Fundamentals of Structural Analysis
Fundamentals of Structural Analysis
5th Edition
ISBN: 9780073398006
Author: Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher: McGraw-Hill Education
Question
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Chapter 11, Problem 27P
To determine

Find the reactions and moments of the members in the frame.

Find the horizontal displacement of joint B.

Expert Solution & Answer
Check Mark

Answer to Problem 27P

Therefore, the horizontal displacement at point B is 0.71in._.

Explanation of Solution

Given information:

The value of Young’s modulus (E) is 3,000kips/in.2.

The moment of inertia (I) of the material is 1,500in.4.

Calculation:

Show the free-body diagram of the sway as in Figure 1.

Fundamentals of Structural Analysis, Chapter 11, Problem 27P , additional homework tip  1

Find the stiffness of each member connected to joint B as follows;

KBA=ILBA=I12KBC=ILBC=I16

Find the total stiffness at joint B as follows;

KB=KBA+KBC=I12+I16=7I48

Find the distribution factors at joint B using the relation;

DFBA=KBAKB=I127I48=47

DFBC=KBCKB=I167I48=37

Find the stiffness of each member connected to joint C as follows;

KCB=ILCB=I16KCD=ILCD=I16KCF=ILCF=1.5I12

Find the total stiffness at joint C as follows;

KC=KCB+KCD+KCF=I16+I16+1.5I12=4I16

Find the distribution factors at joint C using the relation;

DFCB=KCBKC=I164I16=14

DFCD=KCDKC=I164I16=14

DFCF=KCFKC=1.5I124I16=12

Find the stiffness of each member connected to joint D as follows;

KDC=ILDC=I16KDE=ILDE=I12

Find the total stiffness at joint D as follows;

KD=KDC+KDE=I16+I12=7I48

Find the distribution factors at joint D using the relation;

DFDC=KDCKD=I167I48=37

DFDE=KDEKD=I127I48=47

Find the fixed end moments in the member AB as follows:

FEMAB=6EIΔLAB2=6×3000×1500×1(12ft×12in.1ft)2=1,302.083kip-in.×1ft12in.=108.5kips-ft

FEMBA=6EIΔLBA2=6×3000×1500×1(12ft×12in.1ft)2=1,302.083kip-in.×1ft12in.=108.5kips-ft

Find the fixed end moments in the member DE as follows:

FEMDE=6EIΔLDE2=6×3000×1500×1(12ft×12in.1ft)2=1,302.083kip-in.×1ft12in.=108.5kips-ft

FEMED=6EIΔLED2=6×3000×1500×1(12ft×12in.1ft)2=1,302.083kip-in.×1ft12in.=108.5kips-ft

Find the fixed end moments in the member CF as follows:

FEMCF=6EIΔLCF2=6×3000×1.5×1500×1(12ft×12in.1ft)2=1,953.125kip-in.×1ft12in.=162.8kips-ft

FEMFC=6EIΔLFC2=6×3000×1.5×1500×1(12ft×12in.1ft)2=1,953.125kip-in.×1ft12in.=162.8kips-ft

Show the moment distribution computations as in Figure 2.

Fundamentals of Structural Analysis, Chapter 11, Problem 27P , additional homework tip  2

Show the free-body diagram of the members split up as in Figure 3.

Fundamentals of Structural Analysis, Chapter 11, Problem 27P , additional homework tip  3

Due to symmetry the vertical reaction at the joint C will be zero. Therefore consider the member BCD for the calculation purpose.

Consider member BCD:

Take moment about point B.

MBBCD=0VD(32)55.3101.355.3=0VD=6.6kips

Resolve the vertical component of forces.

Fy=0VB+VD=0VB+6.6=0VB=6.6kips

Consider the member AB:

Take moment about point B;

MBBA=08255.3+Ax(12)=0Ax=11.4kipsHB=Ax=11.4kips

Also, Ay=VB=6.6kips.

Consider the member CF:

Take moment about point C;

MCCF=0132101.3+Fx(12)=0Fx=19.4kipsHC=Fx=19.4kips

Consider the member DE:

Take moment about point D;

MDDE=08255.3+Ex(12)=0Ex=11.4kipsHD=Ex=11.4kips

Also, Ey=VD=6.6kips.

Find the value of using the relation:

S=HB+HC+HD=11.4+19.4+11.4=42.2kips()

Find the horizontal displacement at point B using the relation:

ΔHB=3042.2=0.71in.

Therefore, the horizontal displacement at point B is 0.71in._.

Multiply all the values in the Figure 3 by 0.71 in. to get the corrected reaction and moment values.

Show the final reaction and moment values as in Figure 4.

Fundamentals of Structural Analysis, Chapter 11, Problem 27P , additional homework tip  4

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