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General_solutions - The homogeneous aluminum plate ABCD is...

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Unformatted text preview: The homogeneous aluminum plate ABCD is subjected to a biaxial loading as shown. It is known that 0' z = 250 psi and that the change in length of the plate in the x direction must y be zero. Determine the magnitude of the normal stress in the x direction and the normal strain in the z direction. Use an elastic modulus of 10 x 10 6 psi and a Poisson ratio of 0.32 for your calculations. éWw/l 0;- 250 [051.) CX: X/Oé $1 1):. 0132’ €72,” {a 93L + ”b/hxmfl ”,OJFO Fm“ @)62 we gait/14.62% flab/Q’s law“ 214 = #0; «12% +51% >l 01 = m; :03 25%; Problem #5 (20 %) 5.1 Given E =10 x 106psi, G = 3.8 x106 psi, v = 0.32, a= 12.8 6x 0 6x: fer—V?! 5.2 Determine the stress concentration factor for a 3 in. wide axially loaded bar with a 0.6 in. radiushole. y Zr/D : /'% : 0v W27} 6.2 03 0.6 0.8 1-0 Zr/D' 5.3 For the original 1 mm x 1 mm element deformed as shown, find a x , s x , and 7 xy at point C. 5.4 For the stress strain curve shown, determine a) the elastic modulus c 5:9: 200*“) ,wam? (g 6 MW =2006’6L. b) yield strength 25?) m Q Form # Problem #1 (65 points) A rectangular bar is pinned at one end and subjected to an axial load P at the other end as shown. The pin is made of tool L2 steel and has a diameter of 25 mm. The bar 1s made of 6061 — T6 aluminum and had the unloaded dimensions of L= 5 m, w— -— 150 m, t— - 50 mm. élvm- d P.n=25mn-;L= ”snaJW /§0mn\,;t =$mm ForP= 450 kN, calculate: FWD 3) the average normal stress and strain in the bar, q. 6 b) and the factor of safety ifo m. = 255 MPa. F’s " Prfltc ‘Q 3N 0’ Z 1/50 W J=~’ m = 450% = @0200 N/m :Prri'emlfl Pg, I4 [at/SmYOIOSM) -é0 M F " fl. 0/5 @X/OGN/m =0,27/X/o 3”%m- 49, 9 x /0‘i N/m" Lflfim *flb/L b) strain= awn/0'3 "W m z 4.25 c) F.S.= 4.25 c) Calculate the change in the bar’s width, w, due to the applied load P. PAID: 5w . ’5’! =6w gw {EWYW) 3 W ”J 6 = "U6; = '/0'3S)(0.87/X/0'3) =-o.305x/0 pen/55:0». 6’ C v 5w = [mBoSX/O-sX/s'omm) 90,0757 mm. c) 5w: —o,09/57mm Does the width increase or decrease? deem Mb ‘ (A d) What is the average shear stress in the pin? Fl V6,- duA 7L0 Symmulyx K: =//950) ‘ZZS' kid 3 950 m 225m ( 225*") N = 758 ”Mg. ”8:1/0.025m>2- Form # Problem#4 (20 points) Circle the appropriate answer for each question below and briefly explain your answer. a) If a solid aluminum rod is fixed between two walls and the temperature increases A ox%0,ex%0 C9 ox%0 , ex= C cx=0,ex7€0 D 0x: : Ex= b ) If a solid aluminum rod is fixed between two walls and a torque is applied as shown, circle the statement that is true for the stress and strain at point H on the outer surface. 'ch , YaéO B 1940 , Y=0 C ‘t=0 , Y¢0 D ‘c=0 , Y=0 c) The data from a tensile test is show here. Circle the letter that corresponds to the elastic region of the curve. c) Circle the correct definition for the elastic modulus E=cs E=1/(68) @ E=slc Form # Problem #1 (60 points) An A36 steel rod has an initial diameter d o of 25 mm and an initial length L0 of 300 mm. Under an axial load P, the diameter shrinks by 0.002 m. J— lem5 do=Z5Mn~,Lo=3OOMm. 45190.00sz do (:3 TI L I a) Calculate the normal strain in the radial direction. - .002 mm ~60 (g = EA = 0 = -gox;o a": J d 25 mm Mm ' I —6 mm sinus/(1% Ed: ~8OX/O 75,71 b) Calculate the corresponding normal strain in the axial direction (along the length). - ‘6' —o 006. = 0., 0.: £0 =— 0m —~ 0 00-000 ML: V 0132 —e an c) Determine the magnitude of the axial load causing this deformation. _ _ 6 N 0": 56 : wa/09I%(zsox/ 0% = 50 x/O 771 3 5—. F74} P = 0,4 = Scat/06%(WZ—y0025m); = 2%5X/O IV .01? 6.: 75% 5: 6.4 =250X/0‘égg-fsaomf0'075 ’"’"‘ fl 8 = f]: P: 5,45 = 0,075MmaIXO,OZSMf(ZmX/Oq%z p: 27.3 k/V A E L 3mmm. d) Is the load calculate in part c) tensile or compressive? (circle one) C Briefly explain why. . ' i 710 139 d Is éflmn/tl , P0255“. aged 74140 Caffféfdhdé a. $79: 710417 lit / n7 . fl [way/W 267360044 éc. CauMd %/ a 70. ,4. WM. 6) What temperature change would counteract the axial load resulting in no axial deformation? 5": é” :0,o75m% = ALA-r = I2x/o"°/>c (BOOmrnDA'T AT= 20, 2°C, AT= 2018/06 Properties of A36 Steel Modulus of Modulus of Yield Strength (MPa) Ultimate Strength (MPa) Coef. of Them. Elasticity E Rigidity G 0‘), (1'u Poisson’s Expansion 0: (GPa) (GPa) Tens. Comp. Shear Tens. Comp. Shear Ratio (10")/ °C 200 75 250 250 -- 400 400 -- 0.32 12 4.1 Indicate by letter, the region of the o — 8 curve described by each term below. A elastic 0/ strain hardening 4.2 Based on the deformed rectangle shown, the shear strain at point A is A. 0.0357 radians —0. 0357 radians C9). 0135 radians D. —O. 0135 radians 4.3 For rod shown that is fixed on one end and free on the other, which of the following statements regarding the normal stress and normal strain is true when the temperature increases? A. o=0, e=0 B. Caéo, 8=0 0 0 @ 0'=0, 3% D.o#Qe¢ 4.4 The homogenous hollow bar shown is loaded with a single torque on the free end. Circle the one figure below that describes the stresses acting on a cube of material at A near the outer surface. ‘1 3.4 For the problem shown, the applied load F is sufficient to cause the gap to close and. create a compressive stress in material 2. Identify the correct compatibility equation for this b1 . pro em // / if). A) 51 -52+A=0 cum-52:0 1P2 iA D)51=-52 M 3.5 A torque is applied to a shaft as shown. Circle the one stress cube below that correctly displays the stress state at point B. . . I ,. f 3.6 Match each term below with it’s correct number from the graph. A Fracture Strength A Proportional Limit 1 Ultimate Strength 5 Plastic Region i Elastic Region Problem 3 (30 points) 3.1 For the biaxial stress state shown, which one of the following statements correctly describes the sign of the normal strain in the x direction? A) The x direction normal strain is definitely positive. T (TS), T B) The x direction normal strain is definitely negative. <— y —* <— l_X _‘> 0X e x direction normal strain might be positive or ‘— —* egative depending on the values of the two 1 l l normal stress and the material properties. 3.2 Assuming B 1 and [32 are measured in radians, the shear strain at point A of the deformed element is ‘ A)+B1 B)-[31 C) +32 .3. 3.3 If an object undergoes an axial load in the x-direction, for which of the following conditions could the simple equation 5 x = (PL)/(EA) NOT be used? A) P varies linearly with x B) A varies quadratically with x C) o x exceed the proportional limit D) A and B E) A and C F) BandC @, B, and C Problem #4 (20 points) 4.1 Based on the c — 8 graph shown, indicate the letter that corresponds to each term below. E ultimate strength 2 nominal rupture strength L yield strength A elastic modulus __._8_ proportional limit 4.2- Based on the deformed rectangle shown, the shear strain at point A is 0.0246 rad A. 0.0357 radians ,_ ._ _ B. —0.0357 radians C. 0.0135 radians —0.0135 radians 4.3 A rod is rigidly fixed between two walls as shown. Circle the one figure below that describes the stresses acting on a cube of material at A when the temperature is increased. 4.4 For the homogenous hollow bar loaded as shown, which of the following statements regarding the shear strain and the shear stress is true for a point 3 ft from the wall? u— 2 / ’9‘ ®y=0, r=0 B. y#0,r=0 C. y=0,r#0 D. y¢0,t#0 J) €x‘W’ ’57 ' IOMm- WM l9» Problem 3 (30 % ) The bar assembly shown below is subjected to thermal and mechanical loads. The dimensions given below are for the unloaded initial state. After the entire bar is subjected to a uniform temperature increase of 50°C and the concentrated forces. Determine: (a) the axial stress in each material (b) the axial strain in each material (c) the displacement of point D ,_ ik” (d) the final width of concrete (member AB) E] O ' o it D __ a) (Set) =0 1, P951. k” 1, PM!» 300 mm long _P 5mmby5mm 6. $1000!“ TEFy‘O‘j-J/ )8 magnesium allo Am 1004-T61 3‘ 2- P 5-. -3 M 1 kN d 2 26 (’0") 06 (0,005) A a 40 MP.— 5,6 a 3000 30,45 C (0.0! )1' 200 1 b 51113323; > a = 5‘36 gig gun-r 2 kN 2 kN aluminum aleoy 6061—T6 E .3 B 1., .mfi__°;;:i€4i,o.én a: o + (uric-era) nan, > 100 mm long 10 mmby 10 mm 66‘ a 40 (/06) + 2405‘Xso) = L731 (/03) low strength concrete égfi (10" ”-‘d g 'l ('0.6)/f6 ( 6)) 7 [’0‘3‘ 5:22.1(qu)fia 5“ = “3° ’0 + 11(10"X50)= '0'3" 2 we / 22.; (10") on 0 ll I 9 8 q ;\ o. u: S 8 3 3 x/ I! l 9 0 %° cq E i ’ -340“) - MAT ' Véy 5 ”00-619) ”ms/22¢ Io") Displacement of D 0: @5’ mil. : 753, (a flo“) = W '70,”. Final width ofconcrete 10.0067 mm /0 ml” ...
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