{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW17 - PROBLEM 6.3 Using the method of joints determine the...

This preview shows pages 1–5. 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 6.3 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION FBI) Truss: Joint FBDs: Joint C: (2MB = 0:(6 ﬁ)(6 kips)—(9 may = o Cy =4kipsl iZFy =0:B,—6kips-Cy =0 3),:10kips T —’2FI=0:CX=0 17 15 8 FAC = 8.50 kips T 4 FM = 7.50 kips C 4 By inspection: FAE = 12.50 kips C 4 fa = 10 kips 5 4 nL.\ D ‘ 4 6m t. g _| \'. a C Pi m am Lamlgml PROBLEM 6.10 Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression. SOLUTION FED Truss: A rzkﬂ m. 25, = o: A; = o Bysy'mmetry: Ay = Hy =12kN I and FFH = FAB? FGH' = FAG; FFG = FBC FDF = Fan; Far = F35; FEG = Fee 7 = i = 4111 3 4 5 FAG :12.00kN Td FAB =15.00kN C 4 so Fm =15.00kN C 4 By inspection: FGH = 12.00 kN T 4 FHC=0;FCE=12kN FBC=0=FFG{ FCE =12,00 kN T 4 F55 = 12.00 kN T 4 Note: a = tan—1 1 so sina =0.8 3 ,6 = 2am"; so sinﬂ =0.96 \ZFy1= 0:~6stina+FBEsin}3=O FEE 2 5.00m C 4 so FEF =5.00kN C 4 12;;=o:—5kN—%FBD+%FBE+%(151<N)=0 PROBLEM 5.10 CONTINUED -§(5kN)+%(15kN)—6kN FBD=10.00kN C < so FDF = 1000104 C41 1215'), = 0:—6kN+2[%10kN]—FDE = 0 PM = 6.001410 T4 PROBLEM 6.23 C 12 m Determine the force in each member of the truss shown. State whether 3 m each member is in tension or compression. SOLUTION FBD Truss: (EMF = 0: (10 m)Gy — (7.5 mm kN) —(8 m)(30 kN) : 0 Gy=84th aZFX=O:—FX+30kN=0 FX=30kN‘—~ TZFy=0sz+84kN—80kN=O Fy=4kN1 By inspection of joint G: FEG = 0 { FCG=84kNC< 84kN _ FCE _ =FAC 8 J5 J5 =10.5kN FCE :82.0kN T4 FAC = 56.5 kN c 4 2 6 2F=0:—F +—82.0kN —80kN=0 i y J; A5 (—61( ) FAE =19.01312 FAE =19.01kN T‘ 1 5 a2F=0:—F ——19.013kN+-82.0kN =0 x DE ﬁ( ) Ja( ) FDE = 43.99kN FDE = 44.0 kN T 4 iFDF—3OkN=0 FDF : 67.082kN FDF = 67.1 kN T 4 PROBLEM 6.23 CONTINUED 2 2F : 0:— 67.082kN —F —4kN:0 T ,V ﬁ( ) BF F3,r = 56.00kN FBI.- = 56.0 kN C 4 a2F=0-30kN+iF —iF =0 x ' M BD 5 AB 12F—0-56kN—iF —iF ~0 y ' J; EB J5 AB Solving: FED = 42.956 kN FED : 43.0 kN T 4 FAB = 61.929 kN FAB : 61.9kN c 4 1 EF}, = 0: i(42.956 N) + LUV/w — 67.082 N) = 0 J5 J3 FAD = 30.157 kN FAD = 30.2 N T 4 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

HW17 - PROBLEM 6.3 Using the method of joints determine the...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online