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A cantilever beam of length 2 m with a square section of side length 0.1 m is loaded vertically at the free end. The vertical displacement at the free end is 5 mm. The beam is made of steel with Young’s modulus of 2.0 x 1011 N/m2. The maximum bending stress at the fixed end of the cantilver is
- a)20.0 MPa
- b)37.5 MPa
- c)60.0 MPa
- d)75.0 MPa
Correct answer is option 'B'. Can you explain this answer?
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A cantilever beam of length 2 m with a square section of side length 0...
Deflection
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A cantilever beam of length 2 m with a square section of side length 0...
's modulus of 200 GPa. The moment of inertia for a square section is (1/12) * a^4, where 'a' is the side length of the square.
To find the vertical displacement at the free end of the cantilever beam, we can use the formula for deflection in a cantilever beam under a point load:
δ = (F * L^3) / (3 * E * I)
Where:
δ = vertical displacement at the free end
F = load applied at the free end
L = length of the cantilever beam
E = Young's modulus
I = moment of inertia
In this case, F = mg, where m is the mass and g is the acceleration due to gravity. Since the load is applied vertically, F = mg.
Given:
L = 2 m
a = 0.1 m
δ = 5 mm = 0.005 m
E = 200 GPa = 200 * 10^9 Pa
First, let's calculate the moment of inertia for the square section:
I = (1/12) * a^4
I = (1/12) * (0.1^4)
I = 0.0000083333 m^4
Next, we need to convert the load into force:
F = mg
To calculate the mass, we need to find the volume of the square section:
Volume = a^2 * L
Volume = 0.1^2 * 2
Volume = 0.02 m^3
The mass can be calculated using the density of steel, which is approximately 7850 kg/m^3:
m = density * volume
m = 7850 * 0.02
m = 157 kg
Now, we can calculate the force:
F = mg
F = 157 * 9.8
F = 1538.6 N
Finally, we can substitute the values into the formula for deflection:
δ = (F * L^3) / (3 * E * I)
0.005 = (1538.6 * 2^3) / (3 * 200 * 10^9 * 0.0000083333)
0.005 = 6154.4 / (0.05 * 10^9 * 0.0000083333)
0.005 = 6154.4 / (4.1667 * 10^-6)
0.005 = 1476614400
Therefore, the vertical displacement at the free end of the cantilever beam is approximately 1476614400 mm.
To find the vertical displacement at the free end of the cantilever beam, we can use the formula for deflection in a cantilever beam under a point load:
δ = (F * L^3) / (3 * E * I)
Where:
δ = vertical displacement at the free end
F = load applied at the free end
L = length of the cantilever beam
E = Young's modulus
I = moment of inertia
In this case, F = mg, where m is the mass and g is the acceleration due to gravity. Since the load is applied vertically, F = mg.
Given:
L = 2 m
a = 0.1 m
δ = 5 mm = 0.005 m
E = 200 GPa = 200 * 10^9 Pa
First, let's calculate the moment of inertia for the square section:
I = (1/12) * a^4
I = (1/12) * (0.1^4)
I = 0.0000083333 m^4
Next, we need to convert the load into force:
F = mg
To calculate the mass, we need to find the volume of the square section:
Volume = a^2 * L
Volume = 0.1^2 * 2
Volume = 0.02 m^3
The mass can be calculated using the density of steel, which is approximately 7850 kg/m^3:
m = density * volume
m = 7850 * 0.02
m = 157 kg
Now, we can calculate the force:
F = mg
F = 157 * 9.8
F = 1538.6 N
Finally, we can substitute the values into the formula for deflection:
δ = (F * L^3) / (3 * E * I)
0.005 = (1538.6 * 2^3) / (3 * 200 * 10^9 * 0.0000083333)
0.005 = 6154.4 / (0.05 * 10^9 * 0.0000083333)
0.005 = 6154.4 / (4.1667 * 10^-6)
0.005 = 1476614400
Therefore, the vertical displacement at the free end of the cantilever beam is approximately 1476614400 mm.
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A cantilever beam of length 2 m with a square section of side length 0.1 m is loaded vertically at the free end. The vertical displacement at the free end is 5 mm. The beam is made of steel with Young’s modulus of 2.0 x 1011 N/m2. The maximum bending stress at the fixed end of the cantilver isa)20.0 MPab)37.5 MPac)60.0 MPad)75.0 MPaCorrect answer is option 'B'. Can you explain this answer?
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