CHAPTER 8 FIRST AND SECOND MOMENTS OF AREA
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: R1 + R 2 = 3
Resolving vertically gives:
But as the beam is symmetrically loaded, R1 = R 2 Hence,
R= R= 1.5kN 1 2
Bending moment expressions: Range AC At x,
M = R1x = 1.5 x
i.e. a straight line
Range CB At x,
M = R1x − 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:
R 1=
Hence, Resolving vertically gives: Hence,
R1 × 3 = 4 ×1 4 = 1.333kN 3
R1 + R 2 = 4
R 2 = 4 − R1 =4 − 1.333 = 2.667 kN
Bending moment expressions: Range AC At x,
M = R1x = 1.333 x
i.e. a straight line
Range CB At x,
M = R1x − 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 =1× 2 + 4 ×1 R 1=
Hence,
6 = 2kN 3
Resolving vertically gives:
R1 + R 2 =+ 1 4
Hence,
R 2 = 5 − 2 = 3 kN 148 © John Bird & Carl Ross Published by Taylor and Francis
Bending moment expressions: Range AC At x,
M = R 1x = 2 x
i.e. a straight line
Range CB At x,
M = R1x − 1(x − 1) = 2 x – x + 1
i.e.
M=x+1
i.e. a straight line
Range DB At x,
M = R1x − 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 × 2 + 4 × 0 =1× 4 R 1=
Hence,
4 = 2kN 2
Resolving vertically gives:
R1 + R 2 =+ 1 4
Hence,
R 2 = 5 − 2 = 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 + R1 (x − 2) + R 2 (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 × 3 + 6 = 4 × 2
Hence,
R1 =
Resolving vertically gives:
R1 + R 2 = 4
Hence,
R 2 = 4 − 0.667 = 3.333 kN
8−6 = 0.667 kN 3
Bending moment expressions: Range AC At x,
M = R1 x = 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 M = R1 x − 4(x − 1) + 6
At x,
= 0.667 x - 4x + 4 + 6 i.e.
M = - 3.333 x + 10
i.e. a straight line
Shearing Force expressions: Range AC
SF = R 1 = + 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=-4x
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=-2x
i.e. a straight line
M=-2x+6
i.e. a straight line
Range CB At x,
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.
RA × 7 = 3 × 5 + 2 × 3 + 4 × 1
i.e.
R A × 7 = 25
from which,
RA =
Resolving vertically, from which,
25 = 3.57 kN 7
RA + RB = 3 + 2 + 4 = 9 R B = 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
SF = + R A = 3.57 kN
(1) (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
SF = R A - 3 = 3.57 – 3 = 0.57 kN
(3) (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
SF = R A - 3 – 2 = 3.57 – 5 = - 1.43 kN
(5) (6)
Range EB At x,
M = R A × x – 3(x – 2) – 2(x – 4) - 4(x – 6) 155 © John Bird & Carl Ross Published by Taylor and Francis
= 3.57 x – 3 x + 6 – 2x + 8 – 4 x + 24 i.e.
M = - 5.43 x + 38
i.e. a straight line
SF = R A - 3 – 2 – 4 = 3.57 – 5 – 4 = - 5.43 kN
(7) (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
156 © John Bird & Carl Ross Published by Taylor and Francis
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.
Total load = 6
kN × 7 m = 42 kN m
By inspection, R1 = R 2 Hence,
R= R= 1 2
42 = 21kN 2
Bending moment expression: At x,
x2 M = R1 x − 6 × 2
i.e.
M = 21 x - 3 x 2
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.
157 © 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 beams; hence, plot the bending moment and shearing force diagrams.
Total load = 5
kN ×12 m = 60 kN m
By inspection, R1 = R 2 Hence,
R= R= 1 2
60 = 30kN 2
Bending moment expression: x2 2
At x,
M = R1 x − 5 ×
i.e.
M = 30 x – 2.5 x 2
i.e. a parabola
Shearing Force expression: At x,
SF = R1 − 5 x
i.e.
SF = 30 – 5x
i.e. a straight line
The bending moment and shearing force diagrams are shown below.
158 © John Bird & Carl Ross Published by Taylor and Francis
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 2
At x,
M = −6×
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.
159 © John Bird & Carl Ross Published by Taylor and Francis
Bending moment expression: x2 2
At x,
M = − 5×
i.e.
M = – 2.5 x 2
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)
160 © John Bird & Carl Ross Published by Taylor and Francis
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: R1 + R 2 = 3
Resolving vertically gives:
But as the beam is symmetrically loaded, R1 = R 2 Hence,
R= R= 1.5kN 1 2
Bending moment expressions: Range AC At x,
M = R1x = 1.5 x
i.e. a straight line
Range CB At x,
M = R1x − 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:
R 1=
Hence, Resolving vertically gives: Hence,
R1 × 3 = 4 ×1 4 = 1.333kN 3
R1 + R 2 = 4
R 2 = 4 − R1 =4 − 1.333 = 2.667 kN
Bending moment expressions: Range AC At x,
M = R1x = 1.333 x
i.e. a straight line
Range CB At x,
M = R1x − 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 =1× 2 + 4 ×1 R 1=
Hence,
6 = 2kN 3
Resolving vertically gives:
R1 + R 2 =+ 1 4
Hence,
R 2 = 5 − 2 = 3 kN 148 © John Bird & Carl Ross Published by Taylor and Francis
Bending moment expressions: Range AC At x,
M = R 1x = 2 x
i.e. a straight line
Range CB At x,
M = R1x − 1(x − 1) = 2 x – x + 1
i.e.
M=x+1
i.e. a straight line
Range DB At x,
M = R1x − 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 × 2 + 4 × 0 =1× 4 R 1=
Hence,
4 = 2kN 2
Resolving vertically gives:
R1 + R 2 =+ 1 4
Hence,
R 2 = 5 − 2 = 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 + R1 (x − 2) + R 2 (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 × 3 + 6 = 4 × 2
Hence,
R1 =
Resolving vertically gives:
R1 + R 2 = 4
Hence,
R 2 = 4 − 0.667 = 3.333 kN
8−6 = 0.667 kN 3
Bending moment expressions: Range AC At x,
M = R1 x = 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 M = R1 x − 4(x − 1) + 6
At x,
= 0.667 x - 4x + 4 + 6 i.e.
M = - 3.333 x + 10
i.e. a straight line
Shearing Force expressions: Range AC
SF = R 1 = + 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=-4x
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=-2x
i.e. a straight line
M=-2x+6
i.e. a straight line
Range CB At x,
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.
RA × 7 = 3 × 5 + 2 × 3 + 4 × 1
i.e.
R A × 7 = 25
from which,
RA =
Resolving vertically, from which,
25 = 3.57 kN 7
RA + RB = 3 + 2 + 4 = 9 R B = 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
SF = + R A = 3.57 kN
(1) (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
SF = R A - 3 = 3.57 – 3 = 0.57 kN
(3) (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
SF = R A - 3 – 2 = 3.57 – 5 = - 1.43 kN
(5) (6)
Range EB At x,
M = R A × x – 3(x – 2) – 2(x – 4) - 4(x – 6) 155 © John Bird & Carl Ross Published by Taylor and Francis
= 3.57 x – 3 x + 6 – 2x + 8 – 4 x + 24 i.e.
M = - 5.43 x + 38
i.e. a straight line
SF = R A - 3 – 2 – 4 = 3.57 – 5 – 4 = - 5.43 kN
(7) (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
156 © John Bird & Carl Ross Published by Taylor and Francis
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.
Total load = 6
kN × 7 m = 42 kN m
By inspection, R1 = R 2 Hence,
R= R= 1 2
42 = 21kN 2
Bending moment expression: At x,
x2 M = R1 x − 6 × 2
i.e.
M = 21 x - 3 x 2
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.
157 © 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 beams; hence, plot the bending moment and shearing force diagrams.
Total load = 5
kN ×12 m = 60 kN m
By inspection, R1 = R 2 Hence,
R= R= 1 2
60 = 30kN 2
Bending moment expression: x2 2
At x,
M = R1 x − 5 ×
i.e.
M = 30 x – 2.5 x 2
i.e. a parabola
Shearing Force expression: At x,
SF = R1 − 5 x
i.e.
SF = 30 – 5x
i.e. a straight line
The bending moment and shearing force diagrams are shown below.
158 © John Bird & Carl Ross Published by Taylor and Francis
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 2
At x,
M = −6×
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.
159 © John Bird & Carl Ross Published by Taylor and Francis
Bending moment expression: x2 2
At x,
M = − 5×
i.e.
M = – 2.5 x 2
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)
160 © John Bird & Carl Ross Published by Taylor and Francis
