Question

Q1. [15 Marks] The stress field in a thin flat structural member can be described by the following equations: Ox 20x + 30xy2 + 20y Oy = 30x2 – 50xy + 20 Txy = 20x – 100y? 02 = Txz = Tyz = 0 where (x, y, z) are in metres and the stress components are in MPa. a) Determine the body forces required to support the structural member in equilibrium; [2 marks] b) Draw a 2D stress element at the point - P(x, y) = (3 m, 1 m); [2 marks] c) Find the strain state at point P if E =200 GPa and v = 0.3; [3 marks] d) Draw a Mohr's circle of the stress state at point P; [3 marks] e) Determine the principal stresses at point P; [2 marks] f) If the yield strength of the material is Oy = 300 MPa, will yielding occur at point P in accordance to Von-Mises theory? [3 marks]

Answer 1

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Solng Griven, 20 x + 30 х? + 20 у Зох*- 50 xy + 20 Тх 20 x 1 ооу- txz тəz = во а ) The equilibrium equations in x and y directions are given by, Әох + ƏTay ду {x-direction } (0) хь = о + Here, Әо; әх 20 + Зоз- ƏTxy - 2009 ду 4) =) 20 + 3oy*+ -20oy + х = 0 = хь = 2009 – 30 - 20 ( x-component of body force). Әty Әх + доу sy Ү, а (2) {y-direction Here, - 50х. эх 20. доу ә) дх (2) =) 20 + - Sox + Yь У. 5ox -20 : (y-component of body force)

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are b) At point p(x,y) = ( 3m, im), the stress components given by, 2 HOMPa ox 200 + 30 xy +204 20x3 + 30% 3x12 + 20x1 30x²_5oxy Oy + 20 = 30 x 3%-50% 3x1 + 20 = 140 MPa 20x 3 - 1oox 100x12 -40 MPa Toy = 20% -12 -100y? y = 140 MPa Tay = 40MPa Jx = itonpa Txy = •40Mpa 69 2D stress element at point p. Normal Strain in c ) c-direction Ex Given, 7 - Voy E = 200 Gpa = 200x103 MPa (x-165) V = 0.3 (170 – 0.34140 200x103 0.00064

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Normal strain in y-direction A Ey = = (T-05) (140 - 0.3x170) 200x103 0.000445 Abreranstrain May Shear strain Vxy Tuy G Here, E 200x103 G= 76923.077 MPa 2(140) 2(1+0.3) So, Vay - 40 0.00052 76923.077 Therefore, strain state at point p is givens, Ex = 0.00064 Ey tay Ey 0.000 445 rcy 0.00052 Ex Vay

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d) Jy = 14ompa Tay=40 А B r = 170 MPa b c Try = 40mpa The normal and shear stress acting on the surface BC can be expressed as a point in -T plane as, A Coc, tzy) A (170, 40) Tocy is positive since it is giving clockwise rotation) The normal and shear stress acting on the surface A B can be expressed point in as o-plane as, Tocy is negative sinu it B (by, Try) B (140,-40) is giving anti-clockwise rotation. τ GO A(170, 40) 40 20 D с O o 60 80 100 140 120 40 160 20 180 200 -20 -40 Bt Tho, -40) -60 Mohr Arcle

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Maximum Principal stress is given by max Ox+by 2 + + 2 2 2 170 +140 170-140 + + 402 2 2 197.72 MPa 어 Minimal Principal stress is given by 5x+6y 2 + Txy2 159) (1+0=140) 2 ITO + 140 + 402 2 112.28 MPa 2 f) Given, Yield strength of = 300 MPa. According to Von- Mises theory, to prevent failure 0,²+02² - ooz s og

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Here, 197.72 MPa Ta Marepmum principal stress 112.28 MPa. Minimum principal stress Then, 2 < 300. (197.72) + (112. 28) - (197.72)(112:28) 171.756 < 300 =) (Condition is satisfied) So, yielding will not occur at point p according to Von-Moses Theory,