
FBC
P b( )\u22c5
sin \u3c6( ) a b+( )\u22c5:=
FBC 12.01 kN=
Segment BD: 
 + \u3a3Fx=0; ND\u2212 FBC cos \u3c6( )\u22c5\u2212 P cos \u3b8( )\u22c5\u2212 0=
ND FBC\u2212 cos \u3c6( )\u22c5 P cos \u3b8( )\u22c5\u2212:=
ND 15.63\u2212 kN= Ans
+ \u3a3Fy=0; VD FBC sin \u3c6( )\u22c5+ P sin \u3b8( )\u22c5\u2212 0=
VD FBC\u2212 sin \u3c6( )\u22c5 P sin \u3b8( )\u22c5+:=
VD 0 kN= Ans
+ \u3a3\u39cD=0; FBC sin \u3c6( )\u22c5 P sin \u3b8( )\u22c5\u2212( ) d c\u2212sin \u3b8( )\u22c5 MD\u2212 0=
MD FBC sin \u3c6( )\u22c5 P sin \u3b8( )\u22c5\u2212( ) d c\u2212sin \u3b8( )\u22c5:=
MD 0 kN m\u22c5= 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:
\u3b8 atan b
d
\u239b\u239c\u239d
\u239e
\u23a0:= \u3b8 36.87 deg=
\u3c6 atan a b+
d
\u239b\u239c\u239d
\u239e
\u23a0 \u3b8\u2212:= \u3c6 14.47 deg=
Member AB: 
+ \u3a3\u39cA=0; FBC sin \u3c6( )\u22c5 a b+( )\u22c5 P b( )\u22c5\u2212 0=
FBC
P b( )\u22c5
sin \u3c6( ) a b+( )\u22c5:=
FBC 12.01 kN=
Segment BE: 
 + \u3a3Fx=0; NE\u2212 FBC cos \u3c6( )\u22c5\u2212 P cos \u3b8( )\u22c5\u2212 0=
NE FBC\u2212 cos \u3c6( )\u22c5 P cos \u3b8( )\u22c5\u2212:=
NE 15.63\u2212 kN= Ans
+ \u3a3Fy=0; VE FBC sin \u3c6( )\u22c5+ P sin \u3b8( )\u22c5\u2212 0=
VE FBC\u2212 sin \u3c6( )\u22c5 P sin \u3b8( )\u22c5+:=
VE 0 kN= Ans
+ \u3a3\u39cE=0; FBC sin \u3c6( )\u22c5 P sin \u3b8( )\u22c5\u2212( ) e\u22c5 ME\u2212 0=
ME FBC sin \u3c6( )\u22c5 P sin \u3b8( )\u22c5\u2212( ) e\u22c5:=
ME 0 kN m\u22c5= 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
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
+ \u3a3Fx=0; NA 0:= Ans
+
VA w e\u22c5\u2212 P\u2212 0=\u3a3Fy=0;
VA w e\u22c5 P+:= VA 2.17 kN= Ans
+ \u3a3\u39cA=0; MA\u2212 w e\u22c5( ) 0.5 e\u22c5( )\u22c5\u2212 P e( )\u22c5\u2212 0=
MA w e\u22c5( )\u2212 0.5 e\u22c5( )\u22c5 P e( )\u22c5\u2212:= MA 1.654\u2212 kN m\u22c5= Ans
Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point B
+ \u3a3Fx=0; NB 0:= Ans
+
VB w d e+( )\u22c5\u2212 P\u2212 0=\u3a3Fy=0;
VB w d e+( )\u22c5 P+:= VB 3.98 kN= Ans
+ \u3a3\u39cB=0; \u2212MB w d e+( )\u22c5[ ] 0.5 d e+( )\u22c5[ ]\u22c5\u2212 P d e+( )\u22c5\u2212 0=
MB w d e+( )\u22c5[ ]\u2212 0.5 d e+( )\u22c5[ ]\u22c5 P d e+( )\u22c5\u2212:=
MB 9.034\u2212 kN m\u22c5= Ans
Note: Negative sign indicates that MB acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point C
+ \u3a3Fx=0; VC 0:= Ans
+
NC\u2212 w b c+ d+ e+( )\u22c5\u2212 P\u2212 0=\u3a3Fy=0;
NC w\u2212 b c+ d+ e+( )\u22c5 P\u2212:=
NC 5.55\u2212 kN= Ans
+ \u3a3\u39cB=0; MC\u2212 w c d+ e+( )\u22c5[ ] 0.5 c d+ e+( )\u22c5[ ]\u22c5\u2212 P c d+ e+( )\u22c5\u2212 0=
MC w c d+ e+( )\u22c5[ ]\u2212 0.5 c d+ e+( )\u22c5[ ]\u22c5 P c d+ e+( )\u22c5\u2212:=
MC 11.554\u2212 kN m\u22c5= 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:=
\u3b8 30deg:= \u3c6 45deg:=
a 4mm:= b 5.75mm:=
Solution: \u3b1 \u3c6 \u3b8\u2212:=
Equations of equilibrium: For section a -a
 + \u3a3Fx'=0; VA P cos \u3b1( )\u22c5\u2212 0=
VA P cos \u3b1( )\u22c5:=
VA 386.37 N= Ans
+ \u3a3Fy'=0; NA sin \u3b1( )\u2212 0=
NA P sin \u3b1( )\u22c5:=
NA 103.53 N= Ans
+ \u3a3\u39cA=0; MA\u2212 P sin \u3b1( )\u22c5 a\u22c5\u2212 P cos \u3b1( )\u22c5 b\u22c5+ 0=
MA P\u2212 sin \u3b1( )\u22c5 a\u22c5 P cos \u3b1( )\u22c5 b\u22c5+:=
MA 1.808 N m\u22c5= Ans
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
kN
m
:=
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: 
+ \u3a3\u39cA=0; By L1\u22c5 w1 L1\u22c5( ) 0.5 L1\u22c5( )\u2212 0.5w2 L2\u22c5( ) L1 L23+\u239b\u239c\u239d
\u239e
\u23a0\u22c5\u2212 0=
By w1 L1\u22c5( ) 0.5( )\u22c5 0.5w2 L2\u22c5( ) 1 L23 L1\u22c5+
\u239b\u239c\u239d
\u239e
\u23a0
\u22c5+:= By 22.82 kN=
+ \u3a3Fy=0; Ay By+ w1 L1\u22c5\u2212 0.5w2 L2\u22c5\u2212 0=
Ay By\u2212 w1 L1\u22c5+ 0.5w2 L2\u22c5+:= Ay 12.29 kN=
Equations of Equilibrium: For point C 
+ \u3a3Fx=0; NC 0:= Ans
+ \u3a3Fy=0; Ay w1 a b+( )\u22c5\u2212\u23a1\u23a3 \u23a4\u23a6 VC\u2212 0=
VC Ay w1 a b+( )\u22c5\u2212\u23a1\u23a3 \u23a4\u23a6\u2212:= VC 3.92 kN= Ans
+ \u3a3\u39cC=0; MC w1 a b+( )\u22c5\u23a1\u23a3 \u23a4\u23a6 0.5\u22c5 a b+( )\u22c5+ Ay a b+( )\u22c5\u2212 0=
MC w1 a b+( )\u22c5\u23a1\u23a3 \u23a4\u23a6\u2212 0.5\u22c5 a b+( )\u22c5 Ay a b+( )\u22c5+:=
MC 15.07 kN m\u22c5= 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
kN
m
:=
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: 
+ \u3a3\u39cA=0; By L1\u22c5 w1 L1\u22c5( ) 0.5 L1\u22c5( )\u2212 0.5w2 L2\u22c5( ) L1 L23+\u239b\u239c\u239d
\u239e
\u23a0\u22c5\u2212 0=
By w1 L1\u22c5( ) 0.5( )\u22c5 0.5w2 L2\u22c5( ) 1 L23 L1\u22c5+
\u239b\u239c\u239d
\u239e
\u23a0
\u22c5+:= By 22.82 kN=
+ \u3a3Fy=0; Ay By+ w1 L1\u22c5\u2212 0.5w2 L2\u22c5\u2212 0=
Ay By\u2212 w1 L1\u22c5+ 0.5w2 L2\u22c5+:= Ay 12.29 kN=
Equations of Equilibrium: For point D 
+ \u3a3Fx=0; ND 0:= Ans
+ \u3a3Fy=0; Ay w1 a( )\u22c5\u2212\u23a1\u23a3 \u23a4\u23a6 VD\u2212 0=
VD Ay w1 a( )\u22c5\u2212:= VD 4.18 kN= Ans
+ \u3a3\u39cD=0; MD w1 a( )\u22c5\u23a1\u23a3 \u23a4\u23a6 0.5\u22c5 a( )\u22c5+ Ay