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prohojiy [21]
3 years ago
6

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

Engineering
1 answer:
77julia77 [94]3 years ago
4 0

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

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