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The stress as a function of angle is shown below in the figure. Maximum shear stress occurs at 0°±90° and the maximum n

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The stress as a function of angle is shown below in the figure. Maximum shear stress occurs at 0°±90° and the maximum n

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  1. A bar is shown in (a) in pure shear due to torsion. Pure shear means that the only forces acting on the element are shear stresses. The directions of these stresses depend on the direction of the applied torques. Assuming the right hand rule end cb will rotate in the sense shown in (b). The shear stresses shown in (b) are by sign convention positive. A shear stress on a positive face is positive if it acts in the direction of one of the positive axes.

  2. To determine the state of stress as a function of angle we will cut an oblique section at an angle  this results in a triangle with AB unchanged but all other lengths altered. The state of pure shear is shown in (a). In (b) we take an oblique section and resolve the resultant force into two components  and . The resolved components can be rewritten as shown in (c) where we have equated the forces and used the relationship that F= A= .L.t since t is constant for the element A=A0.L (NB Secant=h/a)

  3. The stress as a function of angle is shown below in the figure. Maximum shear stress occurs at 0°±90° and the maximum normal stresses at ±45°.

  4. At 45° there are equal and opposite normal stresses acting on the element. This often leads to torsion failures at 45° to the torsion axis.

  5. To determine the deformation of the bar we need to consider shear strains and normal strains. The shear distortion tends to deform a rectangle to a parallelepiped. The normal stresses at 45° will elongate the section but lateral contraction will also occur due to poisson contraction. Similarly the compressive stress at 135° will lead to a compression of the element but will also extend the element at 45°. The total normal strain in the 45° direction is given by

  6. After deformation Using cosine rule c2=a2+b2-2abcos 

  7. For small strains Earlier we derived Therefore

  8. Work Done = Force x distance = torque x angle moved Power =Rate of Work Done  Is in rads.s-1 Very often rate is given in r.p.m and Power is given in hp 1 hp = 550 ft-lb/s.

  9. A solid steel shaft AB is shown in the Figure. It is to be used to transmit 5 hp from the motor M to which it is attached. If the shaft rotates at =175 rpm and the steel has an allowable shear stress of allow=14.5 ksi, determine the required diameter of the shaft to the nearest 1/8 in.

  10. A solid steel shaft AB is shown in the Figure. It is to be used to transmit 5 hp from the motor M to which it is attached. If the shaft rotates at =175 rpm and the steel has an allowable shear stress of allow=14.5 ksi, determine the required diameter of the shaft to the nearest 1/8 in. Therefore d of the shaft should be greater than 0.858in i.e. 7/8=0.875in

  11. A solid shaft ABC of 50 mm diameter is driven at A by a motor that transmits 50 kW to the shaft at 10 Hz. The gears at B and C drive machinery requiring power equal to 35 kW and 15 kW respectively. Compute the maximum shear stress in the shaft and the angle of twist ac between the motor at A and the gear at C. (Use G=80 GPa).

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