5 ksi the outward normal is in the positive x

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Unformatted text preview: BC ( ρ BC ) max ( 6 π in. · kips ) ( 2 in. ) ( τ BC ) max = ------------------------------- = ------------------------------------------------ = 1.5 ksi 4 J BC ( 8 π in. ) Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm T AB ( ρ AB ) max ( – 2 π in. · kips ) ( 1 in. ) ( τ AB ) max = ----------------------------- = ---------------------------------------------------- = – 4 ksi 4 J AB ( π ⁄ 2 in. ) (E10) T CD ( ρ CD ) max ( – 2.5 π in. · kips ) ( 1 in. ) ( τ CD ) max = ------------------------------- = -------------------------------------------------------- = – 5 ksi 4 J CD ( π ⁄ 2 in. ) (E11) From Equations (E9), (E10), and (E11) we see that the magnitude of maximum torsional shear stress is in segment CD. ANS. τ max = 5 ksi (c) The direction of shear stress at point E can be determined as described below. Shear stress direction using subscripts: In Figure 5.27a we note that τxθ in segment BC is +1.5 ksi. The outward normal is in the positive x direction and the force has to be pointed in the positive θ direction (tangent direction), which at point E is downward. Shear stress direction determined intuitively: Figure 5.27b shows a schematic of segment BC. Consider an imaginary section through E in segment BC. Segment BE tends to rotate clockwise with respect to segment EC. The shear stress will oppose the imaginary clockwise motion of segment BE; hence the direction will be counterclockwise, as shown. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 222 () 8 in kips 1.5 ksi x y B 8.5 in kips 1.5 ksi E (a) x C (c) 8 in kips Figure 5.27 Direction of shear stress in Example 5.6. (b) We complete the rest of the stress cube using the fact that a pair of symmetric shear stresses points either toward the corner or away from the corner, as shown in Figure 5.27c. COMMENTS 1. Suppose that we do not follow the sign convention for internal torque. Instead, we show the internal torque in a direction that counterbalances the external torque as shown in Figure 5.28. Then in the calculation of φ D – φ A the addition and subtraction must be done manually to account for clockwise and counterclockwise rotation. Also, the shear stress direction must now be determined intuitively. 2 in kips 8 in kips TCD 2 in kips 2.5 in kips TAB A TBC D B B 8 A 10 3 rad cw C 3.75 B 10 3 rad ccw D C 12.5 10 3 rad cw Figure 5.28 Intuitive analysis in Example 5.6. 2. An alternative perspective of the calculation of φ D – φ A is as follows: φD – φA = xD ∫x T------ dx = GJ A xB ∫x T AB ----------------- dx + G AB J AB A xC ∫x B T BC ------------------ dx + G BC J BC xD ∫x C T CD ------------------- dx G CD J CD or, written compactly, Δφ = Ti Δ xi ∑ --------------Gi Ji (5.13) i 3. Note that TBC − TAB = 8π is the magnitude of the applied external torque at the section at B. Similarly TCD − TBC = −8.5π, which is the magnitude of the applied external torque at the section at C. In other words, the internal torques jump by the value of the external torque as one crosses the external torque from left to right. We will make use of this observation in the next section when plotting the torque diagram. Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 5.2.6 Torque Diagram A torque diagram is a plot of the internal torque across the entire shaft. To construct torque diagrams we create a small torsion template to guide us in which direction the internal torque will jump. A torsion template is an infinitesimal segment of the shaft constructed by making imaginary cuts on either side of a supposed external torque. (a) (b) Text T1 Text T1 T2 T2 Figure 5.29 Torsion templates and equations. Template Equations T 2 = T 1 – T ext T 2 = T 1 + T ext Figure 5.29 shows torsion templates. The external torque can be drawn either clockwise or counterclockwise. The ends of the torsion templates represent the imaginary cuts just to the left and just to the right of the applied external torque. The internal torques on these cuts are drawn according to the sign convention. An equilibrium equation is written, which we will call the template equation January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 223 If the external torque on the shaft is in the direction of the assumed torque shown on the template, then the value of T2 is calculated according to the template equation. If the external torque on the shaft is opposite to the direction shown, then T2 is calculated by changing the sign of Text in the template equation. Moving across the shaft using the template equation, we can then draw the torque diagram, as demonstrated in the next example. EXAMPLE 5.7 Calculate the rotation of the section at D with respect to the section at A by drawing the torque diagram using the template shown in Figure 5.29. PLAN We can start the process by considering an imaginary extension on the left end. In the imaginary extension the internal torque is zero. Using the template in Figure 5....
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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.

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