A 30-ft (9.1-m) long steel shaft weighing 150 lb/ft (223.2 kg/m) of length has a 500-lb (2224.1-N) concentrated gear load 10 ft (3.0 m) from the left end of the shaft and a 2000-lb (8896.4-N) concentrated pulley load 15 ft (4.6 m) from the right end of the shaft. Determine the end reactions and the maximum bending moment in this shaft.
Calculation Procedure:
Figure 1 a shows a sketch of the shaft. Label the left- and right-hand reactions LR and RR, respectively.
Take moments about RR to determine the magnitude of LR. Since the shaft has a uniform weight per foot of length, assume that the total weight of the shaft is concentrated at its midpoint. Then 30 LR - 500(20) - 150(30)(15) - 2000(15) = 0; LR = 3583.33 lb (15,939.4 N). Take moments about LR to determine RR. Or, 30 RR - 500(10) - 150(80)(15) - 2000(15) = 0; RR = 3416.67 lb (15,198.1 N). Alternatively, the first reaction found could be subtracted from the sum of the vertical loads, or 500 + 30 X 150 + 2000 - 3583.33 = 3416.67 lb (15,198.1 N). However, taking moments about each support permits checking the results, because the sum of the reactions should equal the sum of the vertical loads, including the weight of the shaft.
The maximum bending moment in a shaft occurs where the shear is zero. Find the vertical shear at each point of applied load or reaction by taking the algebraic sum of the vertical forces to the left and right of the load. Use a plus sign for upward forces and a minus sign for downward forces.
Designate each shear force by V with a subscript number showing its location, in feet (meters) along the shaft from the left end. Use L and R to indicate whether the shear is to the left or right of the load. The shear at the left-hand reaction is VLR = + 3583.33 lb (+15,939.5 N); V10L = 3583.33 - 10 X 150 = 2083.33 lb (9267.1 N), where the product 10 X 150 = the weight of the shaft from the point VLR to the 500-lb (2224.1-N) load. At this load, V10R = 2083.33 - 500 = 1583.33 lb (7043.0 N). To the right of the 500-1b (2224.1-N) load, at the 2000-lb (8896.4-N) load, V20L = 1583.33 - 5 X 150 = 833.33 lb (3706.9 N). To the right of the 2000-lb (8896.4-N) load, V20R = 833.33 - 2000 = -1166.67 lb (-5189.6 N). At the left of VR, V 30L = -1166.67 - 15 X 150 = -3416.67 lb (-15,198.1 N). At the right hand end of the shaft V30R = 3416.67 + 3416.67 = 0.
Draw the shear diagram (Fig. 1 b). This diagram shows that zero shear occurs at a point 15 ft (4.6 m) from the left-hand reaction. Hence, the maximum bending moment Mm on this shaft is Mm = 3583.33(15) - 500(5) - 150(15)(7.5) = 34,340 lb·ft (46,558.8 N·m).
Related Calculations: Use this procedure for shafts of any metal--steel, bronze, aluminum, plastic, etc.--if the shaft is of uniform cross section. For nonuniform shafts, use the procedures discussed later in this section.