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Hw22_Solutions

# Hw22_Solutions - Problem 5.9 5.9 and 5.10 Draw the shear...

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Unformatted text preview: Problem 5.9 5.9 and 5.10 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment. 3 bps/ft: Rene-hows 792%: 0 56A +(3)(m -(3)(3o)= o ' A: -6 my: Le. QkiFs 4. so”? {DEMAND '6C-(3)(:83-(ﬁ}(303=01 C= 54*?5 1 ‘A+DC 041.4%” SHIHM ligl Jr"? A *ﬁDM +TES=O —€~3x-V=0 V :‘€“3X kl}; 92mm -6x -<3x)(§).- M = o M: -6x - 1.5xz kip-ﬂ” C408 .C¥+<-x<9_¥+ PP: 92Mu=0 “M. - (CE-XWSO): o M =3 30X -- 270 kip-H From ‘HVC JMaV-mi (OJ lV']h~‘}‘30"o k-‘ps 4 (m [M 1...; = 70.0 WP: Problem 5.10 2 kipsift 15 hips 5.9 “(15.10 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment. A+oC 04x44“ ”213:0 -v—zx = 0 v=-2x [o‘ps DzMJw M+(2x)(—’,§)= o M 3' —X k?“ M C V ‘ g klys M: *Iéklr H A1’ D- Z skips M 3 V +T 23:0 ~8—V=O . =-81<.f,os "DEMO: 0 (mm- M= o Mat-1m, H A+ 8' 3 kirs ’5 Ups FZJF—G +4 V HEP—5:0 -8—15-V=O v = — 23 kc,” ~492M3=o -(1o)(a;)«-(4)(Is)”— M = o M -‘ -.7 ND kc, H. Pram +ke Juana: (a) Wk... = 0.3.0 k5.” 4 '(L) ML...a .— Hopmrql’tq Problem 5.16 5.15 and 5.16 For the beam and loading shown, determine the maximum normd stress due to bending on a transverse section at C. .10 kN .Usinj CB as a-Yree body 3°?” 92MC= o. -—M + (2.2)(3m3Vm) = o ‘ M = ZZZGXID’ N-vm V Sea-Hon moAu/m ‘Fow Vedanjﬂe 5= #5 h‘ = g- (moawoos‘ = scan/oz m.’ ' : 4:65.? x10“ m’ Norma! s‘h‘LSS 6‘: )1 _ mum“ s ‘ easy r104 : 10.37310‘ ?4 1; 6 = 10.8? MPa. 4‘ PROBLEM 9.1 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis. PROBLEM 9.41 Determine the moments of inertia 7x and 7), of the area shown with respect to centroidal axes respectively parallel and perpendicular to side 1.. 3;,» .4 Determination of centroid: 52' = 0 by symmetry. Part :4th -——-- ”mi-— Determination of Tx: Part <0: 73- = 11243 in.)(0.75 in.)3 + (31:1)(075 m)(3.375 in.)2 = 25.734 1114 Part (2: 7x = \$0.75 in.)(6 in.)3 = 13.50 1114 Part 6): (Same as Part @) :7; = 25.734 m4 ' Entire Section: 7.. = (25.734 + 13.50 + 25.734) 1114 = 64.97 m4 I: = 65.0 m4 4 Determination of Ty: Part G): Ty = 65(075 m)(3 inf + (0-75 in)(3 in-)[(1.5 — 0.9375} m]: = 2.3994 1114 Part @2 IV = 115(6 in.)(0.75 1n.)3 + (6 m)(0.75 in.)[(0.9375 — 0.375) in.]2 = 1.6348 1114 Part @; (Same as Part (0) fy = 2.3994 in“ Entire Section: fy = (2.3994 + 1.6348 + 2.3994) in“ . _ _ = 6.434 m4 f = 6.43 in“ 4 PROBLEM 9.54 To form a nonsymmetn'cal girder, two L3 X 3 X %-in. angles and two L6 X 4 X %-in. angles are welded to a 0.8-in. steel plate as shown. Determine the moments of inertia of the combined section with respect to its centroidal x and y axes. SOLUTION l—"i [Jim : 1 Angle: L3 X 3 X Z: A 91.441112 7x = Ty =124m4 1 L6 X 4 >< —: 2 A = 4.75 1112 7x = 6.27 in4 fy = 17.4 m4 Plate: A = (27 in.)(0.8 in.) = 21.6 1112 Z, = {5(03 in.)(27 in.)3 = 1312.2 1114 continued PROBLEM 9.54 CONTINUED X=0 17: W 2A 2[(1.44 m2)(27 in. — 0.84 16.)] + 2[(4.75 m2)(0.987 in.)]+ (21.6 m 2)(13. 5 in.) or )7 = 2(1441112 + 4.75 1112) + 21.6 m2 _ 376.3lin3 33.98 m2 = 11.0745 in. - Now Ix = 2(1):)1 + 2(1).)3 + (17x)2 ’= 2[6.27 + 4.75(11.075 — 0.987)? in“ + 2[1.24 + 1.44(27 — 0.842 — 11.075)"] 1114 +[13122 + 21.6(13.5 —11.075)2]in4 = 2(489.67)1'114 + 2(32884) 1114 + 1439.22 in4 = 3076.24 in4 or ix = 307611144 2.[17 4 + 4. 75( 0. 4 +199) ]m4 + 2[1.24 +1.44(0.4 + 0.842)? m4 + 1.152 m4 2.(44532)1n4 +2(3.4613)in4 +1.1521n4 7.139111 or 7}, = 97.11'n4{ PROBLEM 9.56 A channel and an angle are welded to an a X 0.75-in. steel plate. Knowing that the centroidal y axis is located as shown, determine (a) the width a, (b) the moments of inertia with respect to the centroidal x and y axes. _ ' (a) Using shape data from Fig.9.13A xA = 1.78 in. AA = 8.44 m2 xC = —0.499 in. AC = 3.09 1112 AP = (0.75a) m2 From the condition = 0 (1.78 m.)(8.44 inz) — (0.499 in.)(3.09 inz) + [[6 — g) in](0.75a m2) = 0 a2 — 12a — 35.950 = 0 a = 14.4823 in. a = 14.48 in. 4 and AP = (0.75 in.)(14.4823 m) = 10.8617 1112 Locate centroid continued Now and PROBLEM 9.56 CONTINUED ix = (Ix)A + (Ix)c + (IX)p = [28.2 in4 + (8.44 in2)(1.78 in. — 0.90302 inf] + [15.2 m4 + (3.09 m2)(3.0 in. — 0.90302 inf] + [—1—12—(144823 in.)(0.75 in.)3 + (10.8617 in2)(0..375 in. + 0.90302 inf] = [(28.2 + 6.4912) + (15.2 + 13.5877) + (0.5091 + 177408)] 1114 = 81.729 in“ 01'. Z=8L7in4< y=.A[28 2 m4 + (.8 44 1n )(1.78 inf] + [0.866 m“ + (3.09 m2)(0.499 inf] - . 2 +|:]15(0.751n.)(14.4823 in.)3 + (10.8617 m2)(6 in. — W) ] = [(28.2 + 26.741) + (0.866 + 0.7694) + (189.842 + 167319)] in4 = 263.15 in4 or 7), = 263 in4 < ...
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Hw22_Solutions - Problem 5.9 5.9 and 5.10 Draw the shear...

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