Question

Question 1. The state of plane strain on An element is Ex-300010), tye 0, Yoxy = 150 (10) Determine the equivalent state of strain which represents the principal strains, and the maximum in-plane shear strain and the associated average normal strain. Specify the Orientation of the corresponding elements for these states of strain with respect to the original element dy X Part A-Determine the orientations of the element at which the I principal strains occur. Open Op = -13.3, 76.7° Ans. Part B - Determine the normal strain Ex' of the element with onientation, Op = -13.3° Ex'= -3.18x1014 Ans. Parte - Determine the normal strain Ey of the element. with onentation Op = -13.3 ty = 147x10-5 Ans Part. Determine the shear strain Yx'y of the element worth orientation Op -13.3 Xx'y = 268010-6 Ans. Part 6 Determine the orientations of the element at which the maximum in-plane shear strain occurs. Os,,es = 31.7, 122° Ans. Pact F - Determine the normal strain Ex of the element with orientation As = 31.7° EX"=- 1.50x10-4 Ans.

When working through the equations, please thoroughly specify all the elements within each equation without skipping calculations. I need to know what goes inside the formulas and how they are derived to for each equation. I need it broken down as simple as possible as if teaching a 3rd grader. Thank you for your help!

Answer 1

Given data Ex = -300x106 Ey = 0, Yay = 150X106 A: Oaientations of the element at which the Principal Strain occur We know the formula tan20p = Yzy Ex-by = 150x10-6 -300x16-6-0 tan 2@p -0.5 20p = tañ'6-0-5) 20p = fan''Etan (26-56)] Ok tan" [tan (180-26:56)] 20p, --26.56 | PP, 7-13.289 Any OR, -13-3° and 2082 = 180-26.56 = 153:43 - = 76.70) Ang Part B:- Normal Strain ei'at Op = -13-3° Ez's Ext6y + Ez<by 6820p+ Yay. Sin 20 = (-200x150) + (-300x166-0) Cat (2 xt12:39] + 150x10-6 din (2xl-13-397 = -150x156-150x156x(0-894)+75x106 = -150x156_134.1x156 33.58x166 61 = 317.58 x106] Hy OR tx' = 3.1758 x104 Ex' = 3-12x104And 2 *(-0-447€)

Parts Normal Strain Eg' of the element with orientation Op=-13-30 ty's Extey - Ex-€y CA 2 Op - Yay Sin20p -(-300x1049) - (-300 1560) - 15oxte“ sim (zat1337] x Cos Tzat-br"] = -150x106 + 150xlõ6x60.894) - 75x17% -04478) = -150x166+134.X166 +3358x106 ty' = 17.68x106 ORE'= 1768x10-5 Ey' =1.77x105 thy poort D'- Shear Sitrain Ya's with oriention Bp=-13-3° Yepy' = + En- Ey Sin (2@p] + Key ces (20p) Valg = + (-300x176_o) 8in (2«t-13-39] + 150x156x 126-13)] --150x156x(-0.4478)+75x156x (0.594) Yºu = + 67: 17 xl + 67: 6 5 x16 6 Yz'N! 134.22X166 2 2 Ya'y' = 268.44x156 Parte Orientations of the element at which maximum Shear Strain occurs tan (29-max) = -(em to . Yay = -1-300x166-0)2 150X1567 (20s) = tant (2) = tanlan 63.43] 20s, = 63.43 OR tas[tan (180+ 634297 Bsi = 31:720 Ary 2 Os2 = 243.43 Os₂ =121-72²] And

Normat strain Ex" of the element at Os - 31.7 €x"= Extęy + Ex=69 61288+ Yaxy Gin 28s 6." =(-320x1080) + (-2004 ) Gan/2017 + 156 xto 8n (2x31.7") 6x" = -150X106_150x1566014478)+75x106601894) = -15c xlő6-67-178156+67.05X166 Ex" - -150.12 x106 OR EX"=-150x154 Any Part Normal Strain Ey" of the element with orientation Os = 317 ty"= 6x7 - ERFEY GOs 20s + Yay Sin(2&c] = -300 x15f40-(-300x106.00 ) Gy(2631.7°) -350x15“sin(243170) = -150X106 + 150x166x(0:4478) –75x156x(0-894) = -150X106767. 1 67.05x10-6 = -150X10-6 Ey! 9-.50 X164) And Part (4): Shear Strain Yulya at :31.7° Yog'g= -(EX- 64) Sin 2ės + Vay 66202) = - 4300x769) Sin(2431-79 + 150815 668/2r31:7) = 150x106x10.894)+75x166(0.4478) 134.1x10-6 +33.58x106 Yz's' = 167.68 x106 You'' = 335-37x156 ir Yaly'= 3.35 x 104 Any 2