2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they current exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1-1. Determine the resultant internal normal force acting on the cross section through point A in each column. In (a), segment BC weighs 180 lb/ft and segment CD weighs 250 1b/ft. In (b), the column has a mass of 200 kg/m.
J5 kip
j8 kN
200 mm 200 mm 6kNj j6 kN A
a+2F,=0;
FA 1.0 3 3 - 1.8 5 = 0
Fa=13.8 kip
10'ft Ans. 8 in 3 kip
8in 200 mm 200 mm 3 kip 4.5 kN j4.5 kN V
b+F=0;
FA 4.5 4.5 5.89 6 6 8 = 0
Ans. 4 ft
Fa = 34.9 kN
4 ft
(er
(a)
(b)
1-2. Determine the resultant internal torque acting on the cross sections through points C and D.The support bearings at A and B allow free turning of the shaft.
0 N m
150 N m 400 N m
Mx=0;
Tc =250 = 0
Ans.
200 mm
Tc =250 Nm
150 mm 200mr
M=0; Tp=0
Ans.
50 mn
150 mm
1-3. Determine the resultant internal torque acting on the cross sections through points B and C.
600 lb ft 350 lb ft
M=0;
Tg + 350 - 500 = 0 Tg=150 lbft M=0; Tc = 500 = 0
Ans.
Tc =500 lbft
Ans.
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*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.
0.3 m
0.1'm
80 N
Equations of Equilibrium:
+7EF=0;
NA - 80 cos 15 = 0 NA = 77.3 N VA - 80 sin 15 = 0 VA = 20.7 N
Ans.
+F,,=0;
Ans.
:0=VWZ + e
MA + 80 cos 45(0.3 cos 30)
80 sin 45(0.1 + 0.3 sin 30) = 0
MA=0.555 Nm
Ans.
or
a+M=0;
MA + 80 sin 15(0.3 + 0.1 sin 30)
80 cos 15(0.1 cos 30) = 0
MA=-0.555 Nm
Aus
Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
BO
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1--5. Determine the resultant internal loadings in the beam at cross sections through points D and E. Point E is just to the right of the 3-kip load.
3 kip
1.5 kip/ft
F
Support Reactions: For member AB
a+Mp=