Question 1 |

Shear stress distribution on the cross-section of the coil wire in a helical compression spring is shown in the figure. This shear stress distribution represents

direct shear stress in the coil wire cross-section | |

torsional shear stress in the coil wire cross-section | |

combined direct shear and torsional shear stress in the coil wire cross-section | |

combined direct shear and torsional shear stress along with the effect of stress concentration at inside edge of the coil wire cross-section |

Question 1 Explanation:

Fig. : Direct shear stress distribution across the cross-section of wire

Fig. : Torsional shear stress distribution across the cross-section of wire

Fig. : Resultant shear stress variation across the cross-section of wire (without considering stress concentration effect)

Fig. : Resultant shear stress variation considering stress concentration effect

Question 2 |

If the wire diameter of a compressive helical spring is increased by 2%, the change in spring stiffness (in %) is _____ (correct to two decimal places).

5.25 | |

8.24 | |

12.36 | |

15.47 |

Question 2 Explanation:

Stiffness of helical spring

\begin{aligned} K&=\frac{G d^{4}}{64 R^{3} n} \\ d&= \text{spring wire diameter}\\ R&= \text{mean coil radius}\\ n&= \text{number of turns}\\ \therefore \quad K &\propto d^{4} \\ \frac{K^{\prime}}{K}&=\left(\frac{d^{\prime}}{d}\right)^{4} \\ K^{\prime}&=\left(\frac{1.02 d}{d}\right)^{4} K \\ K^{\prime}&=1.08243 \mathrm{K} \\ \end{aligned}

% increase in stiffness =\frac{K^{\prime}-K}{K} \times 100 \%=8.243 \%

\begin{aligned} K&=\frac{G d^{4}}{64 R^{3} n} \\ d&= \text{spring wire diameter}\\ R&= \text{mean coil radius}\\ n&= \text{number of turns}\\ \therefore \quad K &\propto d^{4} \\ \frac{K^{\prime}}{K}&=\left(\frac{d^{\prime}}{d}\right)^{4} \\ K^{\prime}&=\left(\frac{1.02 d}{d}\right)^{4} K \\ K^{\prime}&=1.08243 \mathrm{K} \\ \end{aligned}

% increase in stiffness =\frac{K^{\prime}-K}{K} \times 100 \%=8.243 \%

Question 3 |

A helical compression spring made of wire of circular cross-section is subjected to a compressive load. The maximum shear stress induced in the cross-section of the wire is 24 MPA. For the same compressive load, if both the wire diameter and the mean coil diameter are doubled, the maximum shear stress (in MPa) induced in the cross-section of the wire is _______

3 | |

6 | |

9 | |

12 |

Question 3 Explanation:

Max. shear stress induced in spring wire of a

helical compression spring is given by,

\begin{array}{c} \tau_{\max }=\frac{8 W D}{\pi d^{2}} k_{W} \quad \text { or } \quad \frac{8 W C}{\pi d^{2}} k_{W} \\ \tau_{\max } \propto \frac{1}{d^{2}} \\ {\left[\because W_{1}=W_{2}=W ; C_{1}=C_{2}\left(\text { i.e., } C=\frac{D}{d}\right)\right.} \\ \left.\left(k_{W}\right)_{1}=\left(k_{W}\right)_{2}=k_{W}\right] \\ \frac{\tau_{2}}{\tau_{1}}=\left(\frac{d_{1}}{d_{2}}\right)^{2} \\ \frac{\tau_{2}}{24}=\left(\frac{d}{2 d}\right)^{2} \\ \Rightarrow \quad \tau_{2}=6 \mathrm{MPa} \end{array}

helical compression spring is given by,

\begin{array}{c} \tau_{\max }=\frac{8 W D}{\pi d^{2}} k_{W} \quad \text { or } \quad \frac{8 W C}{\pi d^{2}} k_{W} \\ \tau_{\max } \propto \frac{1}{d^{2}} \\ {\left[\because W_{1}=W_{2}=W ; C_{1}=C_{2}\left(\text { i.e., } C=\frac{D}{d}\right)\right.} \\ \left.\left(k_{W}\right)_{1}=\left(k_{W}\right)_{2}=k_{W}\right] \\ \frac{\tau_{2}}{\tau_{1}}=\left(\frac{d_{1}}{d_{2}}\right)^{2} \\ \frac{\tau_{2}}{24}=\left(\frac{d}{2 d}\right)^{2} \\ \Rightarrow \quad \tau_{2}=6 \mathrm{MPa} \end{array}

Question 4 |

The spring constant of a helical compression spring DOES NOT depend on

coil diameter | |

material strength | |

number of active turns | |

wire diameter |

Question 4 Explanation:

Stiffness of helical spring

k=\frac{G d^{4}}{64 R^{3} n}

It depends on wire diameter (d), shear modulus (g), Coil radius (R) and no of active turns (n). Does not depend on strength.

k=\frac{G d^{4}}{64 R^{3} n}

It depends on wire diameter (d), shear modulus (g), Coil radius (R) and no of active turns (n). Does not depend on strength.

Question 5 |

A compression spring is made of music wire of 2mm diameter having a shear strength and shear
modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20mm, free length is
40mm and the number of active coils is 10. If the mean coil diameter is reduced to 10mm, the
stiffness of the spring is approximately

decreased by 8 times | |

decreased by 2 times | |

increased by 2 times | |

increased by 8 times |

Question 5 Explanation:

\begin{aligned} k &=\frac{G d^{4}}{8 D^{3} N} \\ \frac{k_{2}}{k_{1}} &=\left(\frac{D_{1}}{D_{2}}\right)^{3}=\left(\frac{20}{10}\right)^{3}=8 \end{aligned}

\therefore Stiffness of spring is increased by 8 times.

\therefore Stiffness of spring is increased by 8 times.

Question 6 |

A weighing machine consists of a 2 kg pan resting on a spring. In this condition, the pan resting on the spring, the length of the spring is 200 mm. When a mass of 20 kg is placed on the pan, the length of the spring becomes 100 mm. For the spring, the un-deformed length l_{o}
and the spring constant k (stiffiness) are

l_{o}= 220mm, k=1862 N/m | |

l_{o}= 210mm, k=1960 N/m | |

l_{o}= 210mm, k=2156 N/m | |

l_{o}= 220mm, k=2156 N/m |

Question 6 Explanation:

\begin{aligned} F &=k \Delta m \\ \Rightarrow 2 g &=k\left(l_{0}-0.2\right) \qquad \cdots(i) \end{aligned}

When mass of 20 \mathrm{kg} is placed on the pan

22 g=k\left(l_{0}-0.1\right) \qquad \cdots(ii)

Solving (i) and (ii), we get

\begin{aligned} l_{0} &=210 \mathrm{mm} \\ k &=1960 \mathrm{N} / \mathrm{m} \end{aligned}

When mass of 20 \mathrm{kg} is placed on the pan

22 g=k\left(l_{0}-0.1\right) \qquad \cdots(ii)

Solving (i) and (ii), we get

\begin{aligned} l_{0} &=210 \mathrm{mm} \\ k &=1960 \mathrm{N} / \mathrm{m} \end{aligned}

Question 7 |

Two helical tensile springs of the same material and also having identical mean coil diameter and weight, have wire diameters d and \frac{d}{2}.
The ratio of their stiffness is

1 | |

4 | |

64 | |

128 |

Question 7 Explanation:

G_1=G_2, D_1=D_2, d_1=d, D-2=\frac{d}{2}, W_1=W_2

\therefore \; k=\frac{Gd^4}{8D^3N}

\frac{k_1}{k_2}=\left ( \frac{d_1}{d_2} \right )^4 \times \left ( \frac{N_2}{N_1} \right )

Weight(W)=\rho Vg

i.e. \; \; W\propto V \;\;\;[\because \rho _1 =\rho _2]

\text{Volume}=\text{Area} \times \text{Length}

=\frac{\pi}{4}d^2 \times (\pi DN) W_1=W_2\Rightarrow V_1=V_2

\begin{aligned} d_1^2N_1 &=d_2^2N_2 \\ \frac{N_2}{N_1} &=\left ( \frac{d_1}{d_2} \right )^2 \Rightarrow 4 \\ \frac{k_1}{k_2} &=\left ( \frac{d_1}{d_2} \right )^4 \times \left ( \frac{N_2}{N_1} \right )\\ \frac{k_1}{k_2}&= (2)^4 \times 4=64 \end{aligned}

\therefore \; k=\frac{Gd^4}{8D^3N}

\frac{k_1}{k_2}=\left ( \frac{d_1}{d_2} \right )^4 \times \left ( \frac{N_2}{N_1} \right )

Weight(W)=\rho Vg

i.e. \; \; W\propto V \;\;\;[\because \rho _1 =\rho _2]

\text{Volume}=\text{Area} \times \text{Length}

=\frac{\pi}{4}d^2 \times (\pi DN) W_1=W_2\Rightarrow V_1=V_2

\begin{aligned} d_1^2N_1 &=d_2^2N_2 \\ \frac{N_2}{N_1} &=\left ( \frac{d_1}{d_2} \right )^2 \Rightarrow 4 \\ \frac{k_1}{k_2} &=\left ( \frac{d_1}{d_2} \right )^4 \times \left ( \frac{N_2}{N_1} \right )\\ \frac{k_1}{k_2}&= (2)^4 \times 4=64 \end{aligned}

There are 7 questions to complete.

For question 7, i think 16 should be there in options.

Please check the solution.