{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW18 - PROBLEM 6.44 «1 «r 111.x‘.111>1 111 1n...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PROBLEM 6.44 «1 ;' «r 111 .x‘ .111 >1 111+ 1n ll'l*>-«rl|I*-' . . 1, , l 11 1.. A parallel chord Howe truss is loaded as shown. Detennine the force D 9"”; l ' i . A}; l in members CE, DE, and DF. 11 1, ° 1: 211 1! "E a ° SOLUTION FBD Truss: a 2F. =02AX :0 Q 2M1», : o: (4 ﬂ)[1(0.3 kip) + 2(0.6 kip) + 3(1kip) +4(0.6 kip) + 5(o.3 kip) + 6(0.3 kip) 7 My] 2 0 A], = 1.7 kips ? Q EMU = o: (4 ft)(3 kips 71.7 kips) + (2 ft)[%1r£be : 0] FUL- = 2.87 kips FCE : 2.87 kips T 4 Q ZME = 0: (8 ﬁ)(3 kips — 1,7 kips) + (4 ﬁ)(0,3 kip) —(2ﬁ)(ﬁFDF] : o \/l7.21 F = — DE 4.1 (—5.125 + 2.87)kips FDF : 75.125 kips Fm. : 5.13 kips C 4 l-7 ((1165 , 4 a 2E : 0: EU” + FCE) + 17 21F” = 0 11335 = 2.28 kips FDE = 2.28 kips T 4 PROBLEM 6.50 A Howe scissors roof truss is loaded as shown. Determine the force in . ‘ members DF, DG, and EG. ' '3 In ‘ 2m SOLUTION FBD Truss: aZFX=0z Ax=0 Bysymmetrysz=L =6kN 1. FEB Section: Notes: y,— : E m 6 2 5 5 W357” y.:g_r1:gm b 3 3 V,.»yD=5m 6 _1/G:lm v 7)} *3m .0 .0’3 PROBLEM 6.50 CONTINUED (2M0 : o: (1 n1)J%FFG + (2m)(2 kN) + (4m)(lkN —6 kN) : 0 Fm : gﬂ kN FEG : 16.22 kN '1‘ 4 c m, = 0: (2m)(2kN) + (4m)<2kN> *(6 m>[7}—Ovaj — (1 110%13.) : o FDO = gdfékN Fm. = 4.22m C 4 2F” :0 Mimi/70F =0 6 3 ‘F ‘——F w_ E EC. m 00 13 13 FDF =13kN FDF =13.00kN C 4 —» 25‘. =0: PROBLEM 6.55 MM - L if]; i. A roof tmss is loaded as shown. Determine the force in members FH, .‘q. u “i7 1 LVL‘YV‘LI, OJ, and G]. a i M, 0] ' i l ” ol' i an ﬁn . a ~ U. 1 n a I.' I“ , o .1 11 \Q a K 0*“ 1 m. w. #61!»LM|~> (VII ‘ ﬁl‘l i SOLUTION Mel/u zit—4,5; A? By symmetiy: A), : M". : 2.45 kips I > ZR: 0: Ar: 0 Jﬁ (ZMJ = 6 ﬁ(0.3 kip) 7 12 M235 kips) + 7 ﬁ[iFG,J : 0 FG, : 3.887 kips F0, 2 3.89 kips T 4 (EM. : (6 ft)FHJ — (4 MM) kips) 7 (10 ﬂ)(0.3 kips) — (16 ft)(0.1kips) 125.: + (16 ﬁ)(2.45 kips) = 0 l 0: E( By inspection: 3.887 kips) + 2.45 kips 71.4 kips — 3 JB F,” = 75.10 = 5.10 kips C FFH: FHJ : 5.10 kipsC‘ FGJ : 0 F6] = 2.40 kips T 4 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

HW18 - PROBLEM 6.44 «1 «r 111.x‘.111>1 111 1n...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online