Question

Question A-36 steel pipe with an outer diameter of 100 mm and an inner diameter of 80 mm subjected to loadings shown in Figure 1. The pipe is rigidly fixed at B and P - 150 kN. Given the yield stress, Oy -250 MPa and factor of safety, F.S. - 1.5 is used against yielding on this entire pipe. (a) For the stress state at the surface, construct the Mohr circle and determine: (1) the total stresses at surface of the pipe and sketch its stress element; (ii) the principles stresses, (ii) the maximum-in-plane shearing stress and the corresponding normal stress; (iv) the maximum torque, T that can be applied according to maximum shear stress theory; the normal and shearing stress after the element has been rotate 10 clockwise. (v) [25 marks] (6) Explain the usefulness of Mohr's Circle in engineering context and discuss the advantages of Mohr's circle. Quote any references. [15 marks] 80 mm 100 mm Figure 1: A-36 steel pipe with loadings.

Answer 1

- A-36 steel DO. Toomm Couten Diameter) De = 80mm (Inner Diameter ) P 2 150kn y= -25omia F-S z 115 P © considen this section (som of shaft ) soss section o Shean stress Toomm goomm Distribution + Uniform Compressine stress due to P On combining Max stress at extreme Point on the surface of the shaft we get Critical element z Que to T) o (due to p) Critical stress elemento (۱) - Compressive) ale 150x103 x(100)?-(8032 1o3_50*)] г (1 53.078 mpa

T is not given, it can be find out by applying Von-misis. theory. According to Non-mises theory, Jose-vj+(3-625+ + Car-monsito (es. vznis 28). Selain 2 J / { -53.078-0)? +lomo)+(0+53.070)+.G (2xy' toto3= 250 15 (5634,548 + 6 Eng?) 166.66 (5634.5487 62ay?) 2777777 २ Zay gia14 Mpa From (i) Stress element 91:214 the ) 3 H 53.078 53-078 91.214 na we can Draw can With the help of this element, Mohr circle and easily we find Max Principal stress. Here need to Find T We get shear stress corresponding to Torque To Just plot in mour Circler no se 91.214 mia.

Mohr Circle = 121.535 प In Mohr Circle, After Plotring R2 max inplane Shearing stress 2ny and a 94.996 Is 91.214=220 We get son Principal stress 2ny = (gland o=68.457 53.678= (ii). Principal stresses on toy t + Zag? Here Tyzo onzoz-53-078 2nya 91.214 -5 3.078 to + (1༠༣ -༠)༸-+ (2) ད་ 2 -26.539 15 726.5393+(9121432 -26-539 94.996 Max Principal stress 2= - 26.539 gy.996 - 121.535.

Min Principal stress - 26.539 + 94.996 Just see the Magnitude 3>T 68.457 mia. Max and Min Principle stress Angle. from the no plane 1 tan 20,2 22ny 7 tan 20p 2x gialy - 53.078 Opa 36.886 (rr) Max inplane stress It is given by "max (2man) २ 02 2 2max 2 68.457 - (-1211535) 2 max {In plone ) 2 2 12 mo gy.gg6 Mpa. This maximum inplane This shear stress Corresponding from -plane to {ustep } Angle • (2max= 94.996) (45tap)

Corresponding Normal strena at (ust op? Bogle, From (2) Q = 45+36-781, = 81.886: antolat on an on-oy Casaco + 2ay sinace ? -53.07840 + २ .(-53-078-o) Cos (2x81.886) + 91214 Sin (2x 87886) for - 26.539 - 26.539 X -0.960 t51.214x 0.279 24.387 Corresponding to Max inglane stress Condn. Normal stress Corresponding to in-plane stress Canda. Civ) Maximum torque 'T' max shear stress stress theory or Tresca theory 2max (incluerd) = 2man (material strength) 2max (material Shem strength ) by og=yield tensile strength. + Zay? 2. Fos 2ny = IGT (salia shaft) na 3

From Torsion ear 7 7 1 = 22 Ipe Polar moment of zny Inertia T2 Torque 22 Shear stress 32x I m ( coa14_809) 2 50 Rz.Outen Radius 0.00000862T=2ay But in (3) ea 2. Version-034 +( OFUNDO086T 250 2x 1.5 704, 318 +2.396815"T? z 6944.44 Ta 9185.400 Nom Torque Max becording to Shear stress theory b) Many Engineering students are introduend to the ideas and Concept of Mohr Circle when studying state of stress due to Various loading Conda. on structures of components. a The Benefit that Mohr's Circle Provids is that it Creates Visual Representation of what is happening and the relative positioning of the Stress states on an element relative to a of Co-ordinate axes.