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ENGR5004 / Week 8
Old spec Statics Session 16 ExampleTutorialOld spec ENGR2162:210 min

ENGR216 (old spec) Statics session 16 worked example (slides 9-12) official answers

An element is in plane stress with σx=100\sigma_x = 100 MPa (tension), σy=20\sigma_y = 20 MPa (compression) and a shear stress of 80 MPa acting upwards on the right face and rightwards on the top face (τxy=+80\tau_{xy} = +80 MPa). Determine:

Element ABCD: 100 MPa tension on the vertical faces, 20 MPa compression on the horizontal faces, 80 MPa shear (upwards on the right face BC, rightwards on the top face AB).
Element ABCD: 100 MPa tension on the vertical faces, 20 MPa compression on the horizontal faces, 80 MPa shear (upwards on the right face BC, rightwards on the top face AB).
Formulas you may need
  • Principal directions: tan⁡2θp=2τxyσx−σy\tan2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}; 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} (learn this)
  • Maximum shear: tan⁡2θs=−σx−σy2τxy\tan2\theta_s = -\dfrac{\sigma_x-\sigma_y}{2\tau_{xy}}, τmax=(σx−σy2)2+τxy2\tau_{max} = \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}, normal stress σ′=σx+σy2\sigma' = \dfrac{\sigma_x+\sigma_y}{2} (learn this)
  • Stress transformation σx′\sigma_{x'}, τx′y′\tau_{x'y'} (to decide which angle gives which stress) (learn this)
  1. (a)
    the principal planes;
  2. (b)
    the principal stresses, and which principal plane carries which;
  3. (c)
    the planes of maximum in-plane shear, the maximum shear stress, and the normal stress on those planes.