Springs

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

A
direct shear stress in the coil wire cross-section
B
torsional shear stress in the coil wire cross-section
C
combined direct shear and torsional shear stress in the coil wire cross-section
D
combined direct shear and torsional shear stress along with the effect of stress concentration at inside edge of the coil wire cross-section
GATE ME 2021 SET-1   Machine Design
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).
A
5.25
B
8.24
C
12.36
D
15.47
GATE ME 2018 SET-1   Machine Design
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 \%
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 _______
A
3
B
6
C
9
D
12
GATE ME 2017 SET-2   Machine Design
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}
Question 4
The spring constant of a helical compression spring DOES NOT depend on
A
coil diameter
B
material strength
C
number of active turns
D
wire diameter
GATE ME 2016 SET-1   Machine Design
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.
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
A
decreased by 8 times
B
decreased by 2 times
C
increased by 2 times
D
increased by 8 times
GATE ME 2008   Machine Design
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.
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
A
l_{o}= 220mm, k=1862 N/m
B
l_{o}= 210mm, k=1960 N/m
C
l_{o}= 210mm, k=2156 N/m
D
l_{o}= 220mm, k=2156 N/m
GATE ME 2005   Machine Design
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}
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
A
1
B
4
C
64
D
128
GATE ME 2001   Machine Design
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}
There are 7 questions to complete.

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