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A horizontal beam is simply supported at the ends and carries a uniformly distributed load of 10 kN/m between the supports placed 10 m apart. Anticlockwise moments of 150 kNm and 100 kNm are applied to the left and right ends of the beam at the supports. The point of contra-flexure from the right hand side support is __________m
- a)7.5
- b)7.7
Correct answer is option ''. Can you explain this answer?
Verified Answer
A horizontal beam is simply supported at the ends and carries a unifo...
Taking moments about the left end A,
Vb × 10 + 100 + 150 = 10 x 10 x 5
∴ Vb = 25 kN ↑
∴ Va = (10 × 10) − 25 = 75 kN ↑
S.F. Diagram. At any section distant x from A,
the shear force is given by,
S = 75 − 10x At x = 0, S = +75 kN,
At x = 10, S = −25 kN
Section of zero shear. Equating the general expression for shear force to zero,
75 − 10x = 0
∴ x = 7.5 m
B.M. Diagram. At any section distant x from the end A the Bending Moment is given by,
M = 75x − 5x2 − 150
At x = 0, M = −150 kNm At x = 10,
M = 750 − 500 − 150 = +100 kNm
At x = 7.5 m,
Mmax = 75 × 7.5 − 5 × 7.52 − 150
= +131.25 kNm
Point of contra-flexure. Equating the general expression for B.M. to zero,
75x − 5x2 − 150 = 0x2 − 15x + 30 = 0
Solving we get x = 2.375 m.
From right hand side support = 10 − 2.375
= 7.625m
Question_Type: 5
Most Upvoted Answer
A horizontal beam is simply supported at the ends and carries a unifo...
Taking moments about the left end A,
Vb × 10 + 100 + 150 = 10 x 10 x 5
∴ Vb = 25 kN ↑
∴ Va = (10 × 10) − 25 = 75 kN ↑
S.F. Diagram. At any section distant x from A,
the shear force is given by,
S = 75 − 10x At x = 0, S = +75 kN,
At x = 10, S = −25 kN
Section of zero shear. Equating the general expression for shear force to zero,
75 − 10x = 0
∴ x = 7.5 m
B.M. Diagram. At any section distant x from the end A the Bending Moment is given by,
M = 75x − 5x2 − 150
At x = 0, M = −150 kNm At x = 10,
M = 750 − 500 − 150 = +100 kNm
At x = 7.5 m,
Mmax = 75 × 7.5 − 5 × 7.52 − 150
= +131.25 kNm
Point of contra-flexure. Equating the general expression for B.M. to zero,
75x − 5x2 − 150 = 0x2 − 15x + 30 = 0
Solving we get x = 2.375 m.
From right hand side support = 10 − 2.375
= 7.625m
Question_Type: 5
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Community Answer
A horizontal beam is simply supported at the ends and carries a unifo...
Taking moments about the left end A,
Vb × 10 + 100 + 150 = 10 x 10 x 5
∴ Vb = 25 kN ↑
∴ Va = (10 × 10) − 25 = 75 kN ↑
S.F. Diagram. At any section distant x from A,
the shear force is given by,
S = 75 − 10x At x = 0, S = +75 kN,
At x = 10, S = −25 kN
Section of zero shear. Equating the general expression for shear force to zero,
75 − 10x = 0
∴ x = 7.5 m
B.M. Diagram. At any section distant x from the end A the Bending Moment is given by,
M = 75x − 5x2 − 150
At x = 0, M = −150 kNm At x = 10,
M = 750 − 500 − 150 = +100 kNm
At x = 7.5 m,
Mmax = 75 × 7.5 − 5 × 7.52 − 150
= +131.25 kNm
Point of contra-flexure. Equating the general expression for B.M. to zero,
75x − 5x2 − 150 = 0x2 − 15x + 30 = 0
Solving we get x = 2.375 m.
From right hand side support = 10 − 2.375
= 7.625m
Question_Type: 5
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A horizontal beam is simply supported at the ends and carries a uniformly distributed load of 10 kN/m between the supports placed 10 m apart. Anticlockwise moments of 150 kNm and 100 kNm are applied to the left and right ends of the beam at the supports. The point of contra-flexure from the right hand side support is __________ma) 7.5b) 7.7Correct answer is option ''. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A horizontal beam is simply supported at the ends and carries a uniformly distributed load of 10 kN/m between the supports placed 10 m apart. Anticlockwise moments of 150 kNm and 100 kNm are applied to the left and right ends of the beam at the supports. The point of contra-flexure from the right hand side support is __________ma) 7.5b) 7.7Correct answer is option ''. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A horizontal beam is simply supported at the ends and carries a uniformly distributed load of 10 kN/m between the supports placed 10 m apart. Anticlockwise moments of 150 kNm and 100 kNm are applied to the left and right ends of the beam at the supports. The point of contra-flexure from the right hand side support is __________ma) 7.5b) 7.7Correct answer is option ''. Can you explain this answer?.
A horizontal beam is simply supported at the ends and carries a uniformly distributed load of 10 kN/m between the supports placed 10 m apart. Anticlockwise moments of 150 kNm and 100 kNm are applied to the left and right ends of the beam at the supports. The point of contra-flexure from the right hand side support is __________ma) 7.5b) 7.7Correct answer is option ''. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A horizontal beam is simply supported at the ends and carries a uniformly distributed load of 10 kN/m between the supports placed 10 m apart. Anticlockwise moments of 150 kNm and 100 kNm are applied to the left and right ends of the beam at the supports. The point of contra-flexure from the right hand side support is __________ma) 7.5b) 7.7Correct answer is option ''. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A horizontal beam is simply supported at the ends and carries a uniformly distributed load of 10 kN/m between the supports placed 10 m apart. Anticlockwise moments of 150 kNm and 100 kNm are applied to the left and right ends of the beam at the supports. The point of contra-flexure from the right hand side support is __________ma) 7.5b) 7.7Correct answer is option ''. Can you explain this answer?.
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A horizontal beam is simply supported at the ends and carries a uniformly distributed load of 10 kN/m between the supports placed 10 m apart. Anticlockwise moments of 150 kNm and 100 kNm are applied to the left and right ends of the beam at the supports. The point of contra-flexure from the right hand side support is __________ma) 7.5b) 7.7Correct answer is option ''. Can you explain this answer?, a detailed solution for A horizontal beam is simply supported at the ends and carries a uniformly distributed load of 10 kN/m between the supports placed 10 m apart. Anticlockwise moments of 150 kNm and 100 kNm are applied to the left and right ends of the beam at the supports. The point of contra-flexure from the right hand side support is __________ma) 7.5b) 7.7Correct answer is option ''. Can you explain this answer? has been provided alongside types of A horizontal beam is simply supported at the ends and carries a uniformly distributed load of 10 kN/m between the supports placed 10 m apart. Anticlockwise moments of 150 kNm and 100 kNm are applied to the left and right ends of the beam at the supports. The point of contra-flexure from the right hand side support is __________ma) 7.5b) 7.7Correct answer is option ''. Can you explain this answer? theory, EduRev gives you an
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