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A cast iron test beam 20mm × 20mm in section and 1m long and supported at the end fails when a central load of 640 N is applied.
What uniformily distributed load will break a cantilever of the same material 50mm wide, 100mm deep and 2m long?
a) 5000 N/m
b) 500 N/mm
c) 50 N/mm
d) 500000 N/mm
Correct answer is option 'A'. Can you explain this answer?
What uniformily distributed load will break a cantilever of the same material 50mm wide, 100mm deep and 2m long?
a) 5000 N/m
b) 500 N/mm
c) 50 N/mm
d) 500000 N/mm
Correct answer is option 'A'. Can you explain this answer?
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A cast iron test beam 20mm ×20mm in section and 1m long andsupported a...
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A cast iron test beam 20mm ×20mm in section and 1m long andsupported a...
Given data:
Cast iron test beam: 20mm x 20mm x 1m, supported at the end, fails at a central load of 640 N
Cantilever beam: 50mm x 100mm x 2m
To find: Uniformly distributed load that will break the cantilever beam
Solution:
1. Calculation of bending stress in the test beam:
The moment of inertia of the beam can be calculated as follows:
I = (b x h^3)/12
where b = 20mm (width of beam), h = 20mm (height of beam)
I = (20 x 20^3)/12 = 26666.67 mm^4
The maximum bending moment that can be sustained by the beam can be calculated as follows:
M = (w x L^2)/8
where w = uniformly distributed load on the beam, L = length of the beam
M = (640 x 1^2)/8 = 80 Nm
The bending stress in the beam can be calculated using the bending moment and the moment of inertia:
σ = M x y / I
where y = distance from the neutral axis to the outer fiber (half the height of the beam)
σ = (80 x 10^6) x (10/20) / 26666.67 = 300 MPa
2. Calculation of maximum bending stress in the cantilever beam:
The moment of inertia of the beam can be calculated as follows:
I = (b x h^3)/12
where b = 50mm (width of beam), h = 100mm (height of beam)
I = (50 x 100^3)/12 = 4.17 x 10^7 mm^4
The maximum bending moment that can be sustained by the beam can be calculated as follows:
M = (w x L^2)/2
where w = uniformly distributed load on the beam, L = length of the beam
M = (w x 2^2)/2 = 2w Nm
The maximum bending stress in the beam can be calculated using the bending moment and the moment of inertia:
σ = M x y / I
where y = distance from the neutral axis to the outer fiber (half the height of the beam)
σ = (2w x 10^6) x (50/2) / 4.17 x 10^7 = 24 MPa
3. Calculation of uniformly distributed load that will break the cantilever beam:
The maximum bending stress that the cantilever beam can sustain is the yield stress of the cast iron material.
From the stress-strain curve of cast iron, the yield stress is approximately 150 MPa.
Therefore, we can equate the maximum bending stress in the cantilever beam to the yield stress of the material:
24 MPa = 150 MPa
2w x 10^6 = (150 x 4.17 x 10^7)/(50/2)
w = 5000 N/m
Therefore, the uniformly distributed load that will break the cantilever beam is 5000 N/m.
Cast iron test beam: 20mm x 20mm x 1m, supported at the end, fails at a central load of 640 N
Cantilever beam: 50mm x 100mm x 2m
To find: Uniformly distributed load that will break the cantilever beam
Solution:
1. Calculation of bending stress in the test beam:
The moment of inertia of the beam can be calculated as follows:
I = (b x h^3)/12
where b = 20mm (width of beam), h = 20mm (height of beam)
I = (20 x 20^3)/12 = 26666.67 mm^4
The maximum bending moment that can be sustained by the beam can be calculated as follows:
M = (w x L^2)/8
where w = uniformly distributed load on the beam, L = length of the beam
M = (640 x 1^2)/8 = 80 Nm
The bending stress in the beam can be calculated using the bending moment and the moment of inertia:
σ = M x y / I
where y = distance from the neutral axis to the outer fiber (half the height of the beam)
σ = (80 x 10^6) x (10/20) / 26666.67 = 300 MPa
2. Calculation of maximum bending stress in the cantilever beam:
The moment of inertia of the beam can be calculated as follows:
I = (b x h^3)/12
where b = 50mm (width of beam), h = 100mm (height of beam)
I = (50 x 100^3)/12 = 4.17 x 10^7 mm^4
The maximum bending moment that can be sustained by the beam can be calculated as follows:
M = (w x L^2)/2
where w = uniformly distributed load on the beam, L = length of the beam
M = (w x 2^2)/2 = 2w Nm
The maximum bending stress in the beam can be calculated using the bending moment and the moment of inertia:
σ = M x y / I
where y = distance from the neutral axis to the outer fiber (half the height of the beam)
σ = (2w x 10^6) x (50/2) / 4.17 x 10^7 = 24 MPa
3. Calculation of uniformly distributed load that will break the cantilever beam:
The maximum bending stress that the cantilever beam can sustain is the yield stress of the cast iron material.
From the stress-strain curve of cast iron, the yield stress is approximately 150 MPa.
Therefore, we can equate the maximum bending stress in the cantilever beam to the yield stress of the material:
24 MPa = 150 MPa
2w x 10^6 = (150 x 4.17 x 10^7)/(50/2)
w = 5000 N/m
Therefore, the uniformly distributed load that will break the cantilever beam is 5000 N/m.
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A cast iron test beam 20mm ×20mm in section and 1m long andsupported at the end fails when acentral load of 640 N is applied.What uniformily distributed loadwill break a cantilever of the same material 50mm wide, 100mm deepand 2m long?a)5000 N/mb)500 N/mmc)50 N/mmd)500000 N/mmCorrect answer is option 'A'. Can you explain this answer?
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