Question

Shaft AB consists of n homogeneous cylindrical elements, which can be solid or hollow. Both of its ends are fixed, and it is subjected to the loading shown. The length of element i is denoted by i,its outer diameter by OD, its inner diameter by ID, its modulus of rigidity by G, and the torque applied to its right end by T, the magnitude T of this torque being assumed to be positive if T is observed as counterclockwise from end B and negative otherwise. Note that ID-0 if the element is solid and also that T1-0. Write a computer program that can be used to determine the reactions at A and B, the maximum shearing stress in each element, and the angle of twist of each element. Use this program (a) to solve Prob. 3.155, (b) to determine the maximum shearing stress in the shaft of Example 3.05. Elementn Element 1 Fig. P3.C3

Answer 1

ANSWER:

The torsion formula for circular shaft is as expressed as,

let it be equation (1)

GO T

And,

let it be equation (2)

Here, G is the modulus of rigidity, is the angle of twist, L is the lenght of the shaft, T is the applied torque on the member, J is the polar moment of inertia, is the shear stree and r is the radius of the shaft.

Polar moment of inertia (J) of a circular shaft is expressed as,

let it be equation (3)

32

Here is the outside diameter, and is the inner diameter of the circular shaft.

General Sign convention for torque: Counter-Clockwise torque is considered as positive, while clockwise torque is considered as negative

Consider the equations written below for the MATLAB program.

Consider the reaction at end of the shaft as redundant and hence the shaft is released at end B.

Calculate the angle of twist with torque
(TB)
at end B or torque at element 1 equivalent to zero.

For given length (L) outer diameter
(Do)
inner diameter modulus of rigidity of the shaft material, and the applied torque on the shaft element (say i)

Compute the polar moment of inertia of the element of the shaft from equation (3) as

th

let it equation (4)

32

let us update the total torque at the subsequent elements on the shaft, to compute the torque on the successive elements as,

let it be equation (5)

Here, is the torque applied on the
th
element of the shaft, and
T. i-1
is the torque applied on the element just beforw the element i

Now for the updated torque at the respective shafts, calculate the shear stress for each shaft element as,

let it be equation (6)

T(Do,/2)

Also, for the updated torque at the elements on the shaft, calculate the angle of twist for each element of the shaft as,

let it be equation (7)

T.L G,J

Now update the angle of twist for the entire shaft, starting with zero angle of twist at the fixed end and then add the angle of twists at successive elements on the shaft as,

let it be equation (8)

Here, i varies through the first element towards the last element (say n) on the shaft such that,

Now compute the angle due to a unit torque applied at end B of the shaft.

Calculate unit-shear stress at each element (i) as,

Funit) Twnk

let it be equation (9)

Do /2) unit

Similarly, calculate unit-twist angle  at each element (i) as,

let it be equation (10)

G,J

Let us update the unit-angle of twist for the entire starting with zero unit-angle of twist at the fixed end and then add the unit-angle of twists at successive elements on the shaft as,

let it be equation (11)

nituni

Here, i varies through the first element towards the last element ( say n) on the shaft.

Also, unit-angle of twist at the end B is updated as,

For the total updated angle at end B equivalent to zero, solve as

let it be equation (12)

mi

Now superimpose the values calculated above, to compute the torque at end B
(TB)
and at end A as,

For torque at end B1

let it be equation (13)

unit,B)

For torque at end A1

let it be equation (14)

Now for each element, calculate the maximum shear stress as,

otal total,i

let it be equation (15)

unit,i

Also, angle of twist for each element is calculated as,

  let it be equation (16)

total,

Type the following code in the MATLAB to obtain the solution for the provided questions

clear;clc; % Take input from the user. ninput(Enter the number of Elements; fprintf("----) for i#1:1:n fprintf('\nEnter length of element %d in [m] ', i) (i) input(''); fprintf( 'Enter outside diameter for the element %d in [m] = ',i) OD(i) input(); fprintf( 'Enter inside diameter for the element %d in [m] = ', i) ID(i) input(); fprintf( 'Enter modulus of rigidity of the element %d in [GPa] ',i) G(1)input(") fprintf(['Enter torque applied at the right end of the",... ' element %d in input(''); [N.m] , ] , i) (1) end fprintf( Elementlt Maximum Stress (MPa)\t Angle of Twist (Degrees)\n) % Calculate shear stress and angle of twist. for i=1:1:n if i--1 Torquei) T() else Torque(1) T(1) + Torque(1-1);

end Tau(i) (((Torque (i))* (OD(i)/2))/3(i))/(10*6); phị(1) (((Torque(i))*(L(1)))/((6(İ))*(10^9)*(J(İ))))*(180/pi); U-Tau(1) = (((00( İ )/2))/7(1))/(10"6); U-phi (1) = (((L(1)))/((6(1))*(10^9)*(J(1))))*(180/pi); % Compute the angle of twist. if i--1 Angle-I(1) phi (1); UAngle TB()U_phi(i); else Angle T(i)Angle_T(i-1) phi(i); UAngle TB()U_Angle_TB(1-1) U_phi(i); end Angle-B Angle. T (1); U_Angle_BU Angle_TB (i); end % Calculate reaction torque. T_B((Angle_B)/(U_Angle_B)); T-A = (sum(Torque))+(T-B); % Compute maximum stress and angle of twist for each element. for k-1:1:n Tau-Total(k) = (Tau(k)) + (T-B" (U-Tau(k))); phiJotal(k)-(phi (k)) + (T-B" (U-phi(k))); fprintf( %2.efit% 2.3f\t\t\t% 2.4fin..k, Tau-Total(k), phi-Total(k))

% Print end-reaction torques. fprintf(["inReaction torque fprintf(['\nReaction torque at rigende support =%2.2f at right end-support = %2.2f ',.. N.mInReaction torque at lefte N·m"],T-B ,TA) " nd-support = %2.2f N·m.],T-B,T-A)

(a)

Use the data and loading conditions to solve problem 3.55P as,

For given 2 shaft elements

At Element 1;

Length, L1 = 0.25 m,

Outer diameter D0.1 = 0.038,

Inner diameter D1.3 = 0 in, and

Modulus of rigidity G1 = 77.2 GPa

At element 2:

Length L2 = 0.2 m,

Outer diameter D0.2 = 0.05 m,

inner diameter D1.2 = o m and

Modulus of rigidity G2 = 77.2 GPa

Consider the torque applied at end of the first element as,

T, 1400 N·m

Now, execute the MATLAB code and enter the input values, to obtain the output for the given problem as,

Enter the number of Elements :2 Enter length of element 1 in [m]0.25 Enter outside diameter for the element 1 in [m] 0.038 Enter inside diameter for the element 1 in [m] 0 Enter modulus of rigidity of the element 1 in [GPa] 77.2 Enter torque applied at the right end of the element 1 in [N·m] = 0 Enter length of element 2 in [m]0.2 Enter outside diameter for the element 2 in [m0.05 Enter inside diameter for the element 2 in [m] 0 Enter modulus f rigidity of the element 2 in [GPa] = 77.2 Enter torque applied at the right end of the element 2 in [N.m] 1400 Element Maximum Stress (MPa) Angle of Twist (Degrees) -27.37 45.02 -0.2673 0.2673 Reaction torque at right end-support -294.94 N.m Reaction torque at left end-support 1105.06 N·m

(b)

Before executing the code to solve sample problem 3.7, modify the code as shown below to incorporate the changes in the units,

Modify line & to !^ in the above MATLAB code as,

fprintf('Elementit Maximum Stress (ksi)t Angle of Twist (Degrees)\n') fprintf ('" % Calculate shear stress and angle of twist. for i-1:1:n n' if i=-1 Torque (i)T(); else Torque (1) =T(1) + Torque (1-1); end Tau (1) ( ( (Torque (1) ) * (OD (1) /2) ) /J (1) ) ; phi (i)(Torque (i))*(L (i)))/((G (1) (J())))(180/pi) UTau (1) (((OD (1)/2))/J(1)); U.phi (i)L (i))/((G(i))(J (i)))) (180/pi) (10 3): (103) -

(b)

Use the data and loading conditions to solve the sample problem 3.7 as,

Consider given 2 shaft element and substitute the following values for each element,

At Element 1:

Length L1 = 60 in,

Outer diameter D0.1 = 2.25 in,

Inner diameter D1.1 = 1.5 in amd

Modulus of rigidity

G, = 1 1.2 x 10° psi

at element 2

Length L2 = 1 in

Outer Diameter D0.2 = 2.25 in

Inner diameter D1.2 = 1.5 and

Modulus of rigidity

G, = 11.2×106 psi

Consider the torque applied at end of the first element a solved in the sample problem

7,-37.7 kip-in

Now, execute the MATLAB code and enter the input values, to obtain the output for the given problem as,

Enter the number of Elements :2 Enter length of element 1 in [in]60 Enter outside diameter for the element 1 in [in] -2.25 Enter inside diameter for the element 1 in [in] 1.5 Enter modulus of rigidity of the element 1 in [psi] -11. 2e6 Enter torque applied at the right end of the element 1 in [kips.in] 0 Enter length of element 2 in [in]1 Enter utside diameter for the element 2 in [in] = 2.25 Enter inside diameter for the element 2 in [in] - 1.5 Enter modulus of rigidity of the element 2 in [psi]-11.2e6 Enter torque applied at the right end of the element 2 in [kips.in] 37.7 Element Maximum Stress (ksi) Angle of Twist (Degrees) -0.34 20.66 -0.0940 0.0940 2 Reaction torque at right end-support -0.62 kips.in Reaction torque at left end-support = 37.08 kips·in

Thus, the maximizing shear stress computed is approximately equal to the yield shear stress given in the problem that is,

Tmax = 20.66 ksi

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