Question

2 ft 5 ft 6 ft 23 points) The beam AB is subjected to the loadings shown?. This beam is supported by a double-sheared pin at A and a rod CB. If the failure normal stress is 60 ksi and the failure shear stress is 50 ksi, determine the minimum required diameter of the rod BC and pin at A, to the nearest 1/8”, if the factor of safety is 3. 630 lb.ft 225 lb/ft A E B 30° 200 lb CA

Answer 1

Sols 225461ft. 630 1ъg КА“ B 30° доод FBc Using Statics Equations, we get 3F 0 : КА Ғве » 30° — (1) 4fy=o : А + goo + FBC sin 30° +1 6xaas А + fse m 30 = Ч7 S — (ii) 4н,-o; (абок 2) + (6 36 ) + (ft , з'x43) ( 1+6 ) 16 aas 1 10 Зо в 6-5 Fься ячas fect 083 - 8s Хь دلا ... 16. 9as Aь КА > 8 Sa, оч дь

for Rod :- 2 allo o failure Goksi N= 3 2 Taitowa 60 aoksi 3 Now, Pallow FBC 2011 983.85 ㅍd2 d= 0.25 in Diameter of Rod 0.25 in for Pin: Pin will fail in shearing As pin is in double shear. Ashen = 2*( 1 & ²) = I do e failure Soksi. N=3 Tallow = so 3 د لبهللما 16.667 ksi

Now, Tallow VA 3 Ashear 16.667 X 10 3 16.gas T d² II a 0.0254 in Nearest to 0.125 in. 1th inch will be : Diameter of pin 0.125 in Completed