Question 7.2: A plane-stress condition exists at a point on the surface of......
A plane-stress condition exists at a point on the surface of a loaded structure, where the stresses have the magnitudes and directions shown on the stress element of Fig. 7-8a. Determine the stresses acting on an element that is oriented at a clockwise angle of 15° with respect to the original element.
The stresses acting on the original element (Fig. 7-8a) have the following values:
σ_x = -46 MPa σ_y = 12 MPa τ_{xy} = -19 MPa
An element oriented at a clockwise angle of 15° is shown in Fig. 7-8b, where the x_1 axis is at an angle θ = -15° with respect to the x axis. (As an alternative, the x_1 axis could be placed at a positive angle θ = 75°.)
We can readily calculate the stresses on the x_1 face of the element oriented at θ = -15° by using the transformation equations (Eqs. 7-4a and 7-4b). The calculations proceed as follows:
\frac{σ_x+σ_y}{2} = -17 MPa \frac{σ_x\ -\ σ_y}{2} = -29 MPa
sin 2θ = sin (-30°) = -0.5 cos 2θ = cos (-30°) = 0.8660
Substituting into the transformation equations, we get
σ_{x_1}=\frac{σ_x+σ_y}{2}+\frac{σ_x\ -\ σ_y}{2}cos\ 2θ+τ_{xy}sin\ 2θ
= -17 MPa + (-29 MPa)(0.8660) + (-19 MPa)(-0.5)
= -32.6 MPa
τ_{x_1y_1}=-\frac{σ_x\ -\ σ_y}{2}sin\ 2θ+τ_{xy}cos\ 2θ
= -(-29 MPa)(-0.5) + (-19 MPa)(0.8660)
= -31.6 MPa
Also, the normal stress acting on the y_1 face (Eq. 7-5) is
σ_{y_1}=\frac{σ_x+σ_y}{2}\ -\ \frac{σ_x\ -\ σ_y}{2}cos\ 2θ\ -\ τ_{xy}sin\ 2θ
= -17 MPa – (-29 MPa)(0.8660) – (-19 MPa)(-0.5)
= -1.4 MPa
This stress can be verified by substituting θ = 75° into Eq. (7-4a). As a further check on the results, we note that σ_{x1}+σ_{y1}=σ_{x}+σ_{y}.
The stresses acting on the inclined element are shown in Fig. 7-8b, where the arrows indicate the true directions of the stresses. Again we note that both stress elements shown in Fig. 7-8 represent the same state of stress.