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The intensity of loading on a simply supported beam of 5 meters span increases uniformly from 8 kN/m at one end to 16 kN/m at the other end. The magnitude of the maximum bending moment is __________ kN-m.
- a)37.3
- b)37.9
Correct answer is option ''. Can you explain this answer?
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The intensity of loading on a simply supported beam of 5 meters span ...
The trapezoidal loading on the beam consists of a uniformly distributed load and a triangular loading as shown in figure.
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Reactions: Taking moments about the end A, we have
Consider any section XX at a distance x from A Load intensity at the section XX
= 8 +x / 5 × 8 = 8 + 1.6x
S.F. Analysis S.F. at the section XX
The S.F. diagram follows a parabolic law At x = 0, i. e. , at A
Section of Zero Shear Equating the general expression for shear force to zero,
80 / 3 − 8x − 0.8x2 = 0
∴ 3x2 + 30x − 100 = 0
Solving, we get = 2.637 m.
B.M. Analysis
B.M. at any section XX
The B.M. diagram follows a cubic law.
Maximum bending occurs at the section of zero shear, i.e., at a distance of 2.637 meters from A.
∴ Maximum bending moment
= 37.615 kNm
Question_Type: 5
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The intensity of loading on a simply supported beam of 5 meters span ...
Given data:
Span of the beam = 5 m
Intensity of loading at one end = 8 kN/m
Intensity of loading at the other end = 16 kN/m
To find: Magnitude of the maximum bending moment
Assuming that the beam is made of a homogeneous material and has a rectangular cross-section, the maximum bending moment occurs at the mid-span of the beam. Let us calculate the reactions at the supports of the beam first.
Calculating reactions:
By taking moments about the left support, we get:
ΣM = 0
RA × 5 = (8 + 16) × (5/2) × (5/2) × (1/2)
RA = 37.5 kN
By taking moments about the right support, we get:
ΣM = 0
RB × 5 = (8 + 16) × (5/2) × (5/2) × (1/2)
RB = 37.5 kN
Calculating bending moment:
The bending moment at any section x (0 ≤ x ≤ 5) from the left support is given by:
Mx = (RA × (5 - x) × x/2) + (wx × x/2 × (x/2 - a))
where w = intensity of loading in kN/m and a = distance of the section from the left end of the beam.
Substituting the given values, we get:
Mx = (37.5 × (5 - x) × x/2) + ((8 + (8/5)x) × x/2 × (x/2))
Differentiating Mx w.r.t x and equating it to zero to find the maximum bending moment, we get:
dMx/dx = 0
(37.5/2) - (8/5) × (x/2) = 0
x = 9.375/2 = 4.6875 m
Therefore, the maximum bending moment occurs at x = 4.6875 m from the left support.
Substituting this value of x in the equation for Mx, we get:
Mmax = (37.5 × (5 - 4.6875) × 4.6875/2) + ((8 + (8/5) × 4.6875) × 4.6875/2 × (4.6875/2))
Mmax = 37.9 kN-m
Therefore, the magnitude of the maximum bending moment is 37.9 kN-m (option b).
Span of the beam = 5 m
Intensity of loading at one end = 8 kN/m
Intensity of loading at the other end = 16 kN/m
To find: Magnitude of the maximum bending moment
Assuming that the beam is made of a homogeneous material and has a rectangular cross-section, the maximum bending moment occurs at the mid-span of the beam. Let us calculate the reactions at the supports of the beam first.
Calculating reactions:
By taking moments about the left support, we get:
ΣM = 0
RA × 5 = (8 + 16) × (5/2) × (5/2) × (1/2)
RA = 37.5 kN
By taking moments about the right support, we get:
ΣM = 0
RB × 5 = (8 + 16) × (5/2) × (5/2) × (1/2)
RB = 37.5 kN
Calculating bending moment:
The bending moment at any section x (0 ≤ x ≤ 5) from the left support is given by:
Mx = (RA × (5 - x) × x/2) + (wx × x/2 × (x/2 - a))
where w = intensity of loading in kN/m and a = distance of the section from the left end of the beam.
Substituting the given values, we get:
Mx = (37.5 × (5 - x) × x/2) + ((8 + (8/5)x) × x/2 × (x/2))
Differentiating Mx w.r.t x and equating it to zero to find the maximum bending moment, we get:
dMx/dx = 0
(37.5/2) - (8/5) × (x/2) = 0
x = 9.375/2 = 4.6875 m
Therefore, the maximum bending moment occurs at x = 4.6875 m from the left support.
Substituting this value of x in the equation for Mx, we get:
Mmax = (37.5 × (5 - 4.6875) × 4.6875/2) + ((8 + (8/5) × 4.6875) × 4.6875/2 × (4.6875/2))
Mmax = 37.9 kN-m
Therefore, the magnitude of the maximum bending moment is 37.9 kN-m (option b).
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The intensity of loading on a simply supported beam of 5 meters span increases uniformly from 8 kN/m at one end to 16 kN/m at the other end. The magnitude of the maximum bending moment is __________ kN-m.a) 37.3b) 37.9Correct answer is option ''. Can you explain this answer?
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