Question

4.) The screw of the clamp exerts a compressive force of 500 lb on the wood blocks. Cross section a-a of the clamp is rectangular, 1.00 in. by 0.50 in. • Draw a free body diagram of the clamp sliced by the plane passing through a-a • What should be the value of parameter L in order to have 52% of the cross-section - in the state of tension? • For this value of L. calculate max tensile and max compressive stress in the cross- section and sketch the stress distribution

Answer 1

Annos ab EBD - 500lba na Plini SAN = 500 db M = r ezohet Us 500L alb.in (6) At tensile 0.02H distance from neutral axis the to tensite stress will be same as axial para because the 0.52H is in tension stress .. who is = 160.02H 500 1x0.5 = 500 L (0.02H) 10.5x139,5 LH= 4.1666 in Lx1 = 4.1666 in 1 L= 4:1666 u 1 Scanned with CamScanner

(obt tre 7 -1 F F - To.o2HF 0:514 0.485 (-35) qxy combined Stress duet so stress due so stress distribution moment axial Sorce. ТА My tuan 500*4.1666% (0.50H), 50 (0.5*13) (12 - 500x4.1666x0,5@xl (0.5*13)/12 (Elmer = 26000 psi. ] b - 0.5 0-50H DO 500x4.16667 Best 0.5*13)/12 ht =1000 - 500 * 4.16667 Bekas 0:50x1 (0.5x13/12 VO) = -24000 pel! CSScanned with Cambčanneres