Question

③ A cantilever beam (as shown) is made from 440 steel. F.S. = 2.5 is required. Detection found no cracks for more than 1/32 in) What is the maximum allowable and force? solutions

Answer 1

solution: CCCC + 1 = 2 ft 0.5" - a a=0.5" Given Data: From Data Provided for 4340 steel, Jy = Yield strength = 470 MPa E = Modulus of Elasticity = 200 G Pa = 200 X 10 MPa FOS - Factor of Safety = 2.5, a=0.5", 1-.2 ft Flexural Formula, M = . E ... where, M = Bending Moment o = Bending stress at a distance y from - Neutral Axis I - Moment of Inertia of cross-section - about Neutral Axis E = Young's Modulus of Elasticity ě = Curvature

• Using Flexural Formula, ..@ I As, 5 = Fos o=470 2.5 o = 188 MPa Now, 25" Neutral Axis 0.5" y = 0.25" - 12.7 mm l = length of beam = 2 ft = 609.6 mm I= a" - 12.74 = 2167.87 mm+ 12 Using flexural formula, .: M = 0. I M = 188 X 2167.87 12:7 M = 32091.30 Nmm

As we know, M = Fid where, M = ending Moment F = Allowable End Force l = length of Beam F = M. F = 32091-30 609.6 F = 52.64 N. Conclusion End force is equal : Thus the allowable to 52.64 N
Conclusion - Thus by using flexural formula for bending we can calculate the allowable end force for the cantilever beam. In the solution provided above the value of yield stress is considered as for the ductile material solution is carried out using yeild properties while for brittle material, the ultimate stress properties are considered for calculation purpose. The value of yield stress can fluctuate some what as different standards provide different values for materials, but more probably the fluctuation will be negligible so the solution provided can be considered accurate without any obligation.