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An RCC beam of size 350mm (width) and 600mm (overall depth) is subjected to shear force of 100 kN, bending moment of 200 kNm and twisting moment of 50 kNm. Effective cover to reinforcement is 25 mm. Find the equivalent bending moment which is contributed by torsional moment?
- a)79.83 kNm
- b)279.83 kNm
- c)228.58 kNm
- d)428.58 kNm
Correct answer is option 'A'. Can you explain this answer?
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An RCC beam of size 350mm (width) and 600mm (overall depth) is subject...
Here we have to calculate the equivalent bending moment contributed by torsional moment.
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Calculation of Equivalent Bending Moment Contributed by Torsional Moment
Given data:
- Width of RCC beam (b) = 350 mm
- Overall depth of RCC beam (D) = 600 mm
- Effective cover to reinforcement = 25 mm
- Shear force (V) = 100 kN
- Bending moment (M) = 200 kNm
- Twisting moment (T) = 50 kNm
Formula used:
- Equivalent bending moment contributed by torsional moment = (T*D)/(2h) where h is the height of the section where torsion occurs.
Solution:
- Depth of RCC beam (d) = overall depth - effective cover to reinforcement = 600 - 25 - 25 = 550 mm
- Moment of inertia of the section about the x-axis (Ixx) = (b*d^3)/12 = 350*(550^3)/12 = 1083125000 mm^4
- Torsional constant (J) = (b*d^3)/3 = 350*(550^3)/3 = 3249375000 mm^4
- Maximum shear stress due to torsion (τmax) = (T*J)/(Ixx*h) where τmax = T/(2*A) = 50/(2*19250) = 0.0013 N/mm^2 (approx.)
- Let the height of the section where torsion occurs (h) = x mm
- Total shear stress at a distance x from the top of the section (τ) = VQ/(Ixx*h) where Q is the first moment of area of the section about the x-axis below x
- Q = (b*(d-x)^2*x) + (b*x^2*(d-x)/2) = 350*(550-x)^2*x + 175*(x^2)*(550-x)
- Equating τmax to τ, we get:
- (T*J)/(Ixx*x) = VQ/(Ixx*x^2)
- Solving for x, we get:
- x = (T*D)/(2V + sqrt((2V)^2 + 4(T^2*D^2)/(Ixx^2)))
- Substituting the given values, we get:
- x = (50*600)/(2*100 + sqrt((2*100)^2 + 4*(50^2*600^2)/(1083125000^2))) = 192.5 mm (approx.)
- Therefore, the height of the section where torsion occurs (h) = 550 - 25 - 192.5 = 332.5 mm
- Equivalent bending moment contributed by torsional moment = (T*D)/(2h) = (50*600)/(2*332.5) = 79.83 kNm
Therefore, the equivalent bending moment contributed by torsional moment is 79.83 kNm.
Given data:
- Width of RCC beam (b) = 350 mm
- Overall depth of RCC beam (D) = 600 mm
- Effective cover to reinforcement = 25 mm
- Shear force (V) = 100 kN
- Bending moment (M) = 200 kNm
- Twisting moment (T) = 50 kNm
Formula used:
- Equivalent bending moment contributed by torsional moment = (T*D)/(2h) where h is the height of the section where torsion occurs.
Solution:
- Depth of RCC beam (d) = overall depth - effective cover to reinforcement = 600 - 25 - 25 = 550 mm
- Moment of inertia of the section about the x-axis (Ixx) = (b*d^3)/12 = 350*(550^3)/12 = 1083125000 mm^4
- Torsional constant (J) = (b*d^3)/3 = 350*(550^3)/3 = 3249375000 mm^4
- Maximum shear stress due to torsion (τmax) = (T*J)/(Ixx*h) where τmax = T/(2*A) = 50/(2*19250) = 0.0013 N/mm^2 (approx.)
- Let the height of the section where torsion occurs (h) = x mm
- Total shear stress at a distance x from the top of the section (τ) = VQ/(Ixx*h) where Q is the first moment of area of the section about the x-axis below x
- Q = (b*(d-x)^2*x) + (b*x^2*(d-x)/2) = 350*(550-x)^2*x + 175*(x^2)*(550-x)
- Equating τmax to τ, we get:
- (T*J)/(Ixx*x) = VQ/(Ixx*x^2)
- Solving for x, we get:
- x = (T*D)/(2V + sqrt((2V)^2 + 4(T^2*D^2)/(Ixx^2)))
- Substituting the given values, we get:
- x = (50*600)/(2*100 + sqrt((2*100)^2 + 4*(50^2*600^2)/(1083125000^2))) = 192.5 mm (approx.)
- Therefore, the height of the section where torsion occurs (h) = 550 - 25 - 192.5 = 332.5 mm
- Equivalent bending moment contributed by torsional moment = (T*D)/(2h) = (50*600)/(2*332.5) = 79.83 kNm
Therefore, the equivalent bending moment contributed by torsional moment is 79.83 kNm.
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An RCC beam of size 350mm (width) and 600mm (overall depth) is subjected to shear force of 100 kN, bending moment of 200 kNm and twisting moment of 50 kNm. Effective cover to reinforcement is 25 mm. Find the equivalent bending moment which is contributed by torsional moment?a)79.83 kNmb)279.83 kNmc)228.58 kNmd)428.58 kNmCorrect answer is option 'A'. Can you explain this answer?
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