Chapter 2, Problem 2.94P

To determine

The compressive load applied in the z-direction according to Tresca yield criteria

Answer to Problem 2.94P

Compressive load applied in the z-direction is $60\text{kN}$ .

Explanation of Solution

Given data:

  $\begin{array}{l}\text{Area of the face of the metal cube, }A=5{\text{cm}}^2\\ \text{Shear yield strength of the metal, }k=140\text{MPa,}\\ \text{Compressive load applied on the metal, }\\ F_x=-40\text{kN,}\\ F_y=-80\text{kN}\text{.}\end{array}$

Calculation:

Calculating the compressive stress due to applied load,

  $\begin{array}{l}{\sigma }_x=\frac{F_x}{A}\\ {\sigma }_x=\frac{-40\times 1000}{5\times {10}^{-4}}\\ {\sigma }_x=-80\text{MPa,}\\ {\sigma }_y=\frac{F_y}{A}\\ {\sigma }_y=\frac{-80\times 1000}{5\times {10}^{-4}}\\ {\sigma }_y=-160\text{MPa}\text{.}\end{array}$

Writing the formula for principal stresses,

  $\begin{array}{l}{\sigma }_{1/2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}\\ \text{Here, }{\tau }_{xy}=0,\\ \text{Pugging the value of }{\sigma }_x\&{\sigma }_y\\ {\sigma }_{1/2}=\frac{-80-160}{2}\pm \sqrt{{\left(\frac{-80-(-160)}{2}\right)}^2+0^2}\\ {\sigma }_{1/2}=\frac{-240}{2}\pm \sqrt{{\left(\frac{-80+160}{2}\right)}^2}\\ {\sigma }_{1/2}=\frac{-240}{2}\pm \frac{80}{2}\\ {\sigma }_{1/2}=-120\pm 40\\ {\sigma }_1=-120+40\\ {\sigma }_1=-80\text{MPa,}\\ {\sigma }_2=-120-140\\ {\sigma }_2=-160\text{MPa}\end{array}$

Let the stress in the z-direction be ${\sigma }_3=-{\sigma }_z$

Now according to Tresca yield criteria,

  $\begin{array}{l}{\tau }_{\max }=\left|\left(\frac{{\sigma }_1-{\sigma }_2}{2}\right),\left(\frac{{\sigma }_3-{\sigma }_3}{2}\right),\left(\frac{{\sigma }_3-{\sigma }_1}{2}\right)\right|\\ k=\left|\left(\frac{{\sigma }_1-{\sigma }_2}{2}\right),\left(\frac{{\sigma }_2-{\sigma }_3}{2}\right),\left(\frac{{\sigma }_3-{\sigma }_1}{2}\right)\right|\\ \text{Plugging the value of }k,{\sigma }_1\&{\sigma }_2\\ 140=\left|\left(\frac{-160-(-80)}{2}\right),\left(\frac{-80-(-{\sigma }_z)}{2}\right),\left(\frac{-{\sigma }_z-(-160)}{2}\right)\right|\\ 140=\left|\left(\frac{-160+80}{2}\right),\left(\frac{-80+{\sigma }_z}{2}\right),\left(\frac{-{\sigma }_z+160}{2}\right)\right|\\ 140=\left|\left(\frac{-80}{2}\right),\left(\frac{-80+{\sigma }_z}{2}\right),\left(\frac{-{\sigma }_z+160}{2}\right)\right|\end{array}$

Now taking the above values as shown below,

  $\begin{array}{l}140=\left|\left(\frac{-80}{2}\right),\left(\frac{-80+{\sigma }_z}{2}\right),\left(\frac{-{\sigma }_z+160}{2}\right)\right|\\ 140=\left|\frac{-80+{\sigma }_z}{2}\right|\\ 140=\frac{-80+{\sigma }_z}{2}\\ -80+{\sigma }_z=140\times 2\\ -80+{\sigma }_z=280\\ {\sigma }_z=280+80\\ {\sigma }_z=360\text{MPa,}\\ 140=\left|\frac{-{\sigma }_z+160}{2}\right|\\ \frac{-{\sigma }_z+160}{2}=140\\ -{\sigma }_z+160=140\times 2\\ -{\sigma }_z+160=280\\ -{\sigma }_z=280-160\\ -{\sigma }_z=120\\ {\sigma }_z=-120\text{MPa}\end{array}$

Since, we have taken stress in the z-direction as negative for compressive load and out of two values from above one is positive and one is negative.

Hence, the compressive stress in z-direction would be ${\sigma }_z=-120\text{MPa}$

Now calculating the compressive load in the z-direction,

  $\begin{array}{l}{\sigma }_z=\frac{F_z}{A}\\ -120\times {10}^6=\frac{F_z}{5\times {10}^{-4}}\\ F_z=-120\times {10}^6\times 5\times {10}^{-4}\\ F_z=-600\times {10}^2\\ F_z=-60\text{kN}\end{array}$

Hence the compressive load applied in the z-direction according to Trescas criteria would be $60\text{kN}$ .