A) Axial Loads are applied at indicated locations on a...

a) Axial Loads are applied at indicated locations on a bronze bar fastened between a steel and an aluminum bar as shown in the below diagram. What should be the highest value of P that will: a) Limit the total elongation to 3 mm b) Limit the stress in the steel to 140 MPa c) Limit the stress in the aluminum to 80 MPa d) Limit the stress in the bronze to 120 MPa The assembly is assumed to be suitably braced so buckling will not occur. Consider Est = 200 GPa, Eal = 70 GPa, and Ebr = 83 GPa.

Transcript

00:01 Hello students, in this question given as a three bar of steel, bronze and aluminum which are connected together and the cross -sectional area of steel is given as 480 mm square and their young modulus is given as 200 into 10 to the power 3 newton per mm square.

00:29 Cross -sectional area of bronze is equal to the 650 mm square and their modulus of elasticity equal to them 83 into 10 to the power 3 newton per mm square and the cross -sectional area of aluminum is 320 mm square and corresponding young modulus is given as 70 into 10 to the power 3 newton per mm square.

01:21 Now, we will make the other forces in various sections that is steel on the forces is p and p on the bronze force is that will be minus 2p opposite direction and in the aluminum the force applied force is 2p.

01:40 Now, in the first part of the question we have to calculate the value of p when elongation total elongation is given as 3 mm.

01:58 Now, we can say that delta 1 elongation in steel elongation in aluminum and elongation in bronze sum it will be equal to the 3 mm.

02:10 Now, we know that the formula of elongation is equal to pl upon ae.

02:18 Now, we can write here on the steel force is p length is 1 meter.

02:28 So, 10 to the power 3 mm upon the value of cross -sectional area is 480 mm square into young modulus is 200 into 10 to the power 3 newton per mm square and plus for second minus 2p into 2 into 10 to the power 3 upon 650 into 83 10 to the power 3 plus in the aluminum 2p into 1 .5 10 to the power 3 the distance of aluminum is 1 .5 meter given upon 320 into 70 into 10 to the power 3 this is equal to 3.

03:35 So, after simplification and calculation p will be equal to the 42 .733 kilo newton.

03:51 In the next part of the question we have to calculate the value of p when stress in steel should be equal to the 140 mega pascal.

04:04 Now, we know that stress in steel is equal to the force upon area...

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