Two wires of diameter 0.25 cm, one made of steel and the...

Two wires of diameter $0.25 \mathrm{~cm}$, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is $1.5 \mathrm{~m}$ and that of brass wire is $1.0 \mathrm{~m}$. Compute the elongations of the steel and the brass wires.

Key Concepts

Elongation Formula

The elongation of a material under a tensile load can be calculated using the formula ?L = FL/(AE), where ?L is the change in length, F is the applied force, L is the original length, A is the cross-sectional area, and E is Young's Modulus. This relation is used to predict how much a material will stretch under load.

Hooke's Law

Hooke's Law states that, within the elastic limit of a material, the deformation (strain) is directly proportional to the applied force (stress). This principle allows for the calculation of elongation when a material is subjected to a tensile force.

Young's Modulus

Young's Modulus is a measure of a material's stiffness and is defined as the ratio of stress (force per unit area) to strain (relative deformation). It is essential in determining how much a material will elongate under a given load.

Stress and Strain

Stress is the force applied per unit area, while strain is the resultant deformation measured as the change in length relative to the original length. These fundamental concepts in mechanics of materials are used to analyze how materials deform under loads.

Cross-sectional Area Calculation

Calculating the cross-sectional area is crucial when determining the stress experienced by a material. For circular cross-sections, the area is computed using the formula A = ?d²/4, where d is the diameter of the wire.

Transcript

00:01 So, this question belongs to the mechanical properties of solids in which we have given two diameters of diameter d equals to 0 .25 centimeter.

00:10 Okay.

00:11 So from here we can calculate that its radius, it will be equals to 0 .25 centimeter divided by 2.

00:19 This will be equals to 0 .125 centimeter.

00:23 Now the two wires are of steel and brass.

00:27 So let us suppose steel is taken as 1 and brass is taken as 2.

00:32 So we can say that from the given diameter, the length of the steel wire that is l1, it is equals to 1 .5 meter and length of the brass wire l2 it is equals to 1 .0 meter.

00:43 Okay.

00:45 And from the diagram, we can calculate that total force exerted on the steel wire f1, it will be equals to 4 plus 6 multiplied by g that is 10, sorry, 9 .8.

00:56 So from here we get 98 newton.

01:01 And similarly, force on the brass wire f2, it will be equals to 6 multiplied by 9 .8, that is equals to 58 .8 newton.

01:10 Now, we can calculate the change in length from the young's modulus.

01:16 So suppose change in length for it is delta l1 and change in length for it is delta l2 that is elongation.

01:23 So from the yons modulus we know that y it is equals to stress divided by strain f by a divided by delta l by l.

01:32 So from here we get delta l it is equals to f multiplied by length l divided by area multiplied by young's modulus.

01:39 Okay...

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