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Strength of Materials Test 5
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Strength of Materials Test 5
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  • Question 1/10
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    If section modulus of a beam is increased, the bending stress in the beam will:
    Solutions

    Concept:

    Bending Moment Equation

    MI=σy=ER

    M=Iyσ=Zσ

    For the constant bending moment: σ ∝ 1/Z

    So, bending stress is inversely proportional to the Section Modulus

    The strength of two beams of the same material can be compared by the section modulus values.

    The beam is stronger when section modulus is more, the strength of the beam depends on section modulus.

  • Question 2/10
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    The bending moment distribution in a beam is a function of distance x and it is given by M = (5x2 + 20x - 7) Nm. Find the shear stress at x = 2m if the shearing area is 2 mm2.

    Solutions

    Concept:

    shearforce,F=dMdx

    Calculation:

    Bending moment (M) = (5x2 + 20x - 7)

    Now,

    shearforce,F=dMdx

    ∴ F = 10x + 20

    Now,

    (F)(x = 2 m) = 10(2) + 20

    F = 40 N

    Now,

    shear stress (τ)=FA=402

    ∴ τ = 20 MPa
  • Question 3/10
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    A circular cantilever beam is subjected to torsion at the tip. Because of tension, the angle of twist is found to be 0.5° as shown in the figure. The length of beam (in mm) is _______. Take (R = 100 mm, ∅ = 0.1°)

    Solutions

    Concept:

    Arc length ‘S’ is given by,

    S = Rθ

    Also, in the case of torsion, we can write,

    S = L∅

    Calculation:

    ∴ L∅ = Rθ

    L=Rθ

    L=100×0.50.1

    ∴ L = 500 mm
  • Question 4/10
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    If a shaft is simultaneously subjected to torque T/2 and a bending moment of 2M, then the ratio of maximum bending stress to maximum shear stress will be______
    Solutions

    Concept:

    Maximum bending across shaft occurs at either top or bottom fiber and it is given by the bending formula:

    σmax=MZ=2Mπ64d4×d2=64Mπd3

    Maximum shear stress is given by the twisting formula:

    τmax=TrI=(12)×(d2)π32d4=8Tπd3

    σmaxτmax=64Mπd38Tπd3=8MT

  • Question 5/10
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    A rectangle beam 60 mm wide and 150 mm deep is simply supported over a span of 4 m. If the beam is subjected to uniformly distributed load of 4.5 kN/m, then the maximum bending stress induced in the beam is _____ MPa.

    Solutions

    Concept:

    Section modules of rectangular section, Z=db26

    Calculation:

    Given:

    b = 60 mm, d = 150 mm

    Z=bd26=60×15026=225×102mm3

    Now,

    The maximum bending moment at the center of a simply supported beam subjected to UDL is, M=wl28

    w = 4.5 kN/m, l = 4 m = 4 × 103 mm

    M=wl28=4.5×(4×103)28=9×106Nmm

    σbmax=MZ=9×106225×103

    ∴ σmax = 40 MPa

    Points to Remember:

    i) For rectangular section Z=bd26

    ii) Maximum Bending moment in a simply supported beam subjected to UDL in at center and M=wl28

    iii) Maximum Bending Stress =MZ

  • Question 6/10
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    A solid circular beam with a radius of 0.25 m and length of 2 m is subjected to a twisting moment of 20 kNm about the z-axis at the free end, which is the only load acting as shown in the figure. The shear stress component τxy at point ‘M’ in cross-section of the beam at a distance of 1 m from the fixed end is

    A picture containing objectDescription automatically generated

    Solutions

    Concept:

    For the given beam:

    Let the point M is displaced from its original position under the action of external torque acting on the bar.

    The torque so applied along the z-axis would displace the particle M in the y-z and x-z plane.

    ∴ The displacement in the y-z and x-z plane induces strain which is the result of stress in that plane.

    As there is no displacement in the x-z plane, no strain and hence no stress is introduced in the x - y plane.

    ∴ The stress produced in the x-y plane is 0 (zero).

  • Question 7/10
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    A solid circular shaft of diameter 80mm is subjected to a bending moment M and twisting moment T. Maximum bending stress due to M alone is equal to 1.5 times maximum shear stress due to T alone. The magnitude of maximum principle stress developed is 100 MPa. Torsion T on the shaft is _____ kN-m.
    Solutions

    Concept:

    Using the flexural formula,

    MI=ER=σby

    And,

    σb=M.yI

    For the circular shaft, y=d2 and I=π64d4

    σb=32Mπd3

    Using torsion equation,

    TJ=GθL=τr

    And,

    τ=T.rJ

    For the circular shaft, τ=τmax at r=d2 and J=π32d4

    τ=16Tπd3

    Calculation:

    Using the given condition,

    σb=1.5×τ

    32Mπd3=1.5×16Tπd3

    M=0.75T

    Maximum principle stress is given as,

    σ1=16πd3[M+M2+T2]

    100×106=16π×0.083[0.75T+(0.75T)2+T2]

    ∴ T = 5.0265 kN-m
  • Question 8/10
    1 / -0

    A cantilever beam is subjected to uniformly distributed torsion throughout its length. Find out the angle of twist at the free end. Where t = torsion per unit length (Nm/m)

    Solutions

    Explanation:

    Consider an element of length ‘dx’ at distance ‘x’ from fixed support of cantilever beam.

    Torsion act at section x-x is,

    Txx=t.x

    Using torsion equation,

    TJ=GθL=τr

    The angle of twist,

    dθ=TxxdxGJ

    Integrating,

    0θdθ=0LTxxdxGJ

    0θdθ=0Lt.xdxGJ

    θ=[tx22GJ]0L

    ∴ θ=tL22GJ

  • Question 9/10
    1 / -0

    A solid shaft of diameter d and length l is subjected to torque T. Another shaft B of length 1.5l with the same material and half the diameter is subjected to torque 3T4.The ratio of the angle of twist of shaft B to that of shaft A is _____.
    Solutions

    Concept:

    Using torsion equation,

    TJ=Gθl=τr

    Angleoftwist,θ=TlGJ

    Calculation:

    Given:

    dB=0.5dA

    lB=1.5lA

    TA=TandTB=0.75T

    θB=TBlBG×π32(d2)4

    θB=0.75T×1.5lG×π32(d2)4=576TlGπd4

    θA=TAlAG×π32d4

    θA=TlG×π32d4=32TlGπd4

    θBθA=576TlGπd432TlGπd4=18

  • Question 10/10
    1 / -0

    A wooden circular log is to be used to make a rectangular cross-section beam of highest strength against bending, what will be the ratio of depth to breadth of the beam?
    Solutions

    Concept:

    Using the flexural formula,

    MI=ER=σby

    And,

    σb=M.yI

    i.e. strength in bending depends on geometrical parameters section modulus, Z=(Iy)

    Z=Iy=bd312d2=bd26

    Also from the geometry,

    D2=b2+d2

    d2=D2b2     ...(1)

    Z=b(D2b2)6

    Differentiating Z w.r.t. b and equating to zero, since rectangular cross-section beam of highest strength against bending is mentioned.

    dZdb=(D23b2)6=0

    D2=3b2     ... (2)

    Put equation (2) in equation (1)

    d2=3b2b2

    (db)2=2

    (db)=2

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