Question

The column illustrated to the right measures 3.25 feet in length with a diameter of 0.750 inches. If there is a required design factor of 3, determine the maximum allowable load. Assume the column is made of 1020 Hot Rolled Steel.

FIXED D=0.750 in L=3.25 ft PINNED 1/

Answer 1

solo given, length of the columm, L= 3.25 ft. = 0.9906 m diameter of the column, d= 0.750 in - 0.01905 m factor of Safety, N = 3 Now, The value of for 1020 Young's modulus of elasticity HOT ROLLED STEEL is E = 205 Gla = 205 x 10 9 N/m2 Now, The Critical load on the column by using EULER'S EQUATION Cutical load, lov - T²E I Le where, E = Young's modules of elasticity I = moment of Inertia Le = effective length of Column. Now, moment of Inertia I - I (d") of rod IF I (0.01905)" I = 6.4614 x 10-9 me 64 64

also, we know that For the Column in which Other end us Hinged le - L one End is fixed and the effective length is 12 de = 0.9906 = 0.7004 m. Now, The the Column Critical load on Por = x² EI lor = 2 x 205 x 109 x 6.4614 x 10-9 (0.7004)² Per - 26622.4187 N So, The Safe load , Po l o - Critical load factor of Safety on the Column N = 26622.4187 P = 8874. 1395 N Hence, The load on Column is maximum allowable 88 74.1395 Newton.