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.

#### Solution

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

_{AE}sin30 - F

_{AC}cos30 = 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 |