Question 7.1: For the state of plane stress shown in Fig. 7.13, determine ...
For the state of plane stress shown in Fig. 7.13, determine (a) the principal planes, (b) the principal stresses, (c) the maximum shearing stress and the corresponding normal stress.
(a) Principal Planes. Following the usual sign convention, we write the stress components as
s _{x}=+50 MPa s _{y}=-10 MPa t _{x y}=+40 MPa
Substituting into Eq. (7.12), we have
\tan 2 u _{p}=\frac{2 t _{x y}}{ s _{x}- s _{y}}=\frac{2(+40)}{50-(-10)}=\frac{80}{60}
2 u _{p}=53.1^{\circ} and 180° + 53.1° = 233.1°
u _{p}=26.6^{\circ} and 116.6°
(b) Principal Stresses. Formula (7.14) yields
s _{\max , \min }=\frac{ s _{x}+ s _{y}}{2} \pm \sqrt{\left(\frac{ s _{x}- s _{y}}{2}\right)^{2}+ t _{x y}^{2}}=20 \pm \sqrt{(30)^{2}+(40)^{2}}
s _{\max }=20+50=70 MPa
s _{\min }=20-50=-30 MPa
The principal planes and principal stresses are sketched in Fig. 7.14. Making u = 26.6° in Eq. (7.5), we check that the normal stress exerted on face BC of the element is the maximum stress:
s _{x^{\prime}}=\frac{ s _{x}+ s _{y}}{2}+\frac{ s _{x}- s _{y}}{2} \cos 2 u + t _{x y} \sin 2 u (7.5)
s _{x^{\prime}}=\frac{50-10}{2}+\frac{50+10}{2} \cos 53.1^{\circ}+40 \sin 53.1^{\circ}
= 20 + 30 cos 53.1° + 40 sin 53.1° = 70 MPa = s_{\max }
(c) Maximum Shearing Stress. Formula (7.16) yields
t _{\max }=\sqrt{\left(\frac{ s _{x}- s _{y}}{2}\right)^{2}+ t _{x y}^{2}}=\sqrt{(30)^{2}+(40)^{2}}=50 MPaSince s_{\max } and s_{\min } have opposite signs, the value obtained for t _{\max } actually represents the maximum value of the shearing stress at the point considered. The orientation of the planes of maximum shearing stress and the sense of the shearing stresses are best determined by passing a section along the diagonal plane AC of the element of Fig. 7.14. Since the faces AB and BC of the element are contained in the principal planes, the diagonal plane AC must be one of the planes of maximum shearing stress (Fig. 7.15). Furthermore, the equilibrium conditions for the prismatic element ABC require that the shearing stress exerted on AC be directed as shown. The cubic element corresponding to the maximum shearing stress is shown in Fig. 7.16. The normal stress on each of the four faces of the element is given by Eq. (7.17):
s ^{\prime}= s _{ ave }=\frac{ s _{x}+ s _{y}}{2}=\frac{50-10}{2}=20 MPa
