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A cantilever 3m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. The deflection at the free end is __________ mm (Take E = 200 GPa, I = 54000 cm4)
(A) 3.9
(B) 4.5
Correct answer is option ''. Can you explain this answer?
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Verified Answer
A cantilever 3m long, and of symmetrical section 250 mm deep carries ...
AB fixed at end B, Free at end A carrying a point load W at L1 from B end UDL of W throughout to length EI = flexural rigidity δ = deflection at free end
= 
where, L1 = 1.2m, L2 = 1.8m, L = 3m,
W = 80 kN and W = 30 kN/m
EI = 200 × 106 kN/m2 × 54000 × 10−8 m4
= 108000 kNm2
δ = 
= 0.42666 × 10−3 + 0.96 × 10−3 + 2.8125 × 10−3
= 4.2 × 10−3m = 4.2 mm
Question_Type: 5
Most Upvoted Answer
A cantilever 3m long, and of symmetrical section 250 mm deep carries ...
Given data:
Length of the cantilever (L) = 3 m
Depth of the cantilever (d) = 250 mm = 0.25 m
Load (w) = 30 kN/m
Point load (W) = 80 kN
Distance of point load from the fixed end (x) = 1.2 m
Modulus of elasticity (E) = 200 GPa
Moment of inertia (I) = 54000 cm4
1. Calculation of reaction force:
The cantilever is fixed at one end, and a load is acting at the other end. Therefore, the reaction force at the fixed end is equal to the total load acting on the cantilever.
Total load = wL + W
= 30 × 3 + 80
= 170 kN
Reaction force = Total load
= 170 kN
2. Calculation of maximum bending moment:
The maximum bending moment occurs at the fixed end of the cantilever.
Maximum bending moment (M) = Reaction force × Length of the cantilever
= 170 × 3
= 510 kNm
3. Calculation of deflection:
Deflection (δ) at the free end of the cantilever can be calculated using the formula:
δ = (Wx/6EI) (L^3 - 2Lx^2 + x^3) + (wl/24EI) (L^3 - 3Lx^2 + 4x^3)
where,
W = point load
x = distance of point load from the fixed end
w = uniformly distributed load
E = modulus of elasticity
I = moment of inertia
L = length of the cantilever
On substituting the given values, we get:
δ = (80 × 1.2 / 6 × 200 × 10^9 × 54000 × 10^-8) (3^3 - 2 × 3 × 1.2^2 + 1.2^3) + (30 × 10^3 / 24 × 200 × 10^9 × 54000 × 10^-8) (3^3 - 3 × 3 × 1.2^2 + 4 × 1.2^3)
= 4.5 mm
Therefore, the deflection at the free end of the cantilever is 4.5 mm.
Length of the cantilever (L) = 3 m
Depth of the cantilever (d) = 250 mm = 0.25 m
Load (w) = 30 kN/m
Point load (W) = 80 kN
Distance of point load from the fixed end (x) = 1.2 m
Modulus of elasticity (E) = 200 GPa
Moment of inertia (I) = 54000 cm4
1. Calculation of reaction force:
The cantilever is fixed at one end, and a load is acting at the other end. Therefore, the reaction force at the fixed end is equal to the total load acting on the cantilever.
Total load = wL + W
= 30 × 3 + 80
= 170 kN
Reaction force = Total load
= 170 kN
2. Calculation of maximum bending moment:
The maximum bending moment occurs at the fixed end of the cantilever.
Maximum bending moment (M) = Reaction force × Length of the cantilever
= 170 × 3
= 510 kNm
3. Calculation of deflection:
Deflection (δ) at the free end of the cantilever can be calculated using the formula:
δ = (Wx/6EI) (L^3 - 2Lx^2 + x^3) + (wl/24EI) (L^3 - 3Lx^2 + 4x^3)
where,
W = point load
x = distance of point load from the fixed end
w = uniformly distributed load
E = modulus of elasticity
I = moment of inertia
L = length of the cantilever
On substituting the given values, we get:
δ = (80 × 1.2 / 6 × 200 × 10^9 × 54000 × 10^-8) (3^3 - 2 × 3 × 1.2^2 + 1.2^3) + (30 × 10^3 / 24 × 200 × 10^9 × 54000 × 10^-8) (3^3 - 3 × 3 × 1.2^2 + 4 × 1.2^3)
= 4.5 mm
Therefore, the deflection at the free end of the cantilever is 4.5 mm.
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A cantilever 3m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. The deflection at the free end is __________ mm (Take E = 200 GPa, I = 54000 cm4)(A) 3.9(B) 4.5Correct answer is option ''. Can you explain this answer?
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A cantilever 3m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. The deflection at the free end is __________ mm (Take E = 200 GPa, I = 54000 cm4)(A) 3.9(B) 4.5Correct answer is option ''. Can you explain this answer? for Class 7 2024 is part of Class 7 preparation. The Question and answers have been prepared according to the Class 7 exam syllabus. Information about A cantilever 3m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. The deflection at the free end is __________ mm (Take E = 200 GPa, I = 54000 cm4)(A) 3.9(B) 4.5Correct answer is option ''. Can you explain this answer? covers all topics & solutions for Class 7 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cantilever 3m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. The deflection at the free end is __________ mm (Take E = 200 GPa, I = 54000 cm4)(A) 3.9(B) 4.5Correct answer is option ''. Can you explain this answer?.
A cantilever 3m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. The deflection at the free end is __________ mm (Take E = 200 GPa, I = 54000 cm4)(A) 3.9(B) 4.5Correct answer is option ''. Can you explain this answer? for Class 7 2024 is part of Class 7 preparation. The Question and answers have been prepared according to the Class 7 exam syllabus. Information about A cantilever 3m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. The deflection at the free end is __________ mm (Take E = 200 GPa, I = 54000 cm4)(A) 3.9(B) 4.5Correct answer is option ''. Can you explain this answer? covers all topics & solutions for Class 7 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cantilever 3m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. The deflection at the free end is __________ mm (Take E = 200 GPa, I = 54000 cm4)(A) 3.9(B) 4.5Correct answer is option ''. Can you explain this answer?.
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