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rTheory of Structures(ES-3)

MODULE 1

SHEAR AND MOMENT IN BEAMS

INTRODUCTION

The basic problems in Theory of Structures is to determine the relations between the

stresses and deformations caused by loads applied to any structure. In axial or torsional

loadings, we had little trouble in applying the stress and deformation relation because in

the majority of cases the loading either remains constant over the entire structure or is

distributed in definite amounts to the component parts.

The study of bending loads, however is complicated by the fact that the loading effects

vary from section to section of the beam, these loading effects take the form of a

shearing force and a bending moment, usually called the shear and moment diagram.

A. SHEAR AND MOMENT EQUATION

Figure 1 shows a simple beam that carries a concentrated load P and is held

equilibrium by the two reaction R1and R2, For the time being neglect the mass of

the beam and consider only the effect of an axial load P. Assume that a cutting

plane a-a at a distance x at R1divides the beam into two segment. To maintain

equilibrium in this segment of the beam section a-a must supply the resisting

forces necessary to satisfy the conditions of the static equilibrium.In this case the

external load is vertical so the condition Fx= 0

Figure 1a

x

R1

P

L

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x

Mr

VrR1

Mr

Vr

L-x

R2

Figure 1b Figure 1c

Figure 1

Is automatically satisfied.

To satisfy Fy= 0, the vertical unbalanced cause by R1requires the fibers in section a-ato create a resisting force. This is shown as V rand is called the resisting shearing force.

B. BENDING MOMENT EQUATION

Bending moment is defined as the summation of moments about the centroidal

axis of any selected section of all the loads acting as either to the left or to the

right side of the section, and is expressed mathematically as

M = (M)L= (M)R

The subscript L indicating the summation of bending moment is computed in

terms of the loads acting to the left of the section and the subscript R referring to

loads acting to the right of the section.

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SAMPLE PROBLEM

Write the shear and moment equation for segment AB and BC for the beam shown

R1 R2

5KN

6.0 m

4.0 m.

Figure 2a

Solution

First step is to solve for the two reaction by taking summation of moment at point R 2

MR2= 0;

6R1- 5(2) = 0; R1= 1.67 KN

Y = 0; R1+ R25 = 0 ; R2= 3.33 KN

Next step is to pass an exploratory section a-a, x distance from R 1or between points A

and B

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a

R1 R2

5KN

6.0 m

a

x

A B C

Figure 2b

Shear at segment AB V = (Fy)L V = R1= 1.67 KN

Moment at segment AB M = (M)L M = R1(x) = 1.67x KN.m

These equations are valid only for values between 0 and 4 m. that is between A and B.

For Segment BC

R1 R2

5KN

6.0 m

A B C

b

b

x

Figure 2c

Shear at segment BC

V = (Fy)L V = R15 = 1.675 = -3.33 KN

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Moment at segment BC

M = (M)L M = R1(x)5(x-4) = 1.67x5x + 20 = - 3.33x + 20

The value of x ranges from 4 to 6 m.

SAMPLE PROBLEM 2

Write the shear and moment equation for segment AB and BC for the beam shown

below.

6 KN/m.

R1 R2

5KN

6.0 m

A B C

x

a

a

2 m.

Figure 3a

Solution

Solving for the two reaction R1and R2

MR2= 0

6R16(6)(6/2)5(2) = 0; R1= 19.67 KN

Fy= 0; R1+ R256(6) = 0; R2= 21.33 KN

Write the shear and moment equation

For segment AB (see Figure 3a)

Shear; V = (Fy)L= 19.676(x) = 19.676x

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Moment; M = (M)L= 19.67(x)6(x)(x/2) = 19.67x3x2

Value of x ranges from 0 to 4 m.

6 KN/m.

R1 R2

5KN

6.0 m

A B C

x

b

b

2 m.

Figure 3b

Segment BC (see Figure 3b)

Shear; V = (Fy)L= 19.6756(x) = 14.676x

Moment; M = (M)L= 19.67(x)5(x4)6(x)(x/2) = 14.67x + 203x2

Value of x ranges from 4 to 6 m.

SHEAR AND MOMENT DIAGRAM

Shear and moment equation are the graphical visualization of the shear and moment

equations plotted on V-x and M-x below the loading diagram.

SAMPLE PROBLEM 3

Draw the shear and moment diagram for the beam shown

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Solution

See problem 2 in solving for the two reaction R1and R2

Figure 4a

6KN/m

R1 = 19.67 KN R2 = 21.33 KN

5KN

6.0 m

A B C2 m.

V

19.67 KN

- 4.33 KN

- 9.33 KN

-21.33 KN

M

Mmax= 32.24 KN. m

A

A

Discussion

In drawing the shear and moment diagram for problem number 3, first step is to

determine the two reaction for the two support, R1and R2.

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R1= 19.67 KN ; R2= 21.33 KN

Next is plot the shear diagram

Segment AB

Consider the first reaction R1 = 19.67 KN,

Vab = 19.676(4) = - 4.33 KN

Vab = - 4.335 = 9.33 KN

Segment BC

Vbc = - 9.336(2) = -21.33 KN

For Moment Diagram

Segment AA

Determine first the distance w/ zero shear (Figure 4a)

Fy = 0; 19.33 6(x) = 0; x = 3.278 m

Solve for maximum moment

MMax= (19.67)(3.278) = 32.24 KN-m.

Segment AB

MMax = 32.24 (4.33)(43.278) = 30.677 KN-m.

Segment BC

Mbc = 30.677( 9.3321.33)(2) = 0

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PROBLEMS

Write the shear and moment equation and draw the shear and moment diagram for the

beam shown below, determine the maximum moment.

1.

W KN/m

L

R1 R2

2.

3 m 5 m

3 KN/m5 KN/m

A

B

C

RA

RB

10 KN

3.

5 KN/m

10 KN

2 m 3 m

A B

C

RA RB

= 10kN/m

= 6.0m

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4.

80 KN

10 KN/m

2 m 8 m

A B C

RARB

5.

W KN/m

L

6.

10 KN

3 m 6 m

A B C

= 5kN/m

= 3.0m

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7.

W KN/m

L

R1 R2

8.

9 KN 4 KN

A

B

C

D

2 m 6 m 3 m

RA RB

4 KN/m

9.

W KN/m W KN/m

L/3 L/3 L/3

RA

RB

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10.

P

A B C

RA RB

L/2 L/2

= 4kN

3.00m 3.00m