
A horizontal rod is supported at both ends and loaded at middle. If L and Y are length and Young’s modulus respectively, then the depression at the middle is directly proportional to
A.\[L\]
B.\[{L^2}\]
C.\[Y\]
D.\[\dfrac{1}{Y}\]
Answer
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Hint: Use the formula for the depression in the beam due to the load attached to it. This formula gives the relation between the depression in the beam, length of the beam and Young's modulus for the material of the beam.
Formula used:
The formula for the depression \[\delta \] in the beam loaded at the middle is given by
\[\delta = \dfrac{{W{L^3}}}{{4Yb{d^3}}}\] …… (1)
Here, \[W\] is the weight of the load, \[L\] is the length of the beam, \[Y\] is the Young’s modulus of the material of the beam, \[b\] is the breadth of the beam and \[d\] is the depth of the beam.
Complete step by step answer:
Rewrite equation (1) for the depression of the beam at the middle.
\[\delta = \dfrac{{W{L^3}}}{{4Yb{d^3}}}\]
From the above equation, it can be concluded that the depression \[\delta \] of the beam is directly proportional to the weight \[W\] and cube of length \[{L^3}\] of the beam and it is inversely proportional to the Young’s modulus \[Y\] of the material of the beam, breadth \[b\] of the beam and depth \[d\] of the beam.
Hence, the depression \[\delta \] in the beam is directly proportional to the inverse of the Young’s modulus \[\dfrac{1}{Y}\] of the material of the beam.
\[\delta \propto \dfrac{1}{Y}\]
So, the correct answer is “Option D”.
Additional Information:
The beam which is supported at one end and subjected to load at the other end is called cantilever.
The depression \[y\] in cantilever is given by the formula
\[y = \dfrac{{W{L^3}}}{{3Y{I_g}}}\]
Here, \[W\] is the weight of the load attached to the end of cantilever, \[L\] is the length of the cantilever, \[Y\] is the Young’s modulus for the material of the cantilever and \[{I_g}\] is the geometric moment of inertia of the cantilever.
Note:
The Young’s modulus of a material is the property of the material given by the ratio of longitudinal stress to longitudinal strain of that material and it remains the same for that particular material.
Formula used:
The formula for the depression \[\delta \] in the beam loaded at the middle is given by
\[\delta = \dfrac{{W{L^3}}}{{4Yb{d^3}}}\] …… (1)
Here, \[W\] is the weight of the load, \[L\] is the length of the beam, \[Y\] is the Young’s modulus of the material of the beam, \[b\] is the breadth of the beam and \[d\] is the depth of the beam.
Complete step by step answer:
Rewrite equation (1) for the depression of the beam at the middle.
\[\delta = \dfrac{{W{L^3}}}{{4Yb{d^3}}}\]
From the above equation, it can be concluded that the depression \[\delta \] of the beam is directly proportional to the weight \[W\] and cube of length \[{L^3}\] of the beam and it is inversely proportional to the Young’s modulus \[Y\] of the material of the beam, breadth \[b\] of the beam and depth \[d\] of the beam.
Hence, the depression \[\delta \] in the beam is directly proportional to the inverse of the Young’s modulus \[\dfrac{1}{Y}\] of the material of the beam.
\[\delta \propto \dfrac{1}{Y}\]
So, the correct answer is “Option D”.
Additional Information:
The beam which is supported at one end and subjected to load at the other end is called cantilever.
The depression \[y\] in cantilever is given by the formula
\[y = \dfrac{{W{L^3}}}{{3Y{I_g}}}\]
Here, \[W\] is the weight of the load attached to the end of cantilever, \[L\] is the length of the cantilever, \[Y\] is the Young’s modulus for the material of the cantilever and \[{I_g}\] is the geometric moment of inertia of the cantilever.
Note:
The Young’s modulus of a material is the property of the material given by the ratio of longitudinal stress to longitudinal strain of that material and it remains the same for that particular material.
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