aHw3Soln_1-3_ - FBD CD: FBD CK: Note: AB is a two-force...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: FBD CD: FBD CK: Note: AB is a two-force member 5 Q EMC = 0: (18 in.)(—FAB 13 FAB =135.21b 12 13 + (7.5 in.)[2FAB] — (24 in.)(78 1b) = 0 a ZFX =0: C ——(135.21b)=0 x13 CX =124.8 1b a 5 1213 = 0: Cy + E(135.21b)—(781b)= 0 Cy =261b 1 , 12 5 2F, = 0: —F + —(124.8 lb) + —(261b)= 0 x 13 5 13 12 F =125.21b 7 22.6° 4 \ 21L}: 0: V — —(124.8 lb) + —(26 1b) = 0 13 13 V = 24.0 lb 35 67.40 4 «Q ZMK = 0: (5 in.)(124.8 1b) — (12 in.)(26 1b) — M = o M = 312151112114 “Q 2MB = 0: (0.8 m)(1.8 kN) — (0.24 m)Ax = 0 FBI) AB: JBJ=AIXI<N Ax _, fl '7 A ::T£:;:<:\f Tr. ~ W ‘ — x .u /‘N :1 r X z 011W * I Mf‘ ’ ’ w\§ Geometry: " 5* ’ ' ‘ 6 g. 2 2 Kat—MM yzkx; at B, 0.24m:k(0.8 m) 7 L5; 1 50 k = 0.375 — m at J, y J = [0.375i](0.48 m)2 = 0.0864 In 1T1 dy slope of parabola E = 2]“ at J, d—y = 2[O.375ij(0.48 m) = 0.36 = tanQJ dx m a, =19.799° \\ ZFx’ = 0: (6 kN)cos19.799° — (1.8 kN)sin19.799° — F = O F = 6.26 kN 3; 19.800 ‘ I 213,: 0: (6 kN)sin19.799° — (1.8 kN)c0s19.799° — V = 0 V = 0.3387 kN V = 339 N :7 702° 4 ‘0 EM] = 0: (0.48 m)(1.8 kN) — (0.0864 m)(6 kN) — M = 0 M=0.3456 kN«m M=346 N‘m 2+ 4 (a) FBDBeam: a=1m —~2Fx=0: 3x20 g,‘2Fy=0: —1.5kN+2kN—4kN+5kN—By=0 By = 1.5 kN, CEMB = 0: a[4(1.5 kN) — 3(2 kN) + 2(4 kN) — 1(5 kN)] — MB = 0 AlongAC: MB =(3 kN)a=3kN.m} /.s‘£N “.1 3x15 V a. ‘sz=0: —1.5kN—V=0 V=—1.5kN / '\2MJ=0: M—x(1.5kN)=0 M=—(1.5kN)x Along CD: M(1m)=—1.5kN.m ‘ZFyzo: —1.5kN+2kN—V=O V=O.5kN I; ZMK20: M+x(1.5kN)—(x—1m)(2kN)=0 M = —2 kN-m + (0.5 kN)x M(2 m) = —1kN-m continued Along EB: \ skid-M Q 9‘: hb'I'EN‘ CL 95 ‘ZFy=0: V—1.5kN=0 V=1.5kN LEA/5:0: —M—x1(1.5kN)—3kN~m=0 M=—3kN-m—(1.5kN)x1, M(1m)=—4.5kN-m AlongDE: M K we C *2 rim l-S'EN ‘ZFy=0: V+5kN—1.5kN=0 V=—3.5kN Also M is linear here |V| = 3.50 kN along DE 4 m ax |M‘ = 4.50 kN-m atE4 max (1?) (a) FBD Beam: .— zFx = 0; A = 0 x g, fly 5”“ 9k“ ‘ZFy:0; Ay+(2 m)(24 kN/m)—48kN—8kN=0 A c D B L fix Ay=8kN AV‘QAV/m ‘— ZM *Arm-r—IH-O‘SM Q EMA = 0: MA + (1 m)(2 m)(24 kN/m) — (3.5 m)(48 kN) —(2 m)(8 kN)=0, MA =152 kNm‘j Along AC: ff” g )4 [/5 fl ‘ZFy=0: 8kN+x(24kN-m)—V=0 V = 8 kN + (24 kN/m)x IQ ZMJ = o; M +152 kN-m — x(8 kN) — 324 kN/m)x = o 'ISL M = (12 kN/m)x2 + (8 kN)x — 152 kN-m “($213)?” XI Along DB: ~ ‘ZFy=0: V—8kN=0 V=8kN (2MK=0: M+x1(8kN)=0, M=—(8kN)x1 continued AlongCD: 3C :5 “ 3"le L 1/; Arm H t—XV—> N ‘ZFy=0: V—48kN—8kN20, V=56kN (XML :0: M +(x1—0.5 m)(48 kN) +x1(8 kN) =0 M224 kN.m—(56 kN)x1 |V| = 56.0 kN along CD 4 max (b) |M| = 152.0 kN'm atA 4 max (M ZMC = 0: 2.7 kN.m — (0.2 m)A L y :0, Ay =1.35 kN fl 2MB = 0; (0.2 m)Dy — 2.5 chm = 0, Dy = 1.25 kN‘ Q ZMC =0; —(2m)(1.35kN)—( m)(0.4 kN/m)(4 m)_(4 m)Q +(6m)(1.25kN)=0, Q=0.4kN Q:400N,4 ‘ ZFy = 0: 1.35 kN — P — (0.4 kN/m)(4 m) — 0.4 kN +1.25 kN = 0 P=0.6kN P=600N,4 Shear Diag: Vis constant at 1.35 kN from A to C, drops 0.6 kN, then decreases with slope —0.4 kN/m (—1.6 kN) to —0.85 kN at D, drops 0.4 kN to -1-25 kN, and is constant to B. V: 0 where 0.75 kN — (0.4 kN/m)x = 0, x = 1.875 m. Moment Diag: From zero at A, M increases with slope 1.35 kN to 2.70 kN-m at C, the slope drops to 0.75 kN and then decreases to zero at E, where M = 2.7 kN-m + 50.75 kN)(1.875 m) = 3.403 kN.m. This curve continues to D where M = 3.403 kN-m — l(0.85 kN)(2.125 m) 2 = 2.50 kN-m, then M decreases with slope —1.25 kN to zero at B. ...
View Full Document

This note was uploaded on 03/06/2011 for the course ENS 204 taught by Professor Hussein during the Spring '11 term at Yildiz Teknik Üniversitesi.

Page1 / 7

aHw3Soln_1-3_ - FBD CD: FBD CK: Note: AB is a two-force...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online