Strength Of Materials Miscellaneous

Strength Of Materials Miscellaneous

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Strength Of Materials

Strength Of Materials Miscellaneous

Direction: A simply supported beam of span length 6 m and 75 mm diameter carries a uniformly distributed load of 1.5 kN/m.

  1. 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 MPa

    Correct 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

  1. The homogenous state of stress for a metal part undergoing plastic deformation is
    T = 5100
    5100
    5100

    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 MPa

    Correct 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

  1. 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
    88

    Correct Option: C

    l = 6m, w = 1.5KN/m

    M =
    wl²
    		 =
    1.5 × 10³ × 6²
    		 = 6.75 kNm
    88

  1. If the two principal strains at a point are 1000 x 10-6 and –600 × 10-6, then the maximum shear strain is








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    τmax =
    φmax
    		,radius
    ∊1 - ∊2
    22

    φmax = ∊1 = ∊2
    = 1000 × 10–6 - (-600 × 10–6)
    φmax = 1600 × 10–6

    Correct Option: C


    τmax =
    φmax
    		,radius
    ∊1 - ∊2
    22

    φmax = ∊1 = ∊2
    = 1000 × 10–6 - (-600 × 10–6)
    φmax = 1600 × 10–6

  1. 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 = 10kN

    Correct 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