A single-bay frame of the type illustrated in Figure 9.5 carries a horizontal load of 5000 lbacting at the upper-left joint. Assume that h = 15 ft and L = 25 ft. Draw shear andmoment diagrams. Indicate numerical values. Use an approximate method of analysis. FIGURE 9.5 Simplified analysis of a single-bay rigid frame carrying a lateral load.(a) Deflected shape of frameB0.45hMAM0.55hRAIRON=MCD (P/2)(0.45h)(b) Free-body diagram for partsof structure separated at pointsof inflection (points of knownzero moment).робраRAVMac (0.45Ph/L)(L/2)= 0.225PhP/2P/20.45Ph/L 0.45Ph/LRovMCB (0.45Ph/L)(L/2)= 0.225Ph0.45Ph/L 0.45Ph/LP/2P/20.45Ph/L0.45Ph/LP/2P/2P/2P/2M = (0.55h)(P/2)= 0.275Ph0.45Ph/L 0.45Ph/L(c) Free-body of individual beam columnand joint elements. Each element mustbe in a state of translatory and rotationalequilibrium.(d) Final moment diagram. Shear and axialforces are present in the members as wellas the moments indicated.MBC= 0.225PhMCB=0.225PhMBA0.225PhMCD = 0.225PhMA=0.275PhM=0.275Ph= 0.225Ph

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Answered: A single-bay frame of the type…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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A single-bay frame of the type illustrated in Figure 9.5 carries a horizontal load of 5000 lb
acting at the upper-left joint. Assume that h = 15 ft and L = 25 ft. Draw shear and
moment diagrams. Indicate numerical values. Use an approximate method of analysis.

FIGURE 9.5 Simplified analysis of a single-bay rigid frame carrying a lateral load.
(a) Deflected shape of frame
B
0.45h
MA
M
0.55h
RA
IRON
=
MCD (P/2)(0.45h)
(b) Free-body diagram for parts
of structure separated at points
of inflection (points of known
zero moment).
робра
RAV
Mac (0.45Ph/L)(L/2)
= 0.225Ph
P/2
P/2
0.45Ph/L 0.45Ph/L
Rov
MCB (0.45Ph/L)(L/2)
= 0.225Ph
0.45Ph/L 0.45Ph/L
P/2
P/2
0.45Ph/L
0.45Ph/L
P/2
P/2
P/2
P/2
M = (0.55h)(P/2)
= 0.275Ph
0.45Ph/L 0.45Ph/L
(c) Free-body of individual beam column
and joint elements. Each element must
be in a state of translatory and rotational
equilibrium.
(d) Final moment diagram. Shear and axial
forces are present in the members as well
as the moments indicated.
MBC= 0.225Ph
MCB=0.225Ph
MBA
0.225Ph
MCD = 0.225Ph
MA=0.275Ph
M=0.275Ph
= 0.225Ph

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