Engineering Mechanics Test 2

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Engineering Mechanics Test 2

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Question 1
1 / -0

For two springs having same stiffness (S) are in series, the equivalent stiffness would be

A
S/4

B
S/2

C
S

D
2S

Solution

Concept:
Equivalent stiffness of spring for series combination:
\(\frac{1}{{{S_{eq}}}} = \frac{1}{{{S_1}}} + \frac{1}{{{S_2}}}\)
\(\frac{1}{{{S_{eq}}}} = \frac{{({S_1} + {S_2})}}{{{S_1}{S_2}}}\)
\({S_{eq}} = \frac{{{S_1}{S_2}}}{{({S_1} + {S_2})}}\)
But here S1 = S2 = S
Therefore, \({S_{eq}} = \frac{{{S}{S}}}{{({S} + {S})}}=\frac{S^2}{2S}\)
\(\Rightarrow S_{eq}=\frac{S}{2}\)
Equivalent stiffness of spring for parallel combination:
S_eq = S₁ + S₂
Question 2
1 / -0

If the volume of a homogeneous body possesses two planes of symmetry, then its centroid must lie along

A
any plane

B
both the planes

C
line of intersection of the two planes

D
anywhere in the space
Solution
Concept:
- When volume V possesses a plane of symmetry, the first moment of V with respect to that plane is zero and the centroid of the volume is located in the plane of symmetry.
- When a volume possesses two planes of symmetry, the centroid of the volume is located on the line of intersection of the two planes.
- When a volume possesses three planes of symmetry that intersect at a well-defined point (i.e. not along a common line), the point of intersection of the three planes coincides with the centroid of volume.
- This property enables us to determine immediately the locations of the centroids of spheres, ellipsoids, cubes, etc.
- The centroids of unsymmetrical volumes or of volumes possessing only one or two planes of symmetry should be determined by integration.
Question 3
1 / -0

Consider a trapezoidal lamina ABCD, with AB parallel to DC, 6 cm apart; AB is 8 cm; CD is 12 cm; CD extends outwards by 1 cm from the foot of the perpendicular from B on DC. The centre of gravity of the lamina will be

A
Along AC at a height of 3 cm form DC

B
Along BD at a height of 3 cm from DC

C
Along the line joining the mid-point of AB to the mid-point of DC; at a height of 2-8 cm from DC

D
At the inersection point of AC and DB

Solution

The centroid of a trapezium is given by
\(C = \frac{h}{3}\left[ {\frac{{b + 2a}}{{b + a}}} \right]\)
Given that, a = 8 cm, b = 12 cm, h = 6 cm
\(C = \frac{6}{3}\left[ {\frac{{12 + 2\left( 8 \right)}}{{12 + 8}}} \right] = 2.8\;cm\)
The center of gravity of the lamina will be along the line joining the mid-point of AB to the mid-point of DC; at a height of 2-8 cm from DC