Compute the force in each member of the Warren truss shown in

 Compute the force in each member of the Warren truss shown in Fig. 

 

Solution

Hide

ΣME=0$\mathrm{\Sigma }{M}_E=0$

20RA=1000(15)+4000(10)+3000(5)$20R_A=1000(15)+4000(10)+3000(5)$

RA=3500lb$R_A=3500\text{lb}$
 

 

ME=0$M_E=0$

20RE=1000(5)+4000(10)+3000(15)$20R_E=1000(5)+4000(10)+3000(15)$

RE=4500lb$R_E=4500\text{lb}$
 

At Joint A
ΣFV=0$\mathrm{\Sigma }{F}_V=0$

FABsin60∘=3500$F_{AB}\sin {60}^{\circ }=3500$

FAB=4041.45lb$F_{AB}=4041.45\text{lb}$   compression
 

ΣFH=0$\mathrm{\Sigma }{F}_H=0$

FAC=FABcos60∘$F_{AC}=F_{AB}\cos {60}^{\circ }$

FAC=4041.45cos60∘$F_{AC}=4041.45\cos {60}^{\circ }$

FAC=2020.72lb$F_{AC}=2020.72\text{lb}$   tension
 

At Joint B
ΣFV=0$\mathrm{\Sigma }{F}_V=0$

FBCsin60∘+1000=4041.45sin60∘$F_{BC}\sin {60}^{\circ }+1000=4041.45\sin {60}^{\circ }$

FBC=2886.75lb$F_{BC}=2886.75\text{lb}$   tension
 

ΣFH=0$\mathrm{\Sigma }{F}_H=0$

FBD=4041.45cos60∘+FBCcos60∘$F_{BD}=4041.45\cos {60}^{\circ }+F_{BC}\cos {60}^{\circ }$

FBD=4041.45cos60∘+2886.75cos60∘$F_{BD}=4041.45\cos {60}^{\circ }+2886.75\cos {60}^{\circ }$

FBD=3464.10lb$F_{BD}=3464.10\text{lb}$   compression
 

At Joint C
ΣFV=0$\mathrm{\Sigma }{F}_V=0$

FCDsin60∘+2886.75sin60∘=4000$F_{CD}\sin {60}^{\circ }+2886.75\sin {60}^{\circ }=4000$

FCD=1732.05lb$F_{CD}=1732.05\text{lb}$   tension
 

 

ΣFH=0$\mathrm{\Sigma }{F}_H=0$

FCE+FCDcos60∘=2020.72+2886.75cos60∘$F_{CE}+F_{CD}\cos {60}^{\circ }=2020.72+2886.75\cos {60}^{\circ }$

FCE+1732.05cos60∘=2020.72+2886.75cos60∘$F_{CE}+1732.05\cos {60}^{\circ }=2020.72+2886.75\cos {60}^{\circ }$

FCE=2598.07lb$F_{CE}=2598.07\text{lb}$   tension
 

At Joint D
ΣFV=0$\mathrm{\Sigma }{F}_V=0$

FDEsin60∘=1732.05sin60∘+3000$F_{DE}\sin {60}^{\circ }=1732.05\sin {60}^{\circ }+3000$

FDE=5196.15lb$F_{DE}=5196.15\text{lb}$   compression
 

ΣFH=0$\mathrm{\Sigma }{F}_H=0$

FDEcos60∘+1732.05cos60∘=3464.10$F_{DE}\cos {60}^{\circ }+1732.05\cos {60}^{\circ }=3464.10$

5196.15cos60∘+1732.05cos60∘=3464.10$5196.15\cos {60}^{\circ }+1732.05\cos {60}^{\circ }=3464.10$

3464.10=3464.10$3464.10=3464.10$       Check!
 

At Joint E
ΣFV=0$\mathrm{\Sigma }{F}_V=0$

5196.15sin60∘=4500$5196.15\sin {60}^{\circ }=4500$

4500=4500$4500=4500$       Check!
 

ΣFH=0$\mathrm{\Sigma }{F}_H=0$

5196.15cos60∘=2598.07$5196.15\cos {60}^{\circ }=2598.07$

2598.07=2598.07$2598.07=2598.07$       Check!
 

Summary
 

 

AB = 4041.45 lb compression
AC = 2020.72 lb tension
BC = 2886.75 lb tension
BD = 3464.10 lb compression
CD = 1732.05 lb tension
CE = 2598.07 lb tension
DE = 5196.15 lb compression