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
bartleby

Concept explainers

Question
Book Icon
Chapter 7, Problem 36P
To determine

Calculate the deflection and the slopes on both sides of the hinge at B using conjugate beam method.

Expert Solution & Answer
Check Mark

Answer to Problem 36P

The slope on the right of hinge at B is θBR=83.3EI_.

The slope on the left of hinge at B is θBL=0_.

The deflection on the both side of the hinge at B is ΔB=0_.

Explanation of Solution

Given information:

Apply the sign conventions for calculating reactions using the three equations of equilibrium as shown below.

  • For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
  • For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
  • For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as positive and the counter clockwise moment as negative.

Sketch the Free Body Diagram of the beam as shown in Figure 1.

Fundamentals of Structural Analysis, Chapter 7, Problem 36P , additional homework tip  1

Refer to Figure 1.

Consider span DEF.

Use equilibrium equations:

Summation of moments about D is equal to 0.

MD=0RF×10+40×5=0RF=20 kN

Summation of forces along y-direction is equal to 0.

+Fy=0RDR40+20=0RDR=20 kN

The vertical reaction at the left of the hinge at D is RDL=20 kN().

Consider span BCD.

Use equilibrium equations:

Summation of moments about B is equal to 0.

MB=020×10RC×5=0RC=40 kN

Summation of forces along y-direction is equal to 0.

+Fy=0RBR+4020=0RBR=20kipsRBL=2020=0

The vertical reaction at the left of the hinge at B is RBL=0.

Consider span AB.

Summation of forces along y-direction is equal to 0.

+Fy=0RA=0

Summation of moments about A is equal to 0.

MA=0MA=0

Calculate the moment at each point as shown below.

MA=0MB=0MC=20×5=100 kNm

MD=20×10+40×5=0ME=20×15+40×10=100 kNmMC=20×20+40×1540×5=0

Sketch the MEI diagram as shown in Figure 2.

Fundamentals of Structural Analysis, Chapter 7, Problem 36P , additional homework tip  2

In the real beam A is the fixed end, F and C are the roller support, and B and D are the internal hinge but in conjugate-beam method the end conditions has to be change.

Change fixed end at A as free end, the roller at F as roller support, the roller at B as internal hinge, and the internal hinges at B and D as roller support.

Sketch the conjugate beam of the loading from M/EI diagram as shown in Figure 3.

Fundamentals of Structural Analysis, Chapter 7, Problem 36P , additional homework tip  3

Refer to Figure 3.

Consider span ABC.

Use equilibrium equations:

Summation of moments about C is equal to 0.

MC=0RB×512×5×100EI×(13×5)=0RB=83.3EI

Summation of forces along y-direction is equal to 0.

+Fy=0RCL+83.3EI12×5×100EI=0RCL=166.7EI

The vertical reaction at the right of the hinge at C is RCR=166.7EI().

Consider span CDEF.

Use equilibrium equations:

Summation of moments about F is equal to 0.

MF=0(166.7EI×15+RD×1012×5×100EI×(23×5+10)+12×5×100EI×(13×5+5)+12×5×100EI×(23×5))=0RD=333.3EI

Summation of forces along y-direction is equal to 0.

+Fy=0RF166.7+333.3EI12×5×100EI+12×10×100EI=0RF=416.7EI()

Calculate the shear (slope) of conjugate beam at each point as shown below.

VA=0VB=83.3EIVC=83.3EI12×5×100EI=166.7EI

V@ left of D=83.3EI12×10×100EI             =416.7EIVD=83.3EI12×10×100EI+333.3EI    =83.3EI

VE=83.3EI12×10×100EI+333.3EI+12×5×100EI   =166.7EIV@ left of E=83.3EI12×10×100EI+333.3EI+12×10×100EI            =416.7EIVE=83.3EI12×10×100EI+333.3EI+12×10×100EI416.7EI   =0

Calculate the moment (deflection) of conjugate beam at each point as shown below.

MA=0MB=0M@ left ofC=83.3EI×5             =416.7EIMC=83.3EI×512×5×100EI×(13×5)     =0

MD=83.3EI×1012×5×100EI×(13×5+5)12×5×100EI×(23×5)=1666.7EIME=(83.3EI×2012×5×100EI×(13×5+15)12×5×100EI×(23×5+10)+12×5×100EI×(13×5+5)+12×5×100EI×(23×5)+333.3EI×10)=0

Sketch the slope (shear) and deflection (moment) of the beam as shown in Figure 4.

Fundamentals of Structural Analysis, Chapter 7, Problem 36P , additional homework tip  4

Refer to Figure 4.

The slope on the right of hinge at B is θBR=83.3EI_.

The slope on the left of hinge at B is θBL=0_.

The deflection on the both side of the hinge at B is ΔB=0_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Knowledge Booster
Background pattern image
Civil Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Structural Analysis
Civil Engineering
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:Cengage,
Text book image
Structural Analysis (10th Edition)
Civil Engineering
ISBN:9780134610672
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Principles of Foundation Engineering (MindTap Cou...
Civil Engineering
ISBN:9781337705028
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Cengage Learning
Text book image
Fundamentals of Structural Analysis
Civil Engineering
ISBN:9780073398006
Author:Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher:McGraw-Hill Education
Text book image
Sustainable Energy
Civil Engineering
ISBN:9781337551663
Author:DUNLAP, Richard A.
Publisher:Cengage,
Text book image
Traffic and Highway Engineering
Civil Engineering
ISBN:9781305156241
Author:Garber, Nicholas J.
Publisher:Cengage Learning