Question 7.7: An element of material in plane strain undergoes the followi......
An element of material in plane strain undergoes the following strains:
\epsilon _x = 340 × 10^{-6}\quad \quad \epsilon _y = 110 × 10^{-6}\quad \quad γ_{xy} = 180 × 10^{-6}
These strains are shown highly exaggerated in Fig. 7-36a, which shows the deformations of an element of unit dimensions. Since the edges of the element have unit lengths, the changes in linear dimensions have the same magnitudes as the normal strains \epsilon _x and \epsilon _y. The shear strain γ_{xy} is the decrease in angle at the lower-left comer of the element. Determine the following quantities: (a) the strains for an element oriented at an angle θ = 30°, (b) the principal strains, and (c) the maximum shear strains. (Consider only the in-plane strains, and show all results on sketches of properly oriented elements.)
(a) Element oriented at an angle θ = 30°. The strains for an element oriented at an angle θ to the x axis can be found from the transformation equations (Eqs. 7-71a and 7-71b). As a preliminary matter, we make the following calculations:
\frac{\epsilon _x+\epsilon _y}{2}=\frac{(340+110)10^{-6}}{2}=225×10^{-6} \\ \frac{\epsilon _x\ -\ \epsilon _y}{2}=\frac{(340\ -\ 110)10^{-6}}{2}=115×10^{-6} \\ \frac{γ_{xy}}{2}=90×10^{-6}
Now substituting into Eqs. (7-71a) and (7-71b), we get
\epsilon _{x_1}=\frac{\epsilon _x+\epsilon _y}{2}+\frac{\epsilon _x\ -\ \epsilon _y}{2}cos\ 2θ+\frac{γ_{xy}}{2}sin\ 2θ \\ \quad =(225×10^{-6})+(115×10^{-6})(cos\ 60°)+(90×10^{-6})(sin\ 60°) \\ \quad =360×10^{-6} \\ \frac{γ_{x_1y_1}}{2}=-\frac{\epsilon _x\ -\ \epsilon _y}{2}sin\ 2θ+\frac{γ_{xy}}{2}cos\ 2θ \\ \quad =-(115×10^{-6})(sin\ 60°)+(90×10^{-6})(cos\ 60°) \\ \quad =-55×10^{-6}Therefore, the shear strain is
γ_{x_1y_1}=-110×10^{-6}
The strain \epsilon _{y_1} can be obtained from Eq. (7-72). as follows:
\epsilon _{y_1}=\epsilon _x+\epsilon _y\ -\ \epsilon _{x_1}=(340+110\ -\ 360)10^{-6}=90×10^{-6}
The strains \epsilon _{x_1},\epsilon _{y_1} and γ_{x_1y_1} are shown in Fig. 7-36b for an element oriented at θ = 30°. Note that the angle at the lower-left comer of the element increases because γ_{x_1y_1} is negative.
(b) Principal strains. The principal strains are readily determined from Eq. (7-74), as follows:
Thus, the principal strains are
\epsilon _1=370×10^{-6}\quad \quad \epsilon _2=80×10^{-6}
in which \epsilon _1 denotes the algebraically larger principal strain and \epsilon _2 denotes the algebraically smaller principal strain. (Recall that we are considering only in-plane strains in this example.)
The angles to the principal directions can be obtained from Eq. (7-73):
tan\ 2θ_p=\frac{γ_{xy}}{\epsilon _x\ -\ \epsilon _y}=\frac{180}{340\ -\ 110} = 0.7826
The values of 2θ_p between 0 and 360° are 38.0° and 218.0°, and therefore the angles to the principal directions are
θ_p = 19.0° and 109.0°
To determine the value of θ_p associated with each principal strain, we substitute θ_p = 19.0° into the first transformation equation (Eq. 7-71a) and solve for the strain:
\epsilon _{x_1}=\frac{\epsilon _x+\epsilon _y}{2}+\frac{\epsilon _x+\epsilon _y}{2}cos\ 2θ+\frac{γ_{xy}}{2}sin\ 2θ \\ \quad =(225×10^{-6})+(115×10^{-6})(cos\ 38.0°)+(90×10^{-6})(sin\ 38.0°)\\ \quad =370×10^{-6}This result shows that the larger principal strain \epsilon _1 is at the angle θ_{p_1} = 19.0°. The smaller strain \epsilon _2 acts at 90° from that direction (θ_{p_2} = 109.0°). Thus,
\epsilon _{1}=370×10^{-6} and θ_{p_1} = 19.0°
\epsilon _{2}=80×10^{-6} and θ_{p_2} = 109.0°
Note that \epsilon _{1}+\epsilon _{2}=\epsilon _{x}+\epsilon _{y}
The principal strains are portrayed in Fig. 7-36c. There are, of course, no shear strains on the principal planes.
(c) Maximum shear strain. The maximum shear strain is calculated from Eq. (7-75):
The element having the maximum shear strains is oriented at 45° to the principal directions; therefore, θ_s = 19.0° + 45° = 64.0° and 2θ_s = 128.0°. By substituting this value of 2θ_s into the second transformation equation (Eq. 7-71b), we can determine the sign of the shear strain associated with this direction. The calculations are as follows:
\frac{γ_{x_1y_1}}{2}=-\frac{\epsilon _x\ -\ \epsilon _y}{2}sin\ 2θ+\frac{γ_{xy}}{2}cos\ 2θ\\ \quad =-(115×10^{-6})(sin\ 128.0°)+(90×10^{-6})(cos\ 128.0°) \\ \quad =-146×10^{-6}This result shows that an element oriented at an angle θ_{s_2} = 64.0° has the maximum negative shear strain.
We can arrive at the same result by observing that the angle θ_{s_1} to the direction of maximum positive shear strain is always 45° less than θ_{p_1}. Hence,
θ_{s_1}=θ_{p_1}\ -\ 45°=19.0°\ -\ 45°=-26.0° \\ θ_{s_2}=θ_{s_1}+90°=64.0°
The shear strains corresponding to θ_{s_1} and θ_{s_2} are γ_{max} = 290 × 10^{-6} and y_{min} = -290 × 10^{-6} respectively.
The normal strains on the element having the maximum and minimum shear strains are
\epsilon _{aver}=\frac{\epsilon _x+\epsilon _y}{2}=225×10^{-6}
A sketch of the element having the maximum in-plane shear strains is shown in Fig. 7-36d.
In this example, we solved for the strains by using the transformation equations. However, all of the results can be obtained just as easily from Mohr’s circle.