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Practice Midterm 1

# Practice Midterm 1 - Nam I{314 F4M[€XAMPLEJ.51 Link 31 is...

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Unformatted text preview: Nam I, {314%, F4M%[€XAMPLEJ ' (.51) Link 31) is made of brass (E = 15 x 106 psi) and has a cross—sectional f 0.40 i112. L' ' ' mk CE 13 made of aluminum (E = 10.4 x 10“ psi) and has a cross—sectional area of 0 50 in2 De ' V . V . , A termme the maxim f C applied vertically at pomtA 1f the deﬂection of A is not toticezgg (if: 0311 be 50 in. A @ . . Two wooden members of 3 X 64111 uniform rectangular cross sec— , tion are ﬁrmed by the simple glued scarf splice shewni Knowing that the max? imum allowable shearing stress in the glued Splice is 90 psi, determine (a) the lar est load P wh' ' . . in fhe splice. 13h can be safdy applied.» (5) The Corresponding tensﬂe stress ® , A ZSO—mm bar of 15 x 30-mm rectangular cross Section consists of two aluminum layers, 5—mm thick, brazed to a center brass layer of the same thickness. If it is subjected to centric forces 0f magnitude P : 30 kN, and knowing that E; = 70 GPa and E, = 105 GPa, determine the normal stress (:2) ix; the aluminum layers, (5) in the brass'layer. 250 mm 5 mm / 5 mm Aluminum 5 mm Brass P Aluminum . A red consisting of two cylindrical portions AB and BC is re— 1L4“ diameter strained at both ends Portion AB is made of steel (E5 : 29 X 106 psi, 4 as = 6.5 X lO'G/DF) and portion BC is made of brass (E, = 15 X 106 psi, ab 2 10.4 X lO-G/DF) Knowing that the rod is initially unstressed, determine '(a) the normal stresses induced in portions AB and BC by a temperature rise of 65°F, (b) the corresponding deﬂection of point B. 2%411. diameter lfoéaebuj, mm; /2 [4 pm 03’ ° 23:54:33; [57,4 M15 P ‘DZMC = DJ M P—Tq F139;? 0 1 an:l-555é P 92M3=o 5P“?E£"O: Raw-5367’ Q _ 55.x FaoLaD w “.5559 PXCLo) ‘ = ,6 J, EDABD (l5yto‘](o_+0) 2.3333rlo P F L Lo.5554P)(e.ol -4. c=3c=‘E“=———-—~——n—-= -e )0 g 5 E62 Ace (lonro‘xaso) 0 W0“ P T From ‘H.e ole-Forma‘lLfom diagram.) 9 5 8.9.: e 9 : \$s+5c = 2.?743vzo'6 F 6 I Se! ﬁ ‘5 P 2‘76; ‘5? A = 0;33‘05xzo" P DELWMA-l'l‘ou Diagram SA = \$3 + 40,33 3 = 2.3333x1o"P+ (5l(0.330.5‘xlo")? = 3.7253 x10“ P APPPY api‘sfjagewxe-«j yin/til— \$5 : 0.0/41}, = 3.9353"£0_‘P p = MF— ‘ 3.5: x103 I5 = 3.51 “if” «a 3. \$353 xm‘ 6 Two wooden members of 3 X 6— in. uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum allowable shearing stress in the glued splice is 90 psi, determine (a) the largest load P which can be safely applied, (b) the corresponding tensile stress in the splice. SOLUTION 9:». cfo°~llo° =50° A0 = (BlCG) = I8 in: ”E ‘ 3 5M 29 ‘ 2A = 221/; _ (\$023010 = 1) Sin 713 ._ sin 10d)" 3290 CG») P: 3290 IL. -I Z 2. o la) 5*: Bias 9 : 32ctoigcos 50 = 75.5. 6—: 75:5,)”, ‘1 ~ P L P L g ,. q _ k _ _ Dﬂ‘FOr/mq‘hon. Cl EAA — EEA qu 2;: M» P:2g+ﬁ,: 33?; saﬁuvﬁ row a M; a) avg—p a - CCU 611:..83 c“; A 2 7E ._ P E x 3 Cm 6;:--.5 3%.; :_3 220 {DH _ @ Fem emk Jaw) chgo)(5)= 5mm»: G ‘ = *SOVEO' m 1.3+ P21 2‘ p004] w'eqailx 0.99m Mama ﬂayem. Pb ? (Pony) 0% km: ﬂan/ew‘t Free ‘quer‘P Bx [M n 5 fan PL? ‘g’ga = MP; Sat-57.! MPG. ‘ﬁ 6;: ='85.7 MPQ é F4272 ha" P ST = Lisa‘s (AT)+ Lgcolbmﬂ -_- (jQXag'wd‘ )(€\$)+05Xlo.4xld‘)CC\$) = A5121 we“ in 5L0r+eninﬂ due. +5 induced wmpwessfve “Po-ICE 'P .. PL PL 3 ‘ _b§_ ___.&g_ P EsAm * Eb A“ = :1 P 15‘ f _ -9 (2?x10‘)(l.17~71) +05xao‘ X3374!) _ 588. 69x10 P F For zero (05+ JE‘FﬂEc—Hm SP = 5T (mammo‘7)? = ISZSUXIO‘a P = 25.3%:103 A ~_ P _ 25.3 3 3 673‘": _‘A% t — W “T - £50 ”03.055 = '650 “5" “‘ \$3 = + PL” ESAAE; d" LABOQ- (AT) : +W (1mm )(Lz'nz) + (22)(_E.Sx¢o“ )(gg) : he. + 3-6‘fvzo‘3 ‘M f 3.64x10'35n r 0-0036‘1ll‘v1 ‘T as ...
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Practice Midterm 1 - Nam I{314 F4M[€XAMPLEJ.51 Link 31 is...

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