Civil Engineering (CE) Exam > Civil Engineering (CE) Questions > A beam of rectangular cross-section 25 × 35 ...
Start Learning for Free
A beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)
Correct answer is '1.93'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A beam of rectangular cross-section 25 × 35 mm is subjected to a torq...
Given:
Cross-sectional dimensions of the beam: 25 × 35 mm
Torque applied: 79 kN-m
Value of α: 1.87
To find:
Maximum torsional stress induced in the beam in MPa
Formula:
Torsional stress (τ) = (T * α) / (J * r)
Where,
T = Torque applied
α = Angle of twist per unit length
J = Polar moment of inertia
r = Distance from the center of the cross-section to the outermost fiber
Calculation:
Step 1: Calculate the polar moment of inertia (J)
J = (b * h^3) / 3
Where,
b = Width of the cross-section
h = Height of the cross-section
Given: b = 35 mm, h = 25 mm
J = (35 * (25^3)) / 3
J = 91,667 mm^4
Step 2: Calculate the distance from the center to the outermost fiber (r)
r = (b / 2) = (35 / 2) = 17.5 mm
Step 3: Convert the torque to N-mm
Torque (T) = 79 kN-m = 79 × 1000 N-mm
Step 4: Calculate the angle of twist per unit length (α)
α = (T * l) / (G * J)
Where,
l = Length of the beam
G = Shear modulus of elasticity
As the length of the beam is not given, we cannot calculate the value of α.
Step 5: Calculate the maximum torsional stress (τ)
τ = (T * α) / (J * r)
To find the value of α, we need the length of the beam.
Hence, the given information is insufficient to calculate the maximum torsional stress induced in the beam.
Cross-sectional dimensions of the beam: 25 × 35 mm
Torque applied: 79 kN-m
Value of α: 1.87
To find:
Maximum torsional stress induced in the beam in MPa
Formula:
Torsional stress (τ) = (T * α) / (J * r)
Where,
T = Torque applied
α = Angle of twist per unit length
J = Polar moment of inertia
r = Distance from the center of the cross-section to the outermost fiber
Calculation:
Step 1: Calculate the polar moment of inertia (J)
J = (b * h^3) / 3
Where,
b = Width of the cross-section
h = Height of the cross-section
Given: b = 35 mm, h = 25 mm
J = (35 * (25^3)) / 3
J = 91,667 mm^4
Step 2: Calculate the distance from the center to the outermost fiber (r)
r = (b / 2) = (35 / 2) = 17.5 mm
Step 3: Convert the torque to N-mm
Torque (T) = 79 kN-m = 79 × 1000 N-mm
Step 4: Calculate the angle of twist per unit length (α)
α = (T * l) / (G * J)
Where,
l = Length of the beam
G = Shear modulus of elasticity
As the length of the beam is not given, we cannot calculate the value of α.
Step 5: Calculate the maximum torsional stress (τ)
τ = (T * α) / (J * r)
To find the value of α, we need the length of the beam.
Hence, the given information is insufficient to calculate the maximum torsional stress induced in the beam.
Free Test
FREE
| Start Free Test |
Community Answer
A beam of rectangular cross-section 25 × 35 mm is subjected to a torq...
Given: b = 25 mm, d = 35 mm,
T = 79 kN-m = 79,000 N-m and α = 1.87
Maximum torsional stress induced in the beam is
= 1.93 N/mm2
= 1.93 MPa
Attention Civil Engineering (CE) Students!
To make sure you are not studying endlessly, EduRev has designed Civil Engineering (CE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Civil Engineering (CE).
|
Explore Courses for Civil Engineering (CE) exam
|
|
Similar Civil Engineering (CE) Doubts
Top Courses for Civil Engineering (CE)View all
A beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. Can you explain this answer?
Question Description
A beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. 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 beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. 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 beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. Can you explain this answer?.
A beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. 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 beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. 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 beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. Can you explain this answer?.
Solutions for A beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE).
Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free.
Here you can find the meaning of A beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. Can you explain this answer?, a detailed solution for A beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. Can you explain this answer? has been provided alongside types of A beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A beam of rectangular cross-section 25 × 35 mm is subjected to a torque of 79 kN-m. The maximum torsional stress induced in the beam is ____________ MPa. (Take α = 1.87 for the beam) (Answer up to two decimal places)Correct answer is '1.93'. Can you explain this answer? tests, examples and also practice Civil Engineering (CE) tests.
|
|
Explore Courses for Civil Engineering (CE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup