Bars of square and circular cross-section with 0.5 m length are made of a material with shear strength of 20 MPa. The square bar cross-section dimension is 4 cm x 4 cm and the cylindrical bar cross-section diameter is 4 cm. The specimens are loaded as shown in the figure.

Which specimen(s) will fail due to the applied load as per maximum shear stress theory?

Correct Answer: Tensile and compressive load specimens

Based on the Maximum Shear Stress Theory (Tresca's criterion), a material fails when the maximum shear stress (${\tau }_{max}$) induced in the specimen exceeds the shear strength (${\tau }_{\text{strength}}$) of the material.
Here, the shear strength of the material is given as:
$${\tau }_{\text{strength}}=20\text{ MPa}$$

Let us evaluate each loading condition shown in the images:

1. Tensile and Compressive Load Specimens:
For these specimens, a square bar of cross-section dimension $4\text{ cm}\times 4\text{ cm}$ (which is $40\text{ mm}\times 40\text{ mm}$) is subjected to an axial load (tensile and compressive, respectively) of $P=80\text{ kN}=80\times {10}^3\text{ N}$.
The normal stress ($\sigma$) on the cross-section is:
$$\sigma =\frac{P}{A}=\frac{80\times {10}^3\text{ N}}{40\text{ mm}\times 40\text{ mm}}=50{\text{ N/mm}}^2=50\text{ MPa}$$
According to the maximum shear stress theory, the maximum shear stress for uniaxial loading is:
$${\tau }_{max}=\frac{\sigma }{2}$$
Comparing this value with the allowable shear strength:
$${\tau }_{max}=25\text{ MPa}>20\text{ MPa}$$
Since the maximum shear stress exceeds the material's shear strength, both the tensile and compressive load specimens will fail.

2. Torsional Load Specimen:
For this specimen, a cylindrical bar of diameter $d=4\text{ cm}=40\text{ mm}$ is subjected to a torque of $T=64\pi \text{ N·m}=64\pi \times {10}^3\text{ N·mm}$.
The maximum shear stress due to torsion is:
$${\tau }_{max}=\frac{16T}{\pi d^3}=\frac{16\times 64\pi \times {10}^3}{\pi \times {40}^3}=\frac{1,024\times {10}^3}{64\times {10}^3}=16\text{ MPa}$$
Comparing this value with the allowable shear strength:
$${\tau }_{max}=16\text{ MPa}<20\text{ MPa}$$
Therefore, the torsional load specimen will not fail.

3. Bending Load Specimen:
For this specimen, a bending moment of $M=320\text{ N·m}=320\times {10}^3\text{ N·mm}$ is applied. Calculating with the square cross-section configuration (side $a=40\text{ mm}$):
The bending stress is:
$$\sigma =\frac{M}{Z}=\frac{M}{a^3/6}=\frac{320\times {10}^3}{{40}^3/6}=30\text{ MPa}$$
The maximum shear stress due to this bending stress is:
$${\tau }_{max}=\frac{\sigma }{2}$$
Comparing this value with the allowable shear strength:
$${\tau }_{max}=15\text{ MPa}<20\text{ MPa}$$
Therefore, the bending load specimen will not fail.

Conclusion:
Only the tensile and compressive load specimens experience a maximum shear stress ($25\text{ MPa}$) that exceeds the material's shear strength ($20\text{ MPa}$), causing them to fail.