Question

Part A - Normal Stresses, Shear, and Angles The stress element shown in the figure below is subjected to the indicated stresses of magnitude 0,1 = 35 MPa, oyl = 57 MPa, and Tryl = 41 MPa Oy Txy Determine the principal normal stresses 01 and 02, the maximum in-plane shear stress Tmax,in-plane, and the angles at which they occur relative to the given stress element. Follow the sign convention. Suppose that when the element is oriented at an angle 0p1 relative to the given stress element, the normal stress that acts along the z' axis of the element is equal to the maximum normal stress 91. When the element is oriented at an angle 0p2 relative to the given stress element, the normal stress that acts along the r' axis of the element is equal to the minimum normal stress 02. When the element is oriented at an angle 9, relative to the given stress element, the shear stress that acts on the element is equal to the maximum in-plane shear stress Tmax,in-plane and is positive according to the sign convention Express your answers, separated by commas, to three significant figures. Express angles in the range 8 € (-180°, 180°). View Available Hint(s) AED J1 vec ? 01 =, 0p1 = .02 =, 0.2 =, Tmax,in plane = , 03 = MPa,, MPa, MPa, Submit Request Answer

Answer 1

Oy Txy Giveno- Stresses Ox=-35 MPa (compressives Cy=-57 MPa (compressive) Txy = = 41 MPa Ox coneider following Stege element ABC D By А Exy Ox Exy o is given by (when a measured from Reference plane for equilibrium of the element ABCD me Normal stress at any Reference plane) Oxt by On ( ox-Oy) Cos2O + Exy sin 20 2 shear stress at any o is given by : Ox-Oy T= COS20 finzo - Exy ☺ + (070) • Find location of principal plane ( plane on which shear Stress is zor tan 20 = txy 3 (ox-Oy) put value of Ox by Ox, dy & txy in equation ③ we get tan 20 = 2 (41) (-35-(-57)] O, = 37. 4-9

Og = 0,+ goo O2 = 37.89 +90 0g = 127.49° On = 0,402 is the plane of principal normal stress (ie, mawimum and minimum principal stress) * find principal normal stress at 0,= 37.49' [from egna] -3541-57+ (+352 657) cos (2x 37.49)+(41) Sin (2x37.4.06 On=-3.55 MPa (compreseive) & principal normal stress at Og = 127.49° [from eam 0] On’= (-3574-57))+(354-57)) cos( 2x125.49)+ (41) 8in (24|2749) ( compressive) On -88.45 mpa

Maximum Normal principal stress maximum of Ton, on'] Oi= -88.45 MPa (compressive) & plane for maximum Normal prinubad stress ( Om Opi 0g = 127.49' measured from reference plane in Anticlock wise direction. Minimum Normal prinubal stress trinimum of [on, on'] Oz=-3.55 MPa (compressive) A location of plame for minimum nomal principal stress (92) Opa = 0,= 37.490 measured from Reference plone When measured in Anticlockwise direction * Find location of in-plane monimum Shear Stress tan 2 Oz = Oy - Ox 2 Exy tan 203 = -57-(-35) 2141) Oz -7.50° [-ve sign shows that Os is measured in clockwise from Reference plane 04 = -7.50+go 04 = 82.5° 03 & Maximum in-plane shear stress at 03 [from equationa? ox-oy Reference plane -73576-57] san [2x6-75)–(41) cos (2267-5)] ( Eman implone -42.44 MPa

(Tmax) in in plane at Oq (inplane tmage will be) Sin 204 – Txy Cos204 -35-(-57) Sin (282-5) – (41) cos (2x625)] Tmax) in plane = 42.44 MPa 2 3 Tmax) in plane Oy A + 42.44 MPa Shear Streys are complementary in nature 4,63 264 ore complementary planes, in complementary planes magnitude of sheor Stresses are equal but of opposite sign. (that is why at 03 (T max in plane is negative and at Oq.(Imam) in plane Le positive) * Sign convention used:- ① BC is taken as Reference plane 6 ox is taken as positive when tensile in nature. (negative for compressives ③ by is taken as positive when tensile in nature. (negative for compressive spess) @ Try is taken 98 positive when anticlockwise, m reference plane. [ie. Opty Bois taken as positive when measured in anticwek write direction from reference plane) 2 Oy on Txy ry couple