Question

The 50-mm-diameter solid shaft is made of A-36 steel and is subjected to the distributed and concentrated torsional loadings shown where T = 230 N-m. 2 kNm 600 Nm B 0.8 m 0.6 m Part A Determine the angle of twist at the free end A of the shaft due to these loadings. Use Gst = 75.0 GPa Express your answer to three significant figures and include appropriate units. μΑ ? O= Value Units

Answer 1

Given:

Diameter of the shaft,d=50 mm=0.05 m

Modulus of rigidity,G=75 GPa=75*10⁹ Pa

Solution:

In the next fig.I have drawn the diagram of the problem in which I have named some points in the fig.

t = 260mm To - 600N.m. be TT-230 Nm 0.8 m * Ooom

[Note:The unit of uniformly distributed twisting moment is given in kN-m.But it should be in kN-m/m.It is similar to uniformly distributed load unit which is given in N/m and concentrated load is given in N.May be there is printing mistake.]

We know that,angle of twist($\theta$) of any rod is given as:

TL rad GJ

Where, T is the twisting moment applied in N-m

L is the length of the rod in m.

G is the moduluds of rigidity in Pa.

J is the polar moment of inertia in m⁴

For circular rod of diameter,d=50 mm=0.05 m,

J= 32

Putting the value

$J=\frac{\pi }{32}*0.05^{4}m^{4}=6.136*10^{-7}m^{4}$

Now I will derive epression of $\theta$ for section BC.

Consider any rod in which uniformly distributed twisting moment(t) is applied throughout its length,L

In the next fig. I have derived expression of  $\theta$ for section BC.

xx t N.m/m I dock fo Х Tack for L length K dx. Tort LtX Nam OS TL GT for doc length, o do = To da txolx set GJ de J. of reag е Ido = sexda gu v t (227 t2² - GJ 2 267 = ) o = t (2 rod. 2GJ

Now $\theta$ due to uniformly distributed twisting moment is for length,L=0.8 m

$\theta_{t} =\frac{tL_{BC}^{2}}{2GJ}$

Putting the values in the above expression

$\theta_{t} =\frac{2*10^{3}*0.8^{2}}{2*75*10^{9}*6.136*10^{-7}}rad=0.0139rad$---Clockwise

Now angle of twist due T_C=600 N-m applied at point C

$\theta _{C}=\frac{T_{C}L_{BC}}{GJ}$

Putting the values in the above expression

$\theta _{C}=\frac{600*0.8}{75*10^{9}*6.136*10^{-7}}rad=0.01043rad$---Anticlockwise

Now angle of twist due T_A=230 N-m applied at point A

$\theta _{AC}=\frac{T_{A}L_{AC}}{GJ}rad$

Putting the values

$\theta _{AC}=\frac{230*0.6}{75*10^{9}*6.136*10^{-7}}rad=0.003rad---Clockwise$

The net angle of twist at free end point A

$\theta _{A}=\theta _{t}+\theta _{C}+\theta _{AC}$

Putting the values by taking care of sign convention.

$\theta _{A}=0.0139-0.01043+0.003=0.00647rad$---[Taking clockwise twisting as +ve and anticlockwise as -ve]

or $\theta _{A}=\frac{180}{\pi }*0.00647=0.3707^{0}$ in clockwise direction.

Angle of twist at free end A($\theta _{A}$)=0.00647 rad=0.3707⁰ clockwise [Ans]

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