This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Since the degree of static redundancy is 1, the simplest approach is often to take the left wall (or the right
wall) reaction as the unknown variable. We can then apply the compatibility equation, as outlined next. 5.3.1 General Procedure for Statically Indeterminate Shafts. Step 1
(right) wall
torque.
Step 2
Step 3
January, 2010 Make an imaginary cut in each segment and draw freebody diagrams by taking the left (or right) part if the left
reaction is carried as the unknown in the problem. Alternatively, draw the torque diagram in terms of the reaction
Write the internal torque in terms of the reaction torque.
Using Equation (5.12) write the relative rotation of each segment ends in terms of the reaction torque. M. Vable Mechanics of Materials: Torsion of Shafts 5 240 Step 4 Add all the relative rotations. Obtain the rotation of the right wall with respect to the left wall and set it equal to
zero to obtain the reaction torque.
Step 5 The internal torques can be found from equations obtained in Step 2, and angle of rotation and stresses calculated. EXAMPLE 5.12
A solid circular steel shaft (Gs = 12,000 ksi, Es = 30,000 ksi) of 4in. diameter is loaded as shown in Figure 5.62. Determine the maximum shear stress in the shaft.
90 in kips A B x
3 ft Figure P5.62 Shaft in Example 5.12. 240 in kips C D 4 ft 7 ft PLAN
We follow the procedure outlined in Section 5.3.1 to determine the reaction torque TA. For the uniform crosssection the maximum shear
stress will occur in the segment that has the maximum internal torque. SOLUTION
The polar moment of inertia and the torsional rigidity for the shaft can be found as
2 π ( 4 in. )
4
J =  = 25.13 in.
32 4 3 GJ = ( 12000 ksi ) ( 25.13 in. ) = 301.6 ( 10 ) kips·in. 2 (E1) Step 1: We draw the reaction torques TA and TD as shown in Figure 5.44a. By making imaginary cuts in sections AB, BC, and CD and taking the left part we obtain the free body diagrams shown in Figures 5.44 b, c, and d.
90 in kips 240 in kips TD TA
A x
3 ft B C D 4 ft
(a) 7 ft 90 in kips 90 in kips
TA TAB A TAB(b) TA TBC TA A
TBC (c) A
T B 240 in kips
TCD TA A B 90 TC D (d)A
T C
150 Figure 5.44 Free body diagrams of (a) entire shaft; (b) section AB; (c) section BC; (d) section CD.
Step 2: By equilibrium of moments in Figures 5.44 b, c, and d. or from Figure 5.45b we obtain the internal torques as
T AB = – T A T BC = (  T A + 90 ) in.· kips T CD = (  T A – 150 ) in.· kips (E2) Step 3: Using Equation (5.12), the relative rotation in each segment ends can be written as
– T A ( 36 )
T AB ( x B – x A )
–3
φ B – φ A =  =  = – 0.1194 ( 10 ) T A
3
G AB J AB
301.6 ( 10 ) (E3) T BC ( x C – x B )
(  T A + 90 ) 48
–3
φ C – φ B =  =  = ( 0.1592 T A + 14.32 ) ( 10 )
3
G BC J BC
301.6 × 10 (E4) (  T A – 150 ) 84
T CD ( x D – x C )
–3
φ D – φ C =  =  = ( 0.2785 T A – 41.78 ) ( 10 )
3
G CD J CD
301.6 × 10 (E5) Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Step 4: We obtain φ D – φ A . by adding Equations (E3), (E4), and (E5), which we equate to zero to obtain TA:
φ D – φ A = ( – 0.1194 T A – 0.1592 T A + 14.32 – 0.2785 T A – 41.78 ) = 0 or
14.32 – 41.78
T A =  = – 49.28 in.·kips
0.1194 + 0.1592 + 0.2785 (E6) Step 5: We obtain the internal torques by substituting Equation (E6) into Equation (E2):
T AB = 49.28 in.·kips T BC = 139.28 in.·kips (E7) T CD = – 100.72 in.·kips For the uniform crosssection, the maximum shear stress will occur in segment BC and can be found using Equation (5.10):
T BC ( ρ BC ) max
( 139.3 in.· kips ) ( 2 in. )
τ max =  = 4
J BC
25.13 in. (E8)
ANS. January, 2010 τ max = 11.1 ksi M. Vable Mechanics of Materials: Torsion of Shafts 5 241 COMMENTS
1. We could have found the internal torques in terms of TA using the template shown in Figure 5.45a and drawing the torque diagram in
Figure 5.45b.
T
T
T1 90 T2
A TA B C D TA
T2 Figure 5.45 T1 T
150 Template and torque diagram in Example 5.12.
(a ) TA (b) 2. We can find the reaction torque at D from equilibrium of moment in the free body diagram shown in Figure 5.44d as:
T D = 90 – 240 – T A = – 60.72 in.·kips
3. Because the applied torque at C is bigger than that at B, the reaction torques at A and D will be opposite in direction to the torque at C.
In other words, the reaction torques at A and D should by clockwise with respect to the x axis. The sign of TA and TD confirms this
intuitive reasoning. EXAMPLE 5.13
A solid aluminum shaft (Gal = 27 GPa) and a solid bronze shaft (Gbr = 45 GPa) are securely connected to a rigid wheel, as shown in
Figure 5.46. The shaft has a diameter of 75 mm. The allowable shear stresses in aluminum and bronze are 100 MPa and 120 MP...
View
Full
Document
This note was uploaded on 04/08/2010 for the course ENGR 232 taught by Professor Smith during the Spring '10 term at Aarhus Universitet.
 Spring '10
 SMITH
 Torsion

Click to edit the document details