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A simply supported rectangular beam of span 20.0 m is subjected to uniformly distributed load (UDL). The minimum effective depth required to check the deflection of this depth, when modification factors for tension and compression are 0.9 and 1.1, respectively, will be _________ m.
- a)2m
- b)1.8m
- c)1.3m
- d)1.0m
Correct answer is option 'A'. Can you explain this answer?
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Given data:
Span of beam (L) = 20 m
Uniformly distributed load (W) = UDL
Modification factor for tension (m_t) = 0.9
Modification factor for compression (m_c) = 1.1
To find: Minimum effective depth required to check the deflection of a rectangular beam
Assumptions:
1. Beam is simply supported
2. Beam is of rectangular cross-section
3. Load is uniformly distributed
Formula used:
Deflection of a simply supported beam with uniformly distributed load (UDL) is given by:
δ = (5/384) * (W * L^4) / (E * I)
where,
δ = deflection of beam
W = uniformly distributed load
L = span of beam
E = modulus of elasticity of beam material
I = moment of inertia of beam cross-section
For a rectangular cross-section, moment of inertia is given by:
I = (b * d^3) / 12
where,
b = breadth of beam
d = effective depth of beam
To check the deflection of beam, the following condition should be satisfied:
δ <= (l="">
Calculation:
We know that,
I = (b * d^3) / 12
Assuming b = d, we get:
I = (d^4) / 12
Effective depth (d) can be calculated using the following formula:
d = (L / (k * (m_t + m_c - 1) * γ * f_ck^(1/3)))
where,
k = 1.5 (for simply supported beam)
γ = 1.5 (partial safety factor for concrete)
f_ck = characteristic compressive strength of concrete
Substituting the given values, we get:
d = (20 / (1.5 * (0.9 + 1.1 - 1) * 1.5 * f_ck^(1/3)))
Assuming f_ck = 20 N/mm^2, we get:
d = 1.99 m
To check the deflection of beam, we need to calculate the value of moment of inertia:
I = (b * d^3) / 12
Assuming b = d, we get:
I = (d^4) / 12
Substituting the value of effective depth (d), we get:
I = (1.99^4) / 12
I = 29.26 * 10^3 mm^4
Substituting the values of load (W), span (L), moment of inertia (I), and modulus of elasticity (E = 2.5 * 10^5 N/mm^2), we get:
δ = (5/384) * (W * L^4) / (E * I)
Assuming δ = (L / 250), we get:
(L / 250) = (5/384) * (W * L^4) / (E * I)
Substituting the values of load (W), span (L), moment of inertia (I), and modulus of elasticity (E = 2.5 * 10^5 N/mm^2), we get:
1 = (W * L^3) / (384 * γ * f_ck * I * (m_t + m_c -
Span of beam (L) = 20 m
Uniformly distributed load (W) = UDL
Modification factor for tension (m_t) = 0.9
Modification factor for compression (m_c) = 1.1
To find: Minimum effective depth required to check the deflection of a rectangular beam
Assumptions:
1. Beam is simply supported
2. Beam is of rectangular cross-section
3. Load is uniformly distributed
Formula used:
Deflection of a simply supported beam with uniformly distributed load (UDL) is given by:
δ = (5/384) * (W * L^4) / (E * I)
where,
δ = deflection of beam
W = uniformly distributed load
L = span of beam
E = modulus of elasticity of beam material
I = moment of inertia of beam cross-section
For a rectangular cross-section, moment of inertia is given by:
I = (b * d^3) / 12
where,
b = breadth of beam
d = effective depth of beam
To check the deflection of beam, the following condition should be satisfied:
δ <= (l="">
Calculation:
We know that,
I = (b * d^3) / 12
Assuming b = d, we get:
I = (d^4) / 12
Effective depth (d) can be calculated using the following formula:
d = (L / (k * (m_t + m_c - 1) * γ * f_ck^(1/3)))
where,
k = 1.5 (for simply supported beam)
γ = 1.5 (partial safety factor for concrete)
f_ck = characteristic compressive strength of concrete
Substituting the given values, we get:
d = (20 / (1.5 * (0.9 + 1.1 - 1) * 1.5 * f_ck^(1/3)))
Assuming f_ck = 20 N/mm^2, we get:
d = 1.99 m
To check the deflection of beam, we need to calculate the value of moment of inertia:
I = (b * d^3) / 12
Assuming b = d, we get:
I = (d^4) / 12
Substituting the value of effective depth (d), we get:
I = (1.99^4) / 12
I = 29.26 * 10^3 mm^4
Substituting the values of load (W), span (L), moment of inertia (I), and modulus of elasticity (E = 2.5 * 10^5 N/mm^2), we get:
δ = (5/384) * (W * L^4) / (E * I)
Assuming δ = (L / 250), we get:
(L / 250) = (5/384) * (W * L^4) / (E * I)
Substituting the values of load (W), span (L), moment of inertia (I), and modulus of elasticity (E = 2.5 * 10^5 N/mm^2), we get:
1 = (W * L^3) / (384 * γ * f_ck * I * (m_t + m_c -
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A simply supported rectangular beam of span 20.0 m is subjected to uniformly distributed load (UDL). The minimum effective depth required to check the deflection of this depth, when modification factors for tension and compression are 0.9 and 1.1, respectively, will be _________ m.a)2mb)1.8mc)1.3md)1.0mCorrect answer is option 'A'. Can you explain this answer?
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A simply supported rectangular beam of span 20.0 m is subjected to uniformly distributed load (UDL). The minimum effective depth required to check the deflection of this depth, when modification factors for tension and compression are 0.9 and 1.1, respectively, will be _________ m.a)2mb)1.8mc)1.3md)1.0mCorrect answer is option 'A'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A simply supported rectangular beam of span 20.0 m is subjected to uniformly distributed load (UDL). The minimum effective depth required to check the deflection of this depth, when modification factors for tension and compression are 0.9 and 1.1, respectively, will be _________ m.a)2mb)1.8mc)1.3md)1.0mCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A simply supported rectangular beam of span 20.0 m is subjected to uniformly distributed load (UDL). The minimum effective depth required to check the deflection of this depth, when modification factors for tension and compression are 0.9 and 1.1, respectively, will be _________ m.a)2mb)1.8mc)1.3md)1.0mCorrect answer is option 'A'. Can you explain this answer?.
A simply supported rectangular beam of span 20.0 m is subjected to uniformly distributed load (UDL). The minimum effective depth required to check the deflection of this depth, when modification factors for tension and compression are 0.9 and 1.1, respectively, will be _________ m.a)2mb)1.8mc)1.3md)1.0mCorrect answer is option 'A'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A simply supported rectangular beam of span 20.0 m is subjected to uniformly distributed load (UDL). The minimum effective depth required to check the deflection of this depth, when modification factors for tension and compression are 0.9 and 1.1, respectively, will be _________ m.a)2mb)1.8mc)1.3md)1.0mCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A simply supported rectangular beam of span 20.0 m is subjected to uniformly distributed load (UDL). The minimum effective depth required to check the deflection of this depth, when modification factors for tension and compression are 0.9 and 1.1, respectively, will be _________ m.a)2mb)1.8mc)1.3md)1.0mCorrect answer is option 'A'. Can you explain this answer?.
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