Solutions 2

# Solutions 2 - 1-49 The open square butt joint is used to...

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Unformatted text preview: 1-49. The open square butt joint is used to transmit a force of 50 kip from one plate to the other. Determine the average normal and average shear stress comments that tth loading creates on the face of the weld. section AB. 84mins: of Equilibrium : yup-o. N-SOcosJO’IO "-43.3059 ﬁts-o. -V+503in30'-0 V-ZSDkip Aver-(c Nor-d Ill Shear Stress : . 2 . - 2 A (“60.)(6) 13.36.. ‘ using-3.1253.“ A” A' U.“ V . 'W 4' as.“ . 50 mp 2-14. The piece of plastic is originally rectangular. Determine the average normal strain that occurs along the diagonals AC and DB. Geometry .- ACSDB=J4003+3003=SNmm DB'SJ40514-304 8506.40“! A'C'=J4011+300‘= 500.8mm Average Nome! Strain : A'C'-AC 5003-500 AC 500 - 0.00m nun/mm - t.60( no") mmlmm ‘Ac = DB'-DB 506.4- 500 DB 500 a 0.0128 nun/mm I I2.8( no") mmlmm Am ‘00 ‘ ‘1 2-19. The square deforms into the position shown by the dashed lines. Determine the shear strain at each of its corners. A. B. C. and D. Side D'B' remains horizontal. Geometry : D’C‘ - Jun 3)’ +(53m ur)‘ - sum nun C'D‘ - J53: + 531 - 2(53)(\$8)eos 9w - 79.5860 mm B'D’ I 50+ Shin |.\$° - 3 I «.3814 m w'D')8 +(8’C‘)’ -(c‘m* 2(8‘D’H8‘C') 43.33743 + 54.: I n8 - 79.3w mass-mm: I m c030- . 4.20328 9- |0l.73° ﬁl- |80‘-0 II 7827' Show Strain : 0 (1a)., ---x(-9-'-‘-’-)--o.ozozm A” .7 ' (Ya).,'£-9-f- %)I-OJOSM All. at x 78.21‘ (76).,""ﬂ"" W)-0.205M AI. 0 (79),, '5-8(£)l0.0262d A. *3—4. A tension test was perfoer on a steel specimen . O 0 having an original diameter of 0.503 in. and gauge length of '1’"th “0.9"” ("U 2.0) in. The data is listed in the table. Plot the stress-strain diagram and determine approximater the modulus of 2‘0 3 m elasticity. the yield stress. the ultimate stress. and the rupture dang stress. Use a scale of l in. = 20 ksi and l in. = 0.05 in./in. am am Redraw the elastic region. using the same stress scale but a 11.00 0.0035 strain scale of l in. = 0.001 in./'in. 11-80 01150 11.80 0.“)80 12.“) 0.0200 16.60 0.0100 (705\$ ) 20.00 0.1000 21.50 0.28“) 19.50 0.4000 18.50 0.460) A = 910.503)z x 0.1901 19 L 8 2.“) it. can) “IL/'1) 0 0 1.55 000025 _ _ _ . _ _ - 23.15 000015 1 \$.26 ONIZS : a 5536 0.00115 : 5."... s m . 32.0(10’1101 A- 593, mm, 1 59.38 0.0000 : 0039 0010 : 83.54 0.020 3 “I165 0.050 I . , I” 0.!” z ‘ (M ’0) 9013 0200 : 93.“) 0.230 3-21. The stress-strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD. both made from this material. and subjected to a load of P = 80 RN. determine the angle of tilt of the beam when the load is applied. The diameter of the strut is 40 mm and the diameter of the post is 80 mm. From me was - snin diagram. 32.2( 10)‘ s- - 3.220059. Thus. 00- 52 - tow)” a 31.3310. Au 340-0")1 0.. 3|.83( I0.) 3 I — — I 0. I u E I 122‘ '0.) (1)9885 mm IIIIII Feb 4000’) a - — - - 7.93M! “’ Am moma P‘ Opp c- -—--——-o. m I (a E 122‘ lo.) M2 mm mm 5“ I (“Lo I I l9." IIII'II Jen I tcpkp I 0.00247l(5a» I '236 llllll Augkoftilta: “53‘; a - 0.708° 1500 All 5. an» 40») 073’" 4-7!” (7 (MP3) 50:40 W I‘NN‘. ' 03.00.0-‘.o -"" “ H'W" II-Sst M #1356 compression ’ tension 0 0.0] 0.02 0.03 0.0-8 ((mm mm) 3-27. The block is made of titanium 'l‘i-6Al-4V and is subjected to a compression of 0.06 in. along the y axis. and its shape is given a tilt of 0 = 89.7”. Dclcrminc ex. 6V. and y“. I ." Normal Strain : e I ill I 4'06 I -0.0ISO in.lin. Ans r l, 4 4 in. Q ..__J_-__-_ Poisson's Ratio : The lateral and longitudinal strain can be inland using Poisson's rauo. ‘ x c, . we, - o-0.36(-0.0|50) 5 in. - i‘ = 0.00540 in. Iin. Ans Sluar Strain : B I 180° - 89.7° I 903° I l.576032 (a! 7., - g-p- ;- 1.576032 - -o.oosz.: no Am ...
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Solutions 2 - 1-49 The open square butt joint is used to...

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