617
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
SOLUTION
Support Reactions. Referring to the FBD of the entire beam shown in Fig. a,
a+ΣMA = 0; By(12) -
1
2
(4)(6)(4) = 0 By = 4 .00 kip
Internal Loadings. Referring to the FBD of the right segment of the beam sectioned
through C, Fig. b,
+ c ΣFy = 0; VC + 4 .00 = 0 VC = -4 .00 kip Ans.
a+ΣMC = 0; 4 .00(6) - MC = 0 MC = 24 .0 kip # ft Ans.
7–7.
Determine the internal shear force and moment acting at
point C in the beam.
6 ft 6 ft
4 kip/ft
A B
C
Ans:
VC = -4 .00 kip
MC = 24 .0 kip # ft
629
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
VC = 2.49 kN
NC = 2.49 kN
MC = 4 .97 kN # m
ND = 0
VD = -2.49 kN
MD = 16.5 kN # m
7–19.
SOLUTION
Entire beam:
a
Segment AC:
Ans.
Ans.
a
Ans.MC = 4.72 kip # ft
+©MC = 0; -1.8 (3) + 0.45 (1.5) + MC = 0
VC = 1.35 kip
+ c©Fy = 0; 1.8 - 0.45 - VC = 0
NC = -4.32 kip
:+ ©Fx = 0; 4.32 + NC = 0
Ay = 1.8 kip
+ c©Fy = 0; Ay - 1.8 = 0
Ax = 4.32 kip
:+ ©Fx = 0; Ax - 4.32 = 0
T = 4.32 kip
+©MA = 0; -1.8 (6) + T (2.5) = 0
Determine the internal normal force, shear force, and
moment at point C.
8 ft
3 ft
4 ft
150 lb/ ft
2 ft
0.5 ft
A
C
B
Ans:
NC = -4 .32 kip
VC = 1.35 kip
MC = 4 .72 kip # ft
jbar
0
