Chapter 8 Bending Moment and Shear Force Diagrams

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CHAPTER 8 BENDING MOMENT AND SHEAR FORCE DIAGRAMS

EXERCISE 51, Page 121

1. Determine expressions for the bending moment and shearing force distributions for the

following simply supported beam; hence, or otherwise, plot the bending moment and shearing

force diagrams.

To calculate the reactions:

Resolving vertically gives: RR12+= 3

But as the beam is symmetrically loaded, RR12=

Hence, R12= R = 1.5kN

Bending moment expressions:

Range AC

At x, M = Rx1 = 1.5 x i.e. a straight line

Range CB

At x, M = R1 x−− 3(x 1) = 1.5 x – 3x + 3 i.e. M = - 1.5 x + 3 i.e. a straight line

Shearing Force expressions:

Range AC SF = + 1.5 kN

Range CB SF = + 1.5 – 3 = - 1.5 kN

The bending moment and shearing force diagrams are shown below. 146 © John Bird & Carl Ross Published by Taylor and Francis

2. Determine expressions for the bending moment and shearing force distributions for the

following simply supported beam; hence, or otherwise, plot the bending moment and shearing

force diagrams.

Taking moments about B gives: R1 ×=× 3 41

4 Hence, R= = 1.333kN 1 3

Resolving vertically gives: RR12+= 4

Hence, R2 = 4−=− R1 4 1.333 = 2.667 kN

Bending moment expressions:

Range AC

At x, M = Rx1 = 1.333 x i.e. a straight line

Range CB

At x, M = R1 x−− 4(x 2) = 1.333 x – 4x + 8

147 © John Bird & Carl Ross Published by Taylor and Francis i.e. M = - 2.667 x + 8 i.e. a straight line

Shearing Force expressions:

Range AC SF = R1 = + 1.333 kN

Range CB SF = R1 - 4 = + 1.5 – 4 = - 2.667 kN

The bending moment and shearing force diagrams are shown below.

3. Determine expressions for the bending moment and shearing force distributions for the

following simply supported beam; hence, or otherwise, plot the bending moment and shearing

force diagrams.

Taking moments about B gives: R1 ×=×+× 3 12 41

6 Hence, R= = 2kN 1 3

Resolving vertically gives: R12+=+ R 14

Hence, R2 = 52− = 3 kN

148 © John Bird & Carl Ross Published by Taylor and Francis

Bending moment expressions:

Range AC

At x, M = Rx1 = 2 x i.e. a straight line

Range CB

At x, M = R1 x−− 1(x 1) = 2 x – x + 1 i.e. M = x + 1 i.e. a straight line

Range DB

At x, M = R1 x− 1(x −− 1) 4(x − 2) = 2 x – x + 1 – 4x + 8 i.e. M = - 3x + 9 i.e. a straight line

Shearing Force expressions:

Range AC SF = R1 = + 2 kN

Range CD SF = R1 - 1 = 2 – 1 = 1 kN

Range DB SF = R1 - 1 - 4 = 2 – 1 – 4 = - 3 kN

The bending moment and shearing force diagrams are shown below.

4. Determine expressions for the bending moment and shearing force distributions for the

following simply supported beam; hence, or otherwise, plot the bending moment and shearing

force diagrams.

149 © John Bird & Carl Ross Published by Taylor and Francis

Taking moments about D gives: R1 ×+×=× 24014

4 Hence, R= = 2kN 1 2

Resolving vertically gives: R12+=+ R 14

Hence, R2 = 52− = 3 kN

Bending moment expressions:

Range CA

At x, M = −1x = - x i.e. a straight line

Range AB

At x, M = −+1x R1 (x − 2) = -1 x + 2 x - 4 i.e. M = x - 4 i.e. a straight line

Range DB

At x, M = −+1x R12 (x −+ 2) R (x −− 4) 4(x − 4)

= -1 x + 2x - 4 + 3x - 12 – 4x + 16 i.e. M = 0

Shearing Force expressions:

Range CA SF = - 1 kN

Range AB SF = - 1 + R1 = - 1 + 2 = + 1 kN

Range DB SF = - 1 + 2 – 4 + 3 = 0 kN

The bending moment and shearing force diagrams are shown below.

150 © John Bird & Carl Ross Published by Taylor and Francis

5. Determine expressions for the bending moment and shearing force distributions for the

following simply supported beam; hence, or otherwise, plot the bending moment and shearing

force diagrams.

Taking moments about B gives: R1 ×+ 36 =× 42

86− Hence, R= = 0.667kN 1 3

Resolving vertically gives: RR12+= 4

Hence, R2 = 4− 0.667 = 3.333 kN

Bending moment expressions:

Range AC

At x, M = Rx1 = 0.667 x i.e. a straight line

Range CD

At x, M = R1 x−− 4(x 1) = 0.667 x - 4 x + 4 i.e. M = - 3.333 x + 4 i.e. a straight line

151 © John Bird & Carl Ross Published by Taylor and Francis

Range DB

At x, M = R1 x− 4(x −+ 1) 6

= 0.667 x - 4x + 4 + 6 i.e. M = - 3.333 x + 10 i.e. a straight line

Shearing Force expressions:

Range AC SF = R1 = + 0.667 kN

Range CD SF = R1 - 4 = 0.667 - 4 = - 3.333 kN

Range DB SF = R1 – 4 = 0.667 – 4 = - 3.333 kN

The bending moment and shearing force diagrams are shown below.

6. Determine expressions for the bending moment and shearing force distributions for the

following simply supported beam; hence, or otherwise, plot the bending moment and shearing

force diagrams.

Bending moment expressions:

Range AC

152 © John Bird & Carl Ross Published by Taylor and Francis

At x, M = - 4 x i.e. a straight line

Range CB

At x, M = - 4 x – 6(x – 2) = - 4 x - 6 x + 12 i.e. M = - 10 x + 12 i.e. a straight line

Shearing Force expressions:

Range AC SF = - 4 kN

Range CB SF = - 4 - 6 = - 10 kN

The bending moment and shearing force diagrams are shown below.

7. Determine expressions for the bending moment and shearing force distributions for the

following simply supported beam; hence, or otherwise, plot the bending moment and shearing

force diagrams.

Bending moment expressions:

Range AC

At x, M = - 2 x i.e. a straight line

Range CB

At x, M = - 2 x + 6 i.e. a straight line

153 © John Bird & Carl Ross Published by Taylor and Francis

Shearing Force expressions:

Range AC SF = - 2 kN

Range CB SF = - 2 kN

The bending moment and shearing force diagrams are shown below.

8. A horizontal beam of negligible mass is of length 7 m. The beam is simply-supported at its ends

and carries three vertical loads, pointing in a downward direction. The first load is of magnitude

3 kN and acts 2 m from the left end, the second load is of magnitude 2 kN and acts 4 m from the

left end, and the third load is of magnitude 4 kN and acts 6 m from the left end. Calculate the

bending moment and shearing force at the points of discontinuity, working from the left support

to the right support.

To determine the reactions R A and R B :

Taking moments about B in the diagram below gives:

Clockwise moment = anticlockwise moment

i.e. R A × 7 = 3 × 5 + 2 × 3 + 4 × 1

i.e. R A × 7 = 25

25 from which, R = = 3.57 kN A 7

Resolving vertically, R A + R B = 3 + 2 + 4 = 9

from which, RB = 9 – 3.57 = 5.43 kN

154 © John Bird & Carl Ross Published by Taylor and Francis

Range AC

At x, M = R A × x = 3.57 x i.e. a straight line (1)

SF = + R A = 3.57 kN (2)

Range CD

At x, M = R A × x – 3(x – 2) = 3.57 x – 3 x + 6 i.e. M = 0.57 x + 6 i.e. a straight line (3)

SF = R A - 3 = 3.57 – 3 = 0.57 kN (4)

Range DE

At x, M = R A × x – 3(x – 2) – 2(x – 4) = 3.57 x – 3 x + 6 – 2x + 8 i.e. M = - 1.43 x + 14 i.e. a straight line (5)

SF = R A - 3 – 2 = 3.57 – 5 = - 1.43 kN (6)

Range EB

= 3.57 x – 3 x + 6 – 2x + 8 – 4 x + 24 i.e. M = - 5.43 x + 38 i.e. a straight line (7)

SF = R A - 3 – 2 – 4 = 3.57 – 5 – 4 = - 5.43 kN (8)

Plotting equations (1) to (8) results in the bending moment and shearing force diagrams shown above.

Summarising, Bending moments (kN m): 0, 7.14, 8.28, 5.42, 0

Shearing forces (kN): 3.57, 3.57/ 0.57, 0.57/ - 1.43, - 1.43/ - 5.43, - 5.43

EXERCISE 52, Page 124

1. Determine expressions for the bending moment and shearing force distributions for the

following simply supported beams; hence, plot the bending moment and shearing force

diagrams.

kN Total load = 6× 7m = 42 kN m

By inspection, RR12=

42 Hence, R= R = = 21kN 122

Bending moment expression:

x2 At x, M = Rx−× 6 1 2 i.e. M = 21 x - 3 x2 i.e. a parabola

Shearing Force expression:

At x, SF = R1 − 6x i.e. SF = 21 – 6x i.e. a straight line

The bending moment and shearing force diagrams are shown below.

2. Determine expressions for the bending moment and shearing force distributions for the

following simply supported beams; hence, plot the bending moment and shearing force

diagrams.

kN Total load = 5× 12m = 60 kN m

By inspection, RR12=

60 Hence, R= R = = 30kN 122

Bending moment expression:

x2 At x, M = Rx−× 5 1 2 i.e. M = 30 x – 2.5 x2 i.e. a parabola

Shearing Force expression:

At x, SF = R1 − 5x i.e. SF = 30 – 5x i.e. a straight line

The bending moment and shearing force diagrams are shown below.

3. Determine expressions for the bending moment and shearing force distributions for the

following cantilevers; hence, or otherwise, plot the bending moment and shearing force

diagrams.

Bending moment expression:

x2 At x, M = −×6 2 i.e. M = – 3 x2 i.e. a parabola

Shearing Force expression:

At x, SF = - 6 x i.e. a straight line

The bending moment and shearing force diagrams are shown below.

4. Determine expressions for the bending moment and shearing force distributions for the

following cantilevers; hence, or otherwise, plot the bending moment and shearing force

diagrams.

Bending moment expression:

x2 At x, M = −×5 2 i.e. M = – 2.5 x2 i.e. a parabola

Shearing Force expression:

At x, SF = - 5 x i.e. a straight line

The bending moment and shearing force diagrams are shown below.

EXERCISE 53, Page 124

Answers found from within the text of the chapter, pages 112 to 123.

EXERCISE 54, Page 124

1. (b) 2. (c) 3. (c) 4. (a) 5. (c) 6. (b)