Question 1

A thin-walled cylindrical tank with a diameter of 200 mm has an internal pressure of 11 MPa. The yield   stress in tension is 250 MPa, the yield stress in shear is 140 MPa, and the factor of safety is 2.     permissible thickness of the tank is approximately:

Accepted Answer

D

Question 2

A cylindrical tank is assembled by welding steel sections in a helical pattern with angle   50&#176;. Tank diameter is 1.

Accepted Answer

D

Question 3

A thin-walled cylindrical tank with a diameter of 2.0 m and wall thickness of 18 mm is open at the top.

Accepted Answer

B

Question 4

A cylindrical tank is assembled by welding steel sections in a helical pattern with angle

50°. Tank diameter is 1.6 m, thickness is 20 mm, and internal pressure is 2.75 MPa. Modulus

210 GPa and Poisson's
Ratio ν 5 0.28. The longitudinal strain in the the wall of the tank is approximately:

Accepted Answer

First, we need to find the hoop stress which can be calculated using the formula below:

$$\sigma_{hoop}=\frac{pD}{2t}$$

where:

p = internal pressure = 2.75 MPa
D = diameter = 1.6 m
t = thickness = 20 mm = 0.02m

$$\sigma_{hoop}=\frac{2.75 \times 1.6}{2 \times 0.02}=220 \mathrm{MPa}$$
Next, we need to find the axial stress which can be calculated using the formula below:

$$\sigma_{axial}=\frac{\sigma_{hoop}}{2}\left(1+\frac{\cos{\alpha}}{1-\nu}\right)$$

where:

ν = Poisson's Ratio = 0.28
α = angle between welds = 50°
$$\cos{\alpha}=\cos{50^{\circ}}=0.643$$
$$\sigma_{hoop}=220 \mathrm{MPa}$$

$$\sigma_{axial}=\frac{220}{2}\left(1+\frac{0.643}{1-0.28}\right)=78 \mathrm{MPa}$$

Therefore, the normal stress acting perpendicular to the weld is approximately 78 MPa, which corresponds to choice C.

Question 5

A cylindrical tank is assembled by welding steel sections in a helical pattern with angle

50°. Tank diameter is 1.6 m, thickness is 20 mm, and internal pressure is 2.75 MPa. Modulus

210 GPa and Poisson's
Ratio ν 5 0.28. The longitudinal strain in the the wall of the tank is approximately:

Accepted Answer

The answer of A cylindrical tank is assembled by welding...

Question 6

A thin-walled spherical tank with a diameter of 1.5 m and wall thickness of 65 mm has an internal pres- sure of 20 MPa. The maximum shear stress in the wall of the tank is approximately:

Accepted Answer

The answer of A thin-walled spherical tank with a diameter...

Question 7

A thin-walled cylindrical tank, under internal pressure p, is compressed by a force F 5 75 kN. Cylinder diameter is   90 and wall thickness   5.5 mm. Allowable normal stress is 110 MPa and allowable shear   stress is 60 MPa. The maximum allowable internal pressure     is approximately:

Accepted Answer

The maximum allowable internal pressure can be determined by analyzing the stresses induced in the cylinder due to force F and internal pressure p. The hoop stress induced in the cylinder by the internal pressure is given by:
$$\sigma_h = \frac{pd}{2t}$$
Substituting the given values:
$$\sigma_h = \frac{p \cdot 90}{2 \cdot 5.5}$$
The axial stress induced in the cylinder by the force F is given by:
$$\sigma_a = \frac{4F}{\pi d^2}$$
Substituting the given values:
$$\sigma_a = \frac{4 \cdot 75,000}{\pi \cdot 90^2}$$
The maximum stress that the cylinder can withstand is the smaller value of the hoop stress and axial stress, i.e.:
$$\sigma_{max} = \text{min}(\sigma_h, \sigma_a)$$
If $\sigma_{max}$ is less than the allowable normal stress (110 MPa), then the cylinder can withstand the given force and internal pressure. Otherwise, the maximum allowable internal pressure is given by:
$$p_{max} = \frac{2t \cdot \text{allowable normal stress} - d \cdot \text{allowable shear stress}}{d}$$
Substituting the given values:
$$p_{max} = \frac{2 \cdot 5.5 \cdot 10^{-3} \cdot 110 \cdot 10^6 - 90 \cdot 10^{-3} \cdot 60 \cdot 10^6}{90 \cdot 10^{-3}} = 13 \text{ MPa}$$
Therefore, the answer is C) 13 MPa.

Question 8

The pressure relief valve is opened on a thin-walled cylindrical tank, with the radius-to-wall thickness ratio of 128, thereby decreasing the longitudinal strain   original internal pressure in the tank was approximately:

Accepted Answer

The answer of The pressure relief valve is opened on...

Question 9

A segment of a drive shaft (   160 mm) is subjected to a torque   allowable shear stress in the shaft is 45 MPa. The maximum permissible compressive load   P is approximately:

Accepted Answer

To find the maximum permissible compressive load P, we need to use the formula:

$$\frac{T}{J} \leq \frac{τ_{allow}}{n}$$

where T is the applied torque, J is the polar moment of inertia of the shaft section, τallow is the allowable shear stress, and n is the factor of safety.

We can simplify this formula for P as follows:

$$P = \frac{T}{τ_{allow}}\cdot\frac{J}{A}\cdot\frac{n}{1 + \frac{J}{A}\cdot\frac{τ_{allow}}{σ_{allow}}}$$

where A is the cross-sectional area of the shaft section.

We can calculate the values of J and A as follows:

$$J = \frac{\pi}{32} \cdot (d_2^4 - d_1^4) = \frac{\pi}{32} \cdot (200^4 - 160^4) = 9.594 	imes 10^6 \mathrm{~mm^4}$$

$$A = \frac{\pi}{4} \cdot (d_2^2 - d_1^2) = \frac{\pi}{4} \cdot (200^2 - 160^2) = 13,089 \mathrm{~mm^2}$$

Substituting the given values into the formula for P, we get:

$$P = \frac{30 \mathrm{~kN\cdot m}}{45 \mathrm{~MPa}}\cdot\frac{9.594 	imes 10^6 \mathrm{~mm^4}}{13,089 \mathrm{~mm^2}}\cdot\frac{3}{1 + \frac{9.594 	imes 10^6 \mathrm{~mm^4}}{13,089 \mathrm{~mm^2}}\cdot\frac{45 \mathrm{~MPa}}{σ_{allow}}}$$

We can solve this equation for σallow = 45 MPa to get:

$$P = 285.6 \mathrm{~kN}$$

Rounding up to the nearest integer gives:

$$P \approx 286 \mathrm{~kN}$$

Therefore, the correct answer is B.

Question 10

A thin-walled spherical tank has a diameter of 0.75 m and an internal pressure of 20 MPa. The yield   stress in tension is 920 MPa, the yield stress in shear is 475 MPa, and the factor of safety is 2.   of elasticity is 210   Poisson's ratio is 0.28, and maximum normal strain is   The minimum permissible thickness of the tank is approximately:

Accepted Answer

The maximum stress produced in the wall of a thin spherical shell is given by
$$\sigma_{max} = \dfrac{P r}{t}$$
Where $P$ is the internal pressure, $r$ is the radius of the sphere, and $t$ is the thickness of the shell.

From the given data, we have
\begin{align*}
P &= 20 \mathrm {~MPa}\
r &= \dfrac{0.75}{2} = 0.375 \mathrm {~m}\
\sigma_{max} &= \dfrac{P r}{t}
\end{align*}

We need to find the minimum permissible thickness of the tank, so we can rearrange the above equation as
$$t = \dfrac{P r}{\sigma_{max}}$$

Now, we need to calculate the maximum stress that the shell can withstand in tension and in shear, and take the minimum of these two values to ensure that the factor of safety is satisfied.

In tension, the yield stress is 920 MPa, but we need to apply a factor of safety of 2, so the maximum permissible stress in tension is
$$\sigma_{max}^{tension} = \dfrac{920}{2} = 460 \mathrm {~MPa}$$

In shear, the yield stress is 475 MPa, and we need to apply a factor of safety of 2, so the maximum permissible stress in shear is
$$\sigma_{max}^{shear} = \dfrac{475}{2} = 237.5 \mathrm {~MPa}$$

The shell is made of an elastic material with Young's modulus of 210 GPa and Poisson's ratio of 0.28. The maximum normal strain in the shell is given by
$$\epsilon_{max} = \dfrac{\sigma_{max}}{E}\left(1-\nu^2\right)$$

Substituting the values, we get
$$\epsilon_{max} = \dfrac{\sigma_{max}}{E}\left(1-\nu^2\right) = \dfrac{\sigma_{max}}{210 \times 10^9}\left(1-0.28^2\right) = 1220 \times 10^{-6}$$

Solving for the maximum permissible stress in tension, we get
\begin{align*}
\sigma_{max}^{tension} &= \epsilon_{max} E \left(\dfrac{1}{1+\nu}\right)\
460 &= 1220 \times 10^{-6} \times 210 \times 10^9 \times \left(\dfrac{1}{1+0.28}\right)
\end{align*}

Solving for the thickness of the shell, we get
\begin{align*}
t &= \dfrac{P r}{\sigma_{max}^{tension}}\
&= \dfrac{20 \times 10^6 \times 0.375}{460}\
&\approx 10.5 \mathrm{~mm}
\end{align*}