- 106 200 N 800 N 1200 N 600 N T 100 Nm T 100 Nm 100 N 1300 N 200 N 1000 N 100 Mm 100 Mm 100 Mm 1000 N 800 N 200 A 1 (370.07 KiB) Viewed 61 times
- 106 200 N 800 N 1200 N 600 N T 100 Nm T 100 Nm 100 N 1300 N 200 N 1000 N 100 Mm 100 Mm 100 Mm 1000 N 800 N 200 A 2 (380.49 KiB) Viewed 61 times
106 200 N 800 N 1200 N 600 N T= 100 Nm T= 100 Nm 100 N 1300 N 200 N -1000 N 100 mm - 100 mm -100 mm 1000 N 800 N 200 (a) Figure 4.22: Shaft with loading considered in Example 4.13. Mummont mat - teritical 1 306 N 100 N gears and (b) Example 4.13: Critical Location in a Beam Given: A shaft is loaded by the forces and torques shown in Fig. 4.22, which result from the actions of helical the shaft's rolling element bearing supports. The shaft has a diameter of 25 mm. Find: Determine the location in the shaft where the stresses are highest. What are the principal stresses at this location? Solution: First of all, it should be noted that with a diameter of 25 mm, the following can be calculated: M 100 N-m 80 N-m 7(0.025) πd? 4 = 4.909 X 104 m2, A= (c) Figure 4.23: (a) Shear force; (b) normal force and (c) bending moment diagrams for the shaft in Fig. 4.22. G 3 dongrer G 6 sheur axi in force moment fingren 6 Toryndiagram (0.025) I= = 1.917 x 108 mº, 64 64 πα4 7(0.025) JE = 3.835 x 10-8 m. 32 32 The shear and moment diagrams are shown in Fig. 4.23. At first, it is not clear where the critical location is; the ab- solute shear force is highest between 0 and 100 mm, and the moment is highest at 2 = 100 mm. However, the torque is highest between 100 and 300 mm, and the axial load is ten- sile between 100 and 200 mm but compressive between 0 and 100 mm. In practice, a designer must analyze all potential critical locations to determine the most critical. This example will analyze the location just to the right of the gear that acts at C= 100 mm, where V = 200 N, P = 300 N, T = 100 Nm, and M = 100 Nm. These result in the stresses in Table 4.4. Consider the cross section of the shaft shown in Fig. 4.24, with the stress element locations shown. At location A, the axial stress is tensile, but the bending stress is compressive; the resultant normal stress is oz = 0.611 - 65.11 = -64.60 MPa. The shear stress distributions for forsion and shear are shown in Fig. 4.25; clearly at the top of the shaft, the shear stress due to the vertical shear force can be ignored, while the shear stress due to torsion is at its maximum value. Thus, Tız = 32.59 MPa. At location B, the bending stress is zero (see Fig. 4.25). Thus, the normal stress is due to axial force only, or 0. 0.611 MPa. The shear stresses due to torsion and vertical shear are additive; thus, azy = 0.543 + 32.59 = 33.13 MPa. At location C, the normal stresses are both tensile; thus, 0x = 0.611+65.21 = 65.82 MPa. The shear stress is the same as at location A, but it is now negative, so that Tre: = -32.59 MPa. At location D, the bending stress is zero, so the normal stress is the same as at location B, or on = 0.611 MPa. The shear stress due to torsion is zero, but the shear stress due to the vertical shear force has its maximum value of 0.543 MPa. From consideration of these stress elements, we conclus that the element at location Cis critical; it has the largest nc mai and shear stresses. It has the stress state of = 65.82 ME Jy = 0z = 0, Taz = -32.59 MPa, and Tay = Tyz = 0. TE 's a two-dimensional stress state; therefore, one of the pri cipal stresses is zero. The other two principal stresses can obtained from Mohr's circle or from Eq. (2.16), using prof subscripts, as 0 σα +σ, (0u - 0.)2 #VT2z+ 2 4 65.82 65.822 + 32.592 + 4 32.91 E 73.45 MPa. Therefore, the principal stresses are, in their proper or according to Eq. (2.18), 01 = 106.4 MPa, 02 = 0, a 03 = 40.54 MPa.
mm كار 107 loo -ansverse Shear Stress and Strain Table 4.4: Stresses obtained in Example 4.13. Loading Resultant maxinium stressReference Value Axial 300 x = Eq. (2.7) 0.611 MPa 4.909 x 10-4 Bending Mc (100X0.0125) Eq. (4.45) 65.21 MPa 1917 x 108 Vertical shear 4V 4(200) Txy = Table 4.3 0.543 MPa ЗА 3(4.909 * 10-4) Torsion (0.0125)(100) Eq. (432) 32.59 MPa 3.835 310-8 Comprised OB Purit SUL Solution: The reaction forces are found through statics to be RA = 1720 N and RB = 2600 N and are shown in Fig. 4.26c. Figure shows the shear and bending moment diagrams. The maximum shear occurs just to the left of the pulley and equals 2880 N. The maximum bending moment is 430 Nm. In addition, there is a torque of 216 Nm between the pulley and the knife blade. Two locations must be analyzed: the location in the shaft where the moment is largest, and the location where the shear is largest i wnetene som (a) Moment. The magnitude of the normal stress in the c- direction at the location of maximum moment is given by Eq. (4.45) as Figure 4.24: Cross section of shaft at x = 100 mm, with iden- tification of stress elements considered in Example 4.13. (430) ( Mc 4380 d3 Ot = 1 πα" 64 The shear stress due to the torque exerted on the pulley is, from 'Eq. (4.33), Z Тc Try= 뚜 1100 d3 (b) (a) A Mohr's circle can be constructed as discussed in De- sign Procedure 2.5. The Mohr's circle for this case is shown in Fig. 4.28 and has a radius of 2450 Nm/d. Set- ting this equal to the maximum allowable shear stress of 40 MPa yields Figure 4.25: Shear stress distributions. (a) Shear stress due to a vertical shear force; (b) shear stress due to torsion. 2450 = 40 x 106. d3 Therefore, Design of a Shaft for a Case Study: Coil Slitter d = 0.0394 m = 39.4 mm. (b) Shear. The maximum shear stress at the location of maximum shear is, from Table 4.1, 4V Tmax = 3.A = 4(2880) πά 3 4890 d2 Given: Flat rolled sheets are produced in wide rolling mills, but many products are manufactured from strip stock. Fig- ure 4.26a depicts a coil slitting line, where large sheets are cut into ribbons or strips. Figure 4.26b shows a shaft supporting the cutting blades. The rubber rollers support the sheet dur- ing cutting and prevent wrinkling. For such slitting lines the shafts that support the slitting knives are a highly stressed and critical component. Figure 4.26c is a free-body diagram of a shaft for a short slitting line where a single blade is placed in the center of the shaft and a motor drives the shaft through a pulley at the far right end. Find: If the maximum shear stress is 40 MPa and the largest gage sheet causes a blade force of 2000 N, what shaft diame- At one end of the shaft the torsion-induced shear stress is subtracted from this shear stress; at the other end, the effects are cumulative. Thus, the total shear is 4890 Ttot = d2 + 1100 = 40 MPa. d3 Solving numerically gives d = 31.5 mm. ter is needed?