   # Use the conjugate-beam method to determine the slopes and deflections at points B and C of the beams shown in Figs. P6.14 through P6.17. FIG. P6.15, P6.41

#### Solutions

Chapter
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Chapter 6, Problem 41P
Textbook Problem
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## Use the conjugate-beam method to determine the slopes and deflections at points B and C of the beams shown in Figs. P6.14 through P6.17. FIG. P6.15, P6.41

To determine

Find the slope θB&θC and deflection ΔB&ΔC at point B and C of the given beam using the conjugate-beam method.

### Explanation of Solution

The Young’s modulus (E) is 70 GPa.

The moment of inertia (I) is 500×106mm4.

Calculation:

Consider elastic modulus E of the beam is constant.

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

Refer Figure (1),

Since the point C is free end there is no support reaction at Point C. Therefore, the reaction at point C is also equal to zero.

Consider upward is positive and downward is negative.

Consider clockwise is negative and counterclockwise is positive.

Determine the reaction at support A using the Equation of equilibrium;

V=0RA100=0RA=100kN

Determine the bending moment at C;

MC=300kNm

Determine the bending moment at B;

MB=300kNm

Determine the bending moment at A;

MA=100×6+300=900kNm

Show the M/EI diagram for the given beam as in Figure (2).

Conjugate-beam method:

In the given beam system, point A is a fixed end and point B is free end. But in the conjugate-beam method the fixed end of a real beam becomes free and the free end of real beam changed into the fixed end.

Show the M/EI diagram for the conjugate-beam as in Figure (3).

Calculation of shear at B in the conjugate-beam:

The shear force at B of the conjugate beam is equal to the slope at B on the real beam.

Consider the external forces acting (left of B) upward on the free body diagram as positive.

Determine the shear force at B using the relation;

SB=(bh)(12×b×h)

Here, b is the width and h is the height of the triangle and rectangle.

Substitute 6 m for b and 150EI and 300EI for h.

SB=(6×150EI)(12×6×300EI)=1,800kNm2EI

Determine the deflection at point B using the relation;

θB=1,800kNm2EI

Substitute 70 GPa for E and 500×106mm4 for I.

Hence, the slope at point B is 0.0514rad_.

The bending moment at B in the conjugate beam is equal to defection at B on the real beam.

Take the clockwise moments of the external forces about B as positive.

Determine the bending moment B using the relation;

MB=(bh)(b2)(12×b×h)(23×b)

Substitute 6 m for b and 150EI and 300EI for h.

MB=(6×150EI)(62)(12×6×300EI)(23×6)=6,300kNm3EI

Determine the deflection at point B using the formula;

ΔB=6,300kNm3EI

Substitute 70 GPa for E and 500×106mm4 for I.

ΔB=6,300kNm3(70GPa×106kN/m21GPa)(500×106mm4×(1m1,000mm)4)=0

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