A plastic rod A B of length L=0.5 m has a diameter d1=30 mm...

A plastic rod $A B$ of length $L=0.5 \mathrm{m}$ has a diameter $d_{1}=30 \mathrm{mm}$ (see figure). A plastic sleeve $C D$ of length $c=0.3 \mathrm{m}$ and outer diameter $d_{2}=45 \mathrm{mm}$ is securely bonded to the rod so that no slippage can occur between the rod and the sleeve. The rod is made of an acrylic with modulus of elasticity $E_{1}=3.1 \mathrm{GPa}$ and the sleeve is made of a polyamide with $E_{2}=2.5 \mathrm{GPa}$ (a) Calculate the elongation $\delta$ of the rod when it is pulled by axial forces $P=12 \mathrm{kN}$ (b) If the sleeve is extended for the full length of the rod, what is the clongation? (c) If the sleeve is removed, what is the elongation?

Transcript

00:01 Here from this diagram, we have the rod with sleeve showing the applied force, and here f is the pulling force on the cross section of the rod d1 and t2 are the diameters of the rod and the sleeve respectively.

00:16 Your c is the length of the sleeve, your b is the length of the rod, and your l is the total length of the rod, and a, b, c, are the point.

00:30 Points on the rod and the sleeve as shown here in this diagram.

00:34 So from this diagram, we have l is equal to b plus c plus b, which is equal to 2b plus c.

00:44 So we will just rearrange this one.

00:47 So we have b is equal to l minus c over 2.

00:51 So substitute 0 .5 meters for l and 295 millimeters for c.

00:58 So we have b is equals to 0 .5m, 1 ,000 millimeters divided by the 1 meter, negative 295 millimeter over 2.

01:14 So we have equal to 205 millimeter divided by 2.

01:19 So we have 102 .5 millimeter.

01:23 So from the diagram, the total change in the length of the rod will be the total change in the length of the rod is equals to the change in the length of the rod between points c and a plus b, c, d plus d b.

01:48 So here, this is the total change in the length of the rod, and this is the change in the length of the road between point c and d, and the change in the length of the road between a and c.

02:02 And the change in the length of the rod with points d and b since ac and d b are equal in length thus your ac is also equivalent to db due to the applied load so to calculate the cross -section area of the rod we have a1 is equal to pi d1 raised to the par of two four so we just need to substitute 30 millimeters for d1 so we have a1 is equal to pi 30 millimeters raised to the part of 2, divide divide 4.

02:36 So we have equal to 706 .86 square millimeter.

02:43 Now we need to calculate the cross -section area of the sleeve.

02:47 So we need to substitute 45 millimeters for g2...

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