Question

Stress analysis question. Please help solve!

4) The circular bar seen in the figure below is made of a brittle material with ultimate strengths in tension and compression of ou = 290 MPa and ou= 650 MPa, respectively. Let P = 30 N and T = 13 Nm. Using a factor of safety of 1.5 and the Maximum Principal Stress Theory, determine the minimum required diameter, d, to resist failure. (13 points) 1.25m 2.5m

Answer 1

Given :- Brittle Mia material material an Circular bara = 290 MPa, come = 6 ult tension = 290 MPa ult P=30N, TË 13 N-AM Factor of safety (N) = 1.5 dmin to resist by M.P.S.T. design w o failure R 1.25 d 1.25 Torsion at A, B, C = T = 13 Nm at Bending moment x-x à Pxxc at B 7 PX 125= 30x1.25 = 37.5 Nam at C- PX 2.5-PX 1.25 = Px 1.25 - 30x1.25= 37.50.m and also ata 2x2.5- Px1.25= 30x). 25= 37.5 Nimi From Right Side is at is conut M constant Bill aso M= 37.5 Num B (AT= 13 N.M T.M= 13 Nm Dei 4T.M tve Right hand thumb rule TM= 37.5 Nom 16T Td3 o Ta3 CSScanned-with-CamScanner T torsional A =

Page-2] = 381.97 Dend = 32*37.5 - 13 As T 13 I . 16x13 Ta3 66.2 - d² N m² 5 = 6 = 399* ; 3 =0 381.97 03 T = ха Maximum por Maximum Principal theory :-70, 793 >} ther * FS more 10. Do 1985 + 2011]. 393.1 - 2 02.12 393 As the given specimen is Goon brittle material and Loaded as B.M and torsion. Bending stresses are bothe Tensile and compressive, so that the both can equally happens. CSScanned with CamScanner

lage-3 . Page-3) Brittie material is more stronger in compressive stress but weaker in tunsile stresses. compression Tension . So Tult allowable in tension Factor of safety (N) = 290 MPa = 290x106 N wit tension m2 + T5 = 393 +N =15 then (m. d3' = 393 x 1.5 in - 2907106 d² 2.03 x106 de 0.0127 m de 12.7 mm minimum required diameter a de 13 mm CSScanned with CamScanner