2. Cantilever Method: The frame shown is subjected to lateral loads as shown. Supports S, T, U V, W and X are rigid supports. Column HMS (400mm x 400 mm) Column ADINT (500 mm x 600 mm) Column BEJOU (600 mm x 600 mm) Column CFKPV (500 mm x 600 mm) Column GLQW (500 mm x 500 mm) Column RX (300mm x 300 mm) Determine the following. a) The shear forces in columns and girders. b) The axial forces in columns and girders. c) The moment at the end of each member. d) Summarize your results by drawing the frame with the internal forces. 5.0 m 5.50 m 6.0 m 7.0m 120 KN 96 KN (100+SN) KN > 144 KN M B E U THE VALUE OF SN IS 34 ROUND OFF THE ANSWER TO 3 DECIMAL PLACES G W 5.0m R 4.0 m 4.50 m 4.20 m 5.0 m

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Author:Segui, William T.
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Chapter6: Beam–columns
Section: Chapter Questions
Problem 6.8.9P
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2. Cantilever Method: The frame shown is subjected to lateral loads as shown. Supports S, T, U V, W and
X are rigid supports. Column HMS (400mm x 400 mm)
Column ADINT (500 mm x 600 mm)
Column BEJOU (600 mm x 600 mm)
Column CFKPV (500 mm x 600 mm)
Column GLQW (500 mm x 500 mm)
Column RX (300mm x 300 mm)
Determine the following.
a) The shear forces in columns and girders.
b) The axial forces in columns and girders.
c) The moment at the end of each member.
d) Summarize your results by drawing the frame with the internal forces.
5.0 m
5.50 m
6.0 m
7.0 m
5.0 m
(1bo+SN) KN >
144 KN
E
120 KN
K
_96 IN
M
N
R.
THE VALUE OF SN IS 34
ROUND OFF THE ANSWER TO 3 DECIMAL PLACES
5.0 m
4.50 m
4.20 m 4.0 m
Transcribed Image Text:2. Cantilever Method: The frame shown is subjected to lateral loads as shown. Supports S, T, U V, W and X are rigid supports. Column HMS (400mm x 400 mm) Column ADINT (500 mm x 600 mm) Column BEJOU (600 mm x 600 mm) Column CFKPV (500 mm x 600 mm) Column GLQW (500 mm x 500 mm) Column RX (300mm x 300 mm) Determine the following. a) The shear forces in columns and girders. b) The axial forces in columns and girders. c) The moment at the end of each member. d) Summarize your results by drawing the frame with the internal forces. 5.0 m 5.50 m 6.0 m 7.0 m 5.0 m (1bo+SN) KN > 144 KN E 120 KN K _96 IN M N R. THE VALUE OF SN IS 34 ROUND OFF THE ANSWER TO 3 DECIMAL PLACES 5.0 m 4.50 m 4.20 m 4.0 m
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