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ENGR5004 / Week 8
Old spec Statics Session 16 ProblemTutorialOld spec ENGR2162:115 min

ENGR216 (old spec) Statics session 16 worked example 2 (Problem, slides 13-16) official answers

The solid axle of an automobile, radius r=16r = 16 mm, is acted upon by the forces and torque shown: an upward 3 kN reaction at the wheel, a downward 3 kN load from the car body applied 0.15 m inboard of the wheel, and a torque of 350 N m. Point H is on the top of the axle, 0.2 m further inboard than the body load, in the central part of the (symmetric) axle where there is no shear force. Use axes with xx along the axle from the wheel towards H and yy normal to xx in the surface plane at H; with these axes the torque produces a negative shear stress τxy\tau_{xy} at H.

Determine:

Axle with the wheel at the left: 3 kN up at the wheel, 3 kN down 0.15 m inboard, torque 350 N m; point H on the top of the axle a further 0.2 m inboard.
Axle with the wheel at the left: 3 kN up at the wheel, 3 kN down 0.15 m inboard, torque 350 N m; point H on the top of the axle a further 0.2 m inboard.
Formulas you may need
  • Bending stress σmax=∣M∣cI\sigma_{max} = \dfrac{|M|c}{I}, solid circle I=πr44I = \dfrac{\pi r^4}{4} (on the formula sheet)
  • Torsion τmax=TrJ\tau_{max} = \dfrac{Tr}{J}, J=πr42J = \dfrac{\pi r^4}{2} (learn this)
  • Principal stresses σmax,min=σx+σy2±(σx−σy2)2+τxy2\sigma_{max,min} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}, tan⁡2θp=2τxyσx−σy\tan2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}, τmax=(σx−σy2)2+τxy2\tau_{max} = \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} (learn this)
  1. (a)
    the principal axes and principal stresses at point H on top of the axle;
  2. (b)
    the maximum shear stress at the same point.