The free-body diagram of the entire truss is shown in figure 1.
Refer to figure 1 and use symmetry.
Dx=BxDy=By (I)
Here, Dx is the x component of the reaction at the point D, Dy is the y component of the reaction at the point D, Bx is the x component of the reaction at the point B and By is the y component of the reaction at the point B.
The x component of the net force must be equal to zero.
ΣF→x=0 (II)
Here, ΣF→x is the x component of the net force.
Write the expression for ΣF→x .
ΣF→x=2Bxi^
Put the above equation in equation (II).
2Bxi^=0Bx=0
Put equation (I) in the above equation.
Dx=0
The z component of the net force must be equal to zero.
ΣF→z=0 (III)
Here, ΣF→z is the z component of the net force.
Write the expression for ΣF→z .
ΣFz=Bz
Here, Bz is the z component of the reaction at the point B.
Put the above equation in equation (III).
Bz=0
Write the equilibrium equations taking the moments about the point C in the z direction.
ΣMcz=0 (IV)
Here, ΣMcz is the sum of the moments about the point C in the z direction.
Write the equation for ΣMcz .
ΣMcz=−2By(2.8 m)+(2184 N)(2 m)
Put the above equation in equation (IV).
−2By(2.8 m)+(2184 N)(2 m)=0By=780 N
Write the expression for the reaction at the point B.
B→=Byj^
Here B→ is the reaction at the point B.
Substitute 780 N for By in the above equation to find B→ .
B→=(780 N)j^
Consider the free-body joint A. The free-body diagram of joint A is shown in figure 2.
Refer to figure (2) and write the expression for the forces.
F→AB=FABAB→AB (V)
Here, F→AB is the force exerted by member AB and AB is the magnitude of the vector AB→ .
Write the expression for AB→ .
AB→=−0.8i^−4.8j^+2.1k^
Find the magnitude of AB→ .
AB=(−0.8)2+(−4.8)2+(2.1)2=5.30
Substitute −0.8i^−4.8j^+2.1k^ for AB→ and 5.30 for AB in equation (V) to find the expression for F→CA .
F→AB=FAB(−0.8i^−4.8j^+2.1k^)5.30 (VI)
Write the expression for F→AC .
F→AC=FACAC→AC
Here, F→AC is the force exerted by member AC and AC is the magnitude of the vector AC→ .
Substitute 2i^−4.8j^ for AC→ and 5.20 for AC in the above equation to find the expression for F→AC .
F→AC=FAC(2i^−4.8j^)5.20 (VII)
Write the expression for F→AD .
F→AD=FADAD→AD
Here, F→AD is the force exerted by member AD and AD is the magnitude of the vector AD→ .
Substitute 0.8i^−4.8j^−2.1k^ for AD→ and 5.30 for AD in the above equation to find the expression for F→AD .
F→AD=FAD(0.8i^−4.8j^−2.1k^)5.30 (VIII)
The net force must be equal to zero.
ΣF→=0 (IX)
Here, ΣF→ is the net force.
Write the expression for ΣF→ .
ΣF→=F→AB+F→AC+F→AD−(2184 N)j^
Put the above equation in equation (IX).
F→AB+F→AC+F→AD−(2184 N)j^=0
Put equations (VI), (VII) and (VIII) in the above equation.
FAB(−0.8i^−4.8j^+2.1k^)5.30+FAC(2i^−4.8j^)5.20+FAD(0.8i^−4.8j^−2.1k^)5.30−(2184 N)j^=0 (X)
Equate the coefficient of i^ in equation (X) to zero.
−0.85.30(FAB+FAD)+25.20FAC=0 (XI)
Equate the coefficient of j^ in equation (X) to zero.
−4.85.30(FAB+FAD)−4.85.20FAC−2184 N=0 (XII)
Equate the coefficient of k^ in equation (X) to zero.
2.15.30(FAB−FAD)=0FAD=FAB (XIII)
Multiply equation (XI) by −6 and add equation (XII).
−(16.85.20)FAC−2184 N=0FAC=−676 NFAC=676 N, compression
Put equation (XIII) in equation (XI).
−0.85.30(FAB+FAB)+25.20FAC=0−(0.85.30)2FAB+25.20FAC=0
Substitute −676 N for FAC in the above equation.
−(0.85.30)2FAB+25.20(−676 N)=0FAB=−861.25 NFAB=861 N, compression
Put the above equation in equation (XIII).
FAD=861 N, compression
Consider the free-body joint B. The free-body diagram of joint B is shown in figure 3.
Refer to figure (3) and write the expression for the forces.
F→AB=FABAB→AB
Substitute 861.25 N for FAB , −0.8i^−4.8j^+2.1k^ for AB→ and 5.30 for AB in the above equation to find the expression for F→AB .
F→AB=(861.25 N)(−0.8i^−4.8j^+2.1k^)5.30=−(130 N)i^−(780 N)j^+(341.25 N)k^ (XIV)
Write the expression for F→BC .
F→BC=FBCBC→BC
Here, F→BC is the force exerted by member BC and BC is the magnitude of the vector BC→ .
Substitute 2.8i^−2.1k^ for BC→ and 3.5 for BC in the above equation to find the expression for F→BC .
F→BC=FBC(2.8i^−2.1k^)3.5=FBC(0.8i^−0.6k^) (XV)
Write the expression for F→BD .
F→BD=−FBDk^ (XVI)
Here, F→BD is the force exerted by member BD
Write the expression for ΣF→ .
ΣF→=F→AB+F→BC+F→BD+B→
Put the above equation in equation (IX).
F→AB+F→BC+F→BD+B→=0
Put equations (XIV), (XV) and (XVI) in the above equation.
−(130 N)i^−(780 N)j^+(341.25 N)k^+FBC(0.8i^−0.6k^)−FBDk^+B→=0
Substitute (780 N)j^ for B→ in the above equation.
−(130 N)i^−(780 N)j^+(341.25 N)k^+FBC(0.8i^−0.6k^)−FBDk^+(780 N)j^=0 (XVII)
Equate the coefficient of i^ in equation (XVII) to zero.
−130 N+0.8FBC=0FBC=162.5 N, tension
Equate the coefficient of k^ in equation (XVII) to zero.
341.25 N−0.6FBC−FBD=0FBD=341.25 N−0.6FBC
Substitute 162.5 N for FBC in the above equation.
FBD=341.25 N−0.6(162.5 N)=244 N
From symmetry,
FCD=FBC
Here, FCD is the force exerted by CD.
Substitute 162.5 N, tension for FBC in the above equation to find FCD .
FCD=162.5 N, tension
Conclusion:
Thus, the force in member AC is 676 N and the member AC is in compression, the force in members AB and AD is 861 N and these members are in compression, the force in members BC and CD is 162.5 N and these members are in tension, the force in member BD is 244 N and it is in tension.