Problem:
Calculate the reactions in the case of the beam shown in Fig. 1.1(a). Construct the bending moment and shearing force diagrams. Determine the location of the maximum bending moment and markit on each of the diagrams.
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Solution:
Taking moment about B, we get
RA × 6+15×2 = 80×4.5
RA ×6+30 = 360
RA = 330÷6 = 55 kN
RB = (80+15) – 55 = 40 kN
S.F. diagram. Consider a section at a distance x from A.
For x to lie between A and C, the shear force is,
F = RA – (80÷3)x
= 55 – (80÷3)x
At. x = 0, F = 55 kN
At. x = 3m, F= – 25 kN
F = 0
When x = 165 ÷ 80
= 2.06 m
Shear force between C and B is constant and equal to 25 kN.
For x to lie between B and D,
F = RA + RB – 80
= 95 – 80
= 15 kN
The S.F. diagram is shown in Fig. 1.1(b). B.M. diagram. Between A and C, the bending moment is,
M = RA × x –(80÷3)x × (x÷2)
= 55x – (40÷3)x²
At. x = 0, M = 0
At. x = 3 m
M = 165 – 120 = 45 kN.m
At. x = 2.06 m
M = 113.3 – 56.6 = 56.7 kN.m
The variation of bending moment between A and C is parabolic.
For x to lie between C and B,
M = RA × x – 80(x – 1.5)
= 55x – 80x + 120
= 120 – 25x
At. x = 3 m
M = 120 – 75 = 45 kN.m
At. x = 6 m
M = 120 – 150 = – 30 kN.m
M = 0 at.
x = 120 ÷ 25
= 4.8 m
The variation of bending moment between C and B is linear. For x to lie between B and D,
M = RA × x – 80(x – 1.5) + RB (x – 6)
= 55x – 80x
= 120 + 40x – 240
= 15x – 120
At x = 6 m
M = 90 – 120
= –30kN.m
At. x = 8 m
M = 120 – 120 = 0
The variation of bending moment between B and D is also linear. The B.M. diagram is shown in fig. 1.1(c)
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