SOM Chap - 3 (Torsion) Question Bank (2)

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Transcript of SOM Chap - 3 (Torsion) Question Bank (2)

  • AARUPADAI VEEDU INSTITUTE OF TECHNOLOGY, PAIYANOOR

    &

    V.M.K.V. ENGINEERING COLLEGE

    (II YEAR / IV SEMESTER)

    (CBCS 2013 REGULATION)

    STRENGTH OF MATERIALS

    QUESTION BANK

    UNIT III

    PART-A

    1. Write torsional equation.

    =

    =

    Where, T Torque N/mm J Polar moment of inertia mm4 Shear stress N/mm2

    R Radius mm C Modulus of rigidity N/mm2 Angle of twist rad l length - mm

    2. Write the polar modulus for solid shaft and circular shaft.

    For solid shaft,

    Polar modulus,

    Where, J =

    D4

    For hollow shaft,

    Polar modulus,

  • Where, J =

    [ ]

    3. Write an expression for the angle of twist for a hollow circular shaft with external

    diameter D, internal diameter d, length l and rigidity modulus G.

    Where,

    J (polar moment of inertia) =

    [ ]

    D External diameter, mm

    d Internal diameter, mm

    T Torque transmitted, N-mm

    C Modulus of rigidity, N/mm2

    Angle of twist, rad

    l Length of shaft, mm

    4. Define torsional rigidity.

    The torsional equation is,

    =

    Since C, l, J are constant for a given shaft, (angle of twist) is directly proportional to T

    (torque). The term CJ is known as torsional rigidity and it is represented by K.

    5. Calculate the maximum torque that a shaft of 125mm diameter can transmit, if the

    maximum angle of the twist is 1 in a length of 1.5m. Take C=70103 N/mm2.

    Given data:

    Diameter, D = 125mm

    Angle of twist, = 1

    Length, l = 1.5 m = 1500 mm

    Modulus of rigidity, C = 70103 N/mm2

    To find: Maximum Torque, Tmax

  • Solution:

    Torsional Equation

    T =

    =

    T = Tmax = 19.01106 N-mm

    6. Differentiate between close-coiled and open-coiled helical springs.

    Close coiled helical springs Open coiled helical springs

    Adjacent are very close to each other Large gap between adjacent coils

    Only tensile load can carry Tensile and compressive loads can carry

    Helix angle is negligible Helix angle is considerable

    7. A helical spring is made of 4mm steel wire with a mean radius of 25mm and number of

    turns of coil 15. What will be deflection of the spring under a load of 6N? Take C=80103

    N/mm2.

    Given:

    d = 4mm

    R = 25mm

    n = 15

    W = 6N

    C = 80103 N/mm2

    Solution:

    Axial deformation,

    =

    =

    4 9 mm

  • 8. Give shear stress and deflection relation for close-coiled helical springs.

    Deflection, =

    Shear Stress, =

    9. The stiffness of spring is 10N/mm. what is the axial deformation in the spring when a load

    is 50N is acting?

    Given:

    K = 10N/mm

    W = 50N

    Solution:

    K =

    =

    =

    = 5mm.

    10. An open coiled helical spring of mean radius of coil of 20cm and helix angle of 12 is

    subjected to an axial load of 10N. What is the bending moment in the coil?

    Given:

    R = 20 cm = 200mm

    = 12

    W = 10N

    Solution:

    Bending Moment, M = WR sin

    = 10200sin 12

    M = 415.82 N-mm

  • 11. The stiffness of a spring is 10N/mm and the axial deflection is 10mm. What is the axial

    load on the spring?

    Given:

    K =10 N/mm

    = 10mm

    Solution:

    K =

    => W = K = 1010 = 100 N.

    12. Write the equation for the deflection of an open coiled helical spring subjected to an axial

    load W.

    Deflection, =

    [

    ]

    13. Write down the equation for shear strain energy of a close coiled spring.

    Maximum Shear Stress, =

    14. Write the expression for vertical deflection of a close coiled helical spring due to vertical

    load W.

    Deflection, =

    15. What is meant by stiffness? What is the formula for the stiffness of a close coiled helical

    spring subjected to an axial load?

    The stiffness of the spring is defined as the load required producing unit deflection.

    Stiffness, K =

    N/mm.

  • 16. What are the assumptions made in torsion equation?

    The material of the shaft is homogeneous, perfectly elastic and obeys Hookes Law.

    Twist is uniform along the length of the shaft.

    The stress does not exceed the limit of proportionality

    The shaft circular in section remains circular after loading

    Strain and deformation are small.

    17. Why hollow circular shafts are preferred when compared to solid circular shafts?

    The torque transmitted by the hollow shaft is greater than the solid shaft.

    For some material, length and given torque, the weight of the hollow shaft will be

    less compared to solid shaft.

    18. Write down the expression for power transmitted by a shaft.

    Power,

    Where, t Torque in kN-m

    N Speed in rpm

    P Power in Kw

    19. Write down the equation for the maximum shear stress of a solid circular section in

    diameter D when subjected to torque T.

    For Solid Shaft

    Torque, T =

    =

    Where, Shear stress in N/mm2

    T Torque in N-mm

    D Diameter in mm

  • 20. Write down the expression for torque transmitted by hollow shaft.

    T =

    [

    ]

    Where, T torque in N-mm

    Shear Stress in N/mm2

    D- Outer diameter in mm

    d Inner diameter in mm

    PART-B

    1. A solid circular shaft transmits 75kW at 200rpm. Find the shaft diameter if the twist in the

    shaft is not to exceed 1 in 2m length of the shaft and the shearing stress is limited to

    50N/mm2. Take G=100GN/mm

    2.

    Given data:

    P = 75 KW

    N = 200 rpm

    = 1

    = 0.017 rad

    L = 2 m = 2000 mm

    = 50 N/mm2

    C or G = 100 GN/m2

    = 100109 N/m

    2 = 10010

    3 N/mm

    2

    To find:

    Diameter of the shaft

    Solution:

    We know that,

    Power, P =

    75 =

    T = 3.58 kN-m = 3.58 103 N-m

  • Torque, T = 3.58 106 N-mm

    Shear stress and angle of twist both are given.

    First Case: Considering the shear stress ()

    Torque, T =

    3.58 106=

    D = 71.4 mm

    Second case: Considering angle of twist ()

    Where, J =

    =

    D = 80.9 mm

    From the above two cases, we find that suitable diameter for the shaft is 80.9mm (i.e.

    greater of two values).

    Result:

    Shaft Diameter, D = 80.9mm

  • 2. A hollow shaft of diameter ratio 3/8 is required to transmit 588kW at 100rpm. The

    maximum torque exceeds the mean by 20%. The shear stress is limited to 63N/mm2 and the

    twist should not be more than 0.0081rad. Calculate the external diameter required

    satisfying both the conditions. Take G=84Gpa. Length 3m.

    Given:

    P = 588kW

    N = 110rpm

    =

    d = 0.375 D

    Tmax = 1.2 Tmean

    = 63 N/mm2

    = 0.0081 rad

    G or C = 84 GPa = 84 109 N/m

    2

    = 84 N/mm2

    l = 3m = 3000mm

    To find:

    External Diameter (D)

    Solution:

    We know that,

    Power, P =

    588 =

    T = 51.04 kN-m

    = 51.04 N-m

    = 51.04 N-mm

    We know that, Tmax = 1.2 Tmean

    = 1.2 51.04 N-mm

    Tmax = 61.2 N-mm

    Shear stress and angle of twist both are given.

    First Case: considering the shear stress ()

    Torque, Tmax =

    [

    ]

  • 61.2 = [

    4 ( 4)

    ]

    61.2106 =

    [ ]

    External Diameter, D = 171.5 mm

    We Know that, d = 0.375D = 0.375 171.5

    Internal Diameter, d = 64.3 mm

    Second Case: Considering angle of twist ()

    Where J =

    [ ]

    [ ]

    =

    D4-d4 = 2.7109

    D4 (0.375 D) 4 = 2.7 109

    D4 [ ] = 2.7 109

    External Diameter, D = 229.09 mm

    We know that, d = 0.375D

    Internal Diameter, d = 85.90 mm

    From the above two cases, we find that external diameter of the shaft is 229.09 mm and

    internal diameter is 85.90 (i.e. greater of the two values)

    Result:

    1. External Diameter, D = 229.09 mm

    2. Internal Diameter, d= 85.90 mm

  • 3. A shaft is required to transmit power of 300kW running at a speed of 120rpm. If the shear

    strength of the shaft material is 70N/mm2. Design a hollow shaft with inner diameter equal

    to 0.75 times the outer diameter.

    Given Data:

    P = 300 kW

    N = 120 rpm

    = 70 N/mm2

    d = 0.75 D

    To find:

    Inner Diameter (d) and outer diameter (D)

    Solution:

    We know that,

    Power, P =

    300 =

    T = 23.87 kN-m = 23.87 103 N-m

    Shear stress is given. For hollow Shaft

    Torque, T =

    [

    ]

    23.87 106 =

    [

    ]

    23.87106 =

    [ ]

    External Diameter, D = 136.45 mm

    We know that, d = 0.75 D

    = 0.75 136.45

    Internal Diameter, d = 102.33 mm

    Result:

    1. External Diameter, D = 136.45 mm

    2. Internal Diameter, d = 102.33 mm

  • 4.