Question

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

Answer 1

Answer:

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;

$L_{BC}^2 = L_{AB}^2 + L_{AC}^2$

$L_{BC}^2 = (3.5+2.5)^2+ 4^2$

$L_{BC}= \sqrt{(6)^2+ 4^2}$

$L_{BC}= \sqrt{36+ 16}$

$L_{BC}= \sqrt{52}$

$L_{BC}= 7.2111 \ m$

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

$A = \dfrac{\pi}{4}d^2$

where d = 4mm

$A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2$

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection  in length $\delta$ ; we can calculate for the force $F_{BC$ by using the formula:

$\delta = \dfrac{F_{BC}L_{BC}}{AE}$

$F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}$

where ;

E = modulus elasticity

$L_{BC}$ = length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for $L_{BC}$ and 0.006 m for $\delta$ ; we have:

$F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}$

$F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN$     ---- (1)

Similarly; we can determine the force $F_{BC}$ using the allowable  maximum stress; we have the following relation,

$\sigma = \dfrac{F_{BC}}{A}$

${F_{BC}}= {A}*\sigma$

where;

$\sigma =$ maximum allowable stress

Replacing 190 × 10⁶ Pa for $\sigma$ ; we have :

${F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \ kN$     ------ (2)

Comparing (1) and  (2)

The magnitude of the force $F_{BC} = 2.09676 \ kN$ since the elongation of the cable should not exceed 6mm

Finally applying the moment equilibrium condition about point A

$\sum M_A = 0$

$3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0$

$3.5 P - 3.328 F_{BC} = 0$

$3.5 P = 3.328 F_{BC}$

$3.5 P = 3.328 *2.09676 \ kN$

$P =\dfrac{ 3.328 *2.09676 \ kN}{3.5 }$

P = 1.9937 kN

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