Question

2. The C-clamp applies a compressive force of 40 lb on the cylindrical block. Determine the maximum normal stress developed in the clamp. Sketch the stress distribution at the (clamp) section. (20 points) in. 0.25 in. 4 in. 4.5 in. 0.75 in.

Answer 1

tbt O sinan 0.25 in Ri=4 hal" b=0.254 14.54 J 0.75" Ginen IF = 40 lb / Ri= 41 Ro Rith = 4"+1'55" for a - rectangular section, Radius of curvature for neutral axis is h m(Re) So MN- Ri 4.484 h (3) And centroidal axis ; Radius of cerna ture is Mota Rith 2 SO, roo = 4 + 4,5" 2 Nowe distance between centroidal axis and neutral axis is → MG-ru e = = 4.5 – 4048 = 0.02"

Now, distance between load and controidal M= 4G+ 4.5" axis => = 4-5 +4.5 = qui Now, Bending axis moment about centroidal M = Exx 408 9 360 lb-in section Direct tensile stress at the given due to 40 lb load F 40 t А 110.25 Os = 160 lblina Now; distance from neutral axis to inner surface Yi = rn Ri= 4.48-4 = 0.4811 Distance from neutral axis to outer surface Yo=Ro-rn=5-4.48= 0.52" stress at inner surface Obi= Myi A.e. Ri 0.25 0.02% 4 = 8,640 lb/in2 reax bending 360X0.48

Maximum bending stress at out or surface = MYO 360 X 0.52 Ae. Ro 0125 x0002 * 5 = 7,488 (blin2 Resultant stresses on inner surface of+ Obi= 160+8640 = 8,800 Ubl in2 (tensile) Result ant stresses on outer surface +- bo = 160-7,488 =-7,328 lblin2 (compressine) stress distribution by obo obi reaximum normal stress at inner surface Omax= 8,800 (blin = 8,800 psi