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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. - Engineering Mechanics

Answer in Brief

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.

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Solution

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

In Δ DEG, ∠EDG=30o, ∠DGE=90o
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
RC = 85 kN
Σ Fy=0
-20 – 50 + RD + RC=0
RD = -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 -FEC x `sqrt3/2x(I/2)-`FEC x`1/2XI/2`=0
-10 x l - FEC x`I/4`-FEC 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 - FAEsin30 - FACcos30 = 0
-50 - 50 x 0.5 = FAC x 0.866
FAC = -86.6025kN
Joint D
FDC + FDEcos30 = 0
-25.9808 + 0.866FDE = 0
FDE = 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  
Concept: Analysis of Plane Trusses by Using Method of Joints
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