Strength Of Materials Miscellaneous
- The homogenous state of stress for a metal part undergoing plastic deformation is
T = 
5 10 0 
5 10 0 5 10 0
Where the stress component values are in MPa. Using Von Mises yield criterion, the value of estimated shear yield stress, in MPa is
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σeq = √1/2{(σ11 - σ22)² + (σ11 - σ22)² + 6[σ²12 + σ²23 + σ²13]}
We know, σ11 = 10, σ22 = 20,
σ33 = – 10;
σ12 = 5;
σ23 = σ13 = 0
∴ σeq = 27.839 MPa
Shear stress at yield, τy = σeq/√3
= 16.07 MPaCorrect Option: B
σeq = √1/2{(σ11 - σ22)² + (σ11 - σ22)² + 6[σ²12 + σ²23 + σ²13]}
We know, σ11 = 10, σ22 = 20,
σ33 = – 10;
σ12 = 5;
σ23 = σ13 = 0
∴ σeq = 27.839 MPa
Shear stress at yield, τy = σeq/√3
= 16.07 MPa
Direction: A simply supported beam of span length 6 m and 75 mm diameter carries a uniformly distributed load of 1.5 kN/m.
- What is the maximum value of bending moment?
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l = 6m, w = 1.5KN/m
M = wl² = 1.5 × 10³ × 6² = 6.75 kNm 8 8 Correct Option: C
l = 6m, w = 1.5KN/m
M = wl² = 1.5 × 10³ × 6² = 6.75 kNm 8 8
- What is the maximum value of bending stress?
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Bending stress = (Maximum load × distance)
σb = or Bending moment Section modulus σb = 9 × 10³(Nm) = 211.26 MPa π/32 × (0.075)³
Hence closest option is 162.98 MPaCorrect Option: A
Bending stress = (Maximum load × distance)
σb = or Bending moment Section modulus σb = 9 × 10³(Nm) = 211.26 MPa π/32 × (0.075)³
Hence closest option is 162.98 MPa
- A cantilever beam OP is connected to another beam PQ with a pin joint as shown in the figure. A load of 10 kN is applied at the mid-point of PQ.
The magnitude of bending moment (in kN–m) at fixed end O is ________
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P-being internal hinge
∑MP = 0 ...(1)
Condition ‘PQ’, reaction at Q i.e. RQ = 10 kN
Now from (1), ∑MP = 0
M0 – RQ × 1 + RC × 0.5 = 0
⇒ M0 = 10kNCorrect Option: C
P-being internal hinge
∑MP = 0 ...(1)
Condition ‘PQ’, reaction at Q i.e. RQ = 10 kN
Now from (1), ∑MP = 0
M0 – RQ × 1 + RC × 0.5 = 0
⇒ M0 = 10kN
- A cantilever beam having square cross-section of side a is subjected to an end load, If a is increased by 19%, the tip deflection decreases approximately by
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For cantilever beam with an end load.
δ = Pl³ 3EI δ ∝ 1 I δ1 = I2 δ2 I1 δ1 = I(1.19a) = 2.005 δ2 a4 δ2 = δ1 = 0.5δ1 0.005 δ1 - δ2 × 100 = 50% δ1 Correct Option: D
For cantilever beam with an end load.
δ = Pl³ 3EI δ ∝ 1 I δ1 = I2 δ2 I1 δ1 = I(1.19a) = 2.005 δ2 a4 δ2 = δ1 = 0.5δ1 0.005 δ1 - δ2 × 100 = 50% δ1
