Question

5. A cantilever is 200mm long. It is subjected to an axial load which varies from 100N compression to 400N tension. It also takes a transverse load at its free end which varies from 70N up to 120N down. The cantilever is of circular cross-section. It is of diameter 2d for the first 50mm and of diameter d for the remaining length. Determine its diameter taking factor of safety as 2. Assume yield strength 320MPa, Endurance limit in reversed loading = 300MPa, correction factors = 0.7 in reversed axial loading and 1.0 in reversed bending. Stress concentration factor = 1.42 for bending and 1.64 for axial load. Assume other factors if required.

Pls solve the question regarding design of machine elements. Also Pls note that the values in the diagram are not to be taken and the values from the question are to be considered.

Answer 1

P load σ :Stess Given that th4) 200mn Ptax 120N find mean azcal load and Shess am15o -150N can, t an LI 250 av ay low, Calculate equivalent aio Stess ad junction due to aciol toad Corection Pactovs

C250x u) x 320 x 1、64 150 x u TT d2 . 41 082 T9-815 986.ST Ma 2 calculate mean bending load and Vaviable bendi ing lood by Nlow, calculate mean adVaniable bd moment at unction Mom = 25X150 We know thal Section modulus Iz) 위ov cfvculav CrOSS Section 32

Now, calculate Mean and Vaiabie bending Stress ress hom Bending Equation | T--y E M 3150 38265 . 306 mb 3 0.0986 Mb 4250 5uo8.163 Calculate cqivalert nomal styess at juncion 10n) due t bondina 38 26S.306 45408.163x 320 x N2 3826S. 306 ウ20244-897 3 2 258S 10.203

calaulate total eguivalert nomal esdunctian 986 .qST 、 貝58510.203 2 Finaily the Teguined diameier TE Fos 只 (486.asd958SIo. 2036o d By Solving above Equation le ji