Question

problem 78

Flate plate (width = 2.00 in 1.25 in 1.50 in -2.25 in- - 4.00 in -4.75 in Beam Beam for Problems 75 and 76 FIGURE P5-75 Design Problems Forcach of the followin to achieve a minimum design dimensions, and the materie tward an efficient desi she following problems, complete the requested design imum design factor of 3.0. Specify the shape, the d the material for the part to be designed. Work efficient design that will have a low weight. ok shown in Figure P5-77 carries a load of 3000 N pplied and released many times. The link is ma- Lined from a square bar, 12.0 mm on a side, from SAE 14 OOT 1100 steel. The ends must remain 12.0-mm are to facilitate the connection to mating parts. It is desired to reduce the size of the middle part of the link to reduce weight. Complete the design. 78. Complete the design of the beam shown in Figure P5-78 to carry a large hydraulic motor. The beam is attached to the two side rails of the frame of a truck. Because of the vertical accelerations experienced by the truck, the load on the beam varies from 1200 lb upward to 5000 lb downward. One-half of the load is applied to the beam by each foot of the motor. 79. A tensile member in a truss frame is subjected to a load that varies from 0 to 6500 lb as a traveling crane moves across the frame. Design the tensile member. 80. A hanger for a conveyor system extends outward from two supports, as shown in Figure P5-80. The load at Middle section to be designed 12-mm square FIGURE P5-77 Link for Problem 77 Hydraulic motor Side rail of truck Beam to be designed -16 in - 16 in --50 in POURE P5-78 be P5-78 Beam for Problem 78

Answer 1

solution AR Since the forces are symmetric, RA = R = F = FA Bending moment would be in along Fection (). (About A Fr=0 Case 1 :D Mx = 16F 2F = 5000 lb (downward hence tre) F = 5000/2 = 2500 lb Mamax = 16* 2500 = 40,000 lb. in Case 2 - 247-1200 db Cubaord, loence -ve) F = -600 lb Mamin = 16 * (-1800) = -9600&bin o= my ) omax = 40000 ("/) Omin=-9600 (35) Scanned with CamScanner

As ber modified Godman (sitesia , Sa = 5a = | Tmax Ipin | = 24200 (*/z) Suzas - = 1520-19%) Imax t min Sn = am - - 15200 For 0.4% Corbon Steed Su = 8x109 psi , se = 4X10% bsi , Fos = 2 24800(83) + 15200(*13) - 4 x 104 8 x 104 < 0.6173 For So for American Standard Beams s-beam , S3x5.7 , y = 6/2 = 3/2 = 1.5 In = 2.52 15 =0.59 <0.61 7.52 Scanned with CamScanner