Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m. on shaft AB. Knowing that each shaft is solid, determine the maximum shearing stress in a) shaft AB b) shaft BC c) shaft CD (25 points) Given that the torque at B

Explanation:

The image attached to the question is shown in the first diagram below.

From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.

IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :

From the diagram; we can determine the length of BC by using pyhtagoras theorem;

SO;

The cross -sectional of the cable is calculated by the formula :

where d = 4mm

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection  in length ; we can calculate for the force by using the formula:

where ;

E = modulus elasticity

= length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for and 0.006 m for ; we have:

    ---- (1)

Similarly; we can determine the force using the allowable  maximum stress; we have the following relation,

where;

maximum allowable stress

Replacing 190 × 10⁶ Pa for ; we have :

    ------ (2)

Comparing (1) and  (2)

The magnitude of the force since the elongation of the cable should not exceed 6mm

Finally applying the moment equilibrium condition about point A

P = 1.9937 kN

Hence; the maximum load P that can be applied is 1.9937 kN

twist boundary

explanation:

answer:

the answer is d

explanation: