Question

Referring to the truss shown in the figure. Find :
(a) Reaction at D and C
(b)Zero force members.
(c)Forces in member FE and DC by method of section.
(d)Forces in other members by method of joints.

Answer 1

By Geometry:
In Δ ADC, ∠ADC = ∠CAD = 30^o
AC = CD = l
Similarly, in Δ EDC,
ED = EC
Δ DEG and Δ CEG are congruent
DG = GC =`1/2`

In Δ DEG, ∠EDG=30^o, ∠DGE=90^o
tan30= `(EG)/(DG)`

EG = DG.tan30 =`I/2x1/sqrt3=I/(2sqrt3)`
In Δ ACH,
CH =`(AC)/2=I/2`

DH = DC + CH = l +`I/2=(3I)/2`
No horizontal force is acting on the truss, so no horizontal reaction will be present at point A The truss is in equilibrium Applying the conditions of equilibrium
Σ MD = 0
-20 x DG -50 x DH + RC x DC = 0
-20 x`1/(2)50X(3I)/2+RCX1=0`

-10 - 75 + RC = 0
R_C = 85 kN
Σ Fy=0
-20 – 50 + R_D + R_C=0
R_D = -15kN
Loading at point B and F is shown

As per the rule,member BF will have zero force and is a zero force number. Similarly,Member CF will have zero force
Method of sections :

Applying the conditions of equilibrium to the section shown
Σ MD = 0
-20 x DG – FECcos 30 x EG - FECsin30 x DG = 0
-20 x `I/2` x - FECcos30 x EG - FECsin30 x DG = 0
-20 x`I/2`x -F_EC x `sqrt3/2x(I/2)-`F_EC x`1/2XI/2`=0
-10 x l - F_EC x`I/4`-F_EC x`I/4`=0
`-(2I)/4F_(EC)=10L`
FEC = -20kN

RD – 20 - FECsin30 + FEAsin30 = 0
-15 - 20 + 20 x 0.5 + FEA x 0.5 = 0
FECcos30 + FEAcos30 + FDC = 0
-20 x 0.866 + 50 x 0.866 + FDC = 0
FDC = - 25.9808kN

Method of joints:

Joint A
-50 - F_AEsin30 - F_ACcos30 = 0
-50 - 50 x 0.5 = FAC x 0.866
F_AC = -86.6025kN
Joint D
F_DC + FDEcos30 = 0
-25.9808 + 0.866FDE = 0
F_DE = 30kN
Final answer :

Member	Magnitude (in kN)	Nature
AE (AF and EF)	50	Tension
AC (AB and BC)	86.6025	Compression
EC	20	Compression
DE	30	Tension
DC	25.9808	Compression
FB	0
FC	0