Structural Analysis
Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Chapter 13, Problem 1P
To determine

Find the reactions and sketch the shear and bending moment diagrams for the given beam using method of consistent deformation.

Expert Solution & Answer
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Answer to Problem 1P

The reactions acting on the beam are Ax=0_, Ay=99.26kN()_, Dy=60.74kN()_, and MA=233.34kNm_.

Explanation of Solution

Given information:

Apply the sign conventions for calculating reactions, forces, and moments 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 negative and the counter clockwise moment as positive.

Calculation:

Show the free body diagram of the given beam as shown in Figure 1.

Structural Analysis, Chapter 13, Problem 1P , additional homework tip  1

Refer Figure 1.

Consider the horizontal and vertical reaction at A are denoted by Ax and Ay.

Consider the moment at A is denoted by MA.

Consider the vertical reaction at D is denoted by Dy.

The reactions acting in the beam is 4.

The number of Equilibrium reaction is 3. Then,

The degree of indeterminacy of the beam is 1.

Take the vertical reaction at D as the redundant.

Modify the Figure 1 as shown in Figure 2.

Structural Analysis, Chapter 13, Problem 1P , additional homework tip  2

Refer Figure 2.

The deflection at D due to the external moment is denoted by ΔD.

The deflection coefficient representing the deflection at D due to unit value of redundant Dy is fDD.

The deflection at D due to unknown redundant Dy is fDDDy.

Show the deflection at the free end of the cantilever beam with a point load P acting at a distance from the fixed end as follows:

Δ=Pa26EI(3la) (1)

Here, l is the length of the beam, E is modulus of Elasticity of the beam, I is the moment of inertia.

Find the deflection at the free end of the beam due to the load of 60kN as follows:

Substitute 3m for a, 9m for l, and 60kN for P in Equation (1).

Δ1=60×326EI(3×93)=2160EI

Find the deflection at the free end of the beam due to the load of 100kN as follows:

Substitute 6m for a, 9m for l, and 100kN for P in Equation (1).

Δ2=100×626EI(3×96)=12600EI

Calculate the deflection at D due to the loading on the beam as follows:

ΔD=Δ1+Δ2=2160EI12600EI=14760kNm3EI (2)

Calculate the value of fDD as follows:

Substitute 9m for a, 9m for l, and 1kN for P in Equation (1).

fDD=1×926EI(3×99)=243kNm3/kNEI (3)

Show the compatibility Equation as follows:

ΔD+fDD×Dy=0

Modify the above Equation using Equation (2) and (3).

ΔD+fDD×Dy=014760kNm3EI+243kNm3/kNEI×Dy=0Dy=(EI243kNm3/kN×14760kNm3EI)Dy=60.74kN()

Refer Figure 1.

Take the sum of the forces in the vertical direction as zero.

Fy=0Ay+Dy60100=0Ay+60.74160=0Ay=99.26kNAy=99.26kN()

Take the sum of the forces in the horizontal direction as zero.

Fx=0Ax=0

Calculate the moment at A as follows:

MA=Dy×9(100×6)(60×3)=(60.74×9)(100×6)(60×3)=233.34kN

Take the moment at D as zero.

MD=0MAAy×9+60×6+100×3=0

Substitute 99.26kN for Ay.

MA99.26×9+60×6+100×3=0MA233.34=0MA=233.34kNm

Thus, the reactions acting on the beam are Ax=0_, Ay=99.26kN()_, Dy=60.74kN()_, and MA=233.34kNm_.

Show the loading on the beam as shown in the Figure 3.

Structural Analysis, Chapter 13, Problem 1P , additional homework tip  3

Refer Figure 3.

Find the shear force at support A as follows:

VA=Ay=99.26kN

Find the shear force at support B as follows:

VB=Ay60=99.2660=39.26kN

Find the shear force at support C as follows:

VC=Ay60100=99.2660100=60.74kN

Find the shear force at support D as follows:

VD=Dy=60.74kN

Calculate the bending moment at A as follows:

MA=233.34kNm

Calculate the bending moment at B as follows:

MB=MAAy×3=233.3499.26×3=64.44kNm

Calculate the bending moment at C as follows:

MC=MAAy×6+60×3=233.3499.26×6+180=182.22kNm

Calculate the bending moment at D as follows:

MD=MAAy×9+60×6+100×3=233.3499.26×9+360+300=0kNm

Sketch the shear force and bending moment diagram as shown in Figure 4.

Structural Analysis, Chapter 13, Problem 1P , additional homework tip  4

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