Maximum Shear Stress Theory

Q. At a point in a steel member, the major principal stress is 200 MPa (tensile) and the minor principal stress is compressive. If the uniaxial tensile yield stress is 250 MPa, then according to the maximum shear stress theory, the magnitude of the minor principal stress (compressive) at which yielding will commence is

1. 200 MPa
2. 100 MPa
3. 50 MPa
4. 25 MPa

Q. Steel has proportionality limit of 300 MPa in simple tension. It is subjected to principal stresses of 120 MPa (tensile), 60 MPa (tensile) and 30 MPa (compressive). What is the factor of safety according to maximum shear stress theory?

1. 1.5
2. 1.75
3. 1.8
4. 2.0

Q. The Mohr's circle of plane stress for a point in a body is shown. The design is to be done on the basis of the maximum shear stress theory for yielding. Then, yielding will just begin if the designer chooses a ductile material whose yield strength is

1. 45 MPa
2. 50 MPa
3. 90 MPa
4. 100 MPa

Q. Principal stresses at a point are 80 $N/m{m}^2$ and 40 $N/m{m}^2$, both tensile. The yield stess in simple tension for this material is 200 $N/m{m}^2$. The value of factors of safety according to maximum principal stress theory and maximum shear stress theory, respectively, are

1. 2.5 and 2.5
2. 2.5 and 5
3. 5 and 5
4. 5 and 1.67

Q. A certain steel has proportionallity limit of 3000 $kg/c{m}^2$ in simple tension. It is subjected to principal stresss of 1200 $kg/c{m}^2$ (tensile), 600 $kg/c{m}^2$ (tensile) and 300 $kg/c{m}^2$ (compressive). The factor of safety according to maximum shear stress theory is

1. 1.50
2. 1.75
3. 1.80
4. 2.00

Q. According to maximum shear stress failure criterion yielding in material occurs when

1. $maximumshearstress=\frac{1}{2}yieldstress$

2. maximum shear stress $=\sqrt{2}$ yield stress

3. maximum shear stress $=\sqrt{3}\times$ yield stress
4. maximum shear stress = 2 × yield stress

Q. In a strained body, the three principal stresses at a point are denoted by ${\sigma }_1,{\sigma }_{2}and{\sigma }_3$ such that ${\sigma }_1>{\sigma }_2>{\sigma }_3.$. If ${\sigma }_0$ denoted yield stress, then according to the maximum shear stress theory

1. ${\sigma }_1-{\sigma }_2={\sigma }_0$

2. ${\sigma }_1-{\sigma }_3={\sigma }_0$

3. ${\sigma }_2-{\sigma }_3={\sigma }_0$

4. $\frac{{\sigma }_1-{\sigma }_3}{2}={\sigma }_0$

Q. A member is made of structural steel;. When it is subjected to simple tension, the limit of proportionality is 280 $N/m{m}^2$. If the principal stresses $p_{1}and{p}_2$ developed in the member are 100 $N/m{m}^2$ (tensile) and 40 $N/m{m}^2$ (compressive) respectively and Poisson's ratio is 0.30, then the factor of safety according to maximum shear stress theory would be

1. 2.75
2. 2.5
3. 2.25
4. 2.0

Q. The principal stresses at a point in a critical section of a machine component are ${\sigma }_1=60MPa.$
${\sigma }_2=5MPa.$ and ${\sigma }_3=-40MPa.$ For the material of the component, the tensile yield strength is ${\sigma }_y=200MPa$. According to the maximum shear stress theory, the factor of safety is

1. 1.67
2. 2.00
3. 3.60
4. 4.00

Q. The principal stresses at a point are $2\sigma$ (tensile) and $\sigma$ (compressive), and the stress at elastic limit for the material in simple tension is 210 $N/m{m}^2$. According to maximum shear strain theory, the value of $\sigma$ at failure is

1. 70 $N/m{m}^2$
2. 105 $N/m{m}^2$
3. 140 $N/m{m}^2$
4. 210 $N/m{m}^2$

Q. A failure theory postulated for metals is shown in a two dimensional stress plane. The theory is called

1. Maximum distortion energy theory
2. Maximum normal stress theory
3. Maximum shearing stress theory
4. Maximum strain theory

Q. In a two dimensional stress system, it is assumed that the principal stress ${\sigma }_{1}and{\sigma }_2$ are such that ${\sigma }_1>{\sigma }_2$; then according to the maximum shear stress theory, the failure occurs when (where ${\sigma }_y$ is the yield stress, $\mu$ is the Poisson's ratio and E is the modulus of elasticity)

1. $\frac{1}{E}({\sigma }_1-\mu {\sigma }_2)\ge \frac{{\sigma }_y}{E}$

2. ${\sigma }_1^2+{\sigma }_2^2+2\mu {\sigma }_1{\sigma }_2\ge {\sigma }_y^2$

3. ${\sigma }_1-{\sigma }_2\ge {\sigma }_y$

4. ${\sigma }_1^2+{\sigma }_2^2-{\sigma }_1{\sigma }_2\ge {\sigma }_y$

Q. The details of the principal stresses at a certain point in a steel member are as follows :
Major principal stress ${\sigma }_1=180N/m{m}^2$ (Tensile)
Minor principal stress ${\sigma }_2$ is Compressive.
If the uniaxial tensile yield stress is 240 $N/m{m}^2$, according to maximum shear stress theory, what would be the value of ${\sigma }_2$ in $N/m{m}^2$ at which yielding will commence?

1. 120 tension
2. 90 tension
3. 80 compression
4. 60 compression

Q. The Von-Mises stress at a point in a body subjected to forces is proportional to the square root of the

1. distortional strain energy per unit volume
2. dilatational strain energy per unit volume
3. plastic strain energy per unit volume
4. total strain energy per unit volume

Q. According to maximum shear stress failure theory, yielding occurs in the material when

1. max. shear stress = yield stress
2. max. shear stress = 2 times yield stress
3. max. shear stress = 1/2 of yield stress
4. max. shear stress = $\sqrt{2}$ times yield stress

Q. According to maximum shear stress criterion, at what ratio of maximum shear stress to yield stress of material does the yielding of material take place?

1. 2
2. $2/\sqrt{3}$
3. $1/\sqrt{3}$
4. 1/2

Q. At a critical point in a component, the state of stress is given as ${\sigma }_{xx}=100MPa,{\sigma }_{yy}=220MPa,{\sigma }_{xy}={\sigma }_{yx}=80$ MPa and all other stress components are zero. The yield strength of the material is 468 MPa. The factor of safety on the basis of maximum shear stress theory is
_____ (round off to one decimal place).

1. 1.8

Q. At what value of the ultimate shear strength, shall the material under the action of uniform axial tension fail due to shear?

1. < 0.5 times the ultimate tensile strength
2. < 0.7 times the ultimate tensile strength
3. = ultimate tensile strength
4. > the ultimate tensile strength