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SSC JE CE Previous Year Paper 3 (Held On: 25 Jan 18 Morning)

Option 4 : m = m_{1} + m_{2} + 3

ST 1: Building Material and Concrete Technology

18425

20 Questions
20 Marks
12 Mins

__Concept:__

A compound truss having m members, when formed by n number of simple trusses will be rigid and determinate if,

\({\rm{m\;}}\left( {{\rm{number\;of\;equilibrium\;equations}}} \right) = \mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{m}}_{\rm{i}}}\)

Where,

n is the number of members of an ith simple truss.

For this case, n = 2.

So, right hand side becomes m_{1 }+ m_{2}

We have 3 available equation of equilibrium which are

\({\rm{\;}}\sum {{\rm{F}}_{\rm{x}}} = 0,\sum {{\rm{F}}_{\rm{y}}} = 0,{\rm{and\;}}\sum {{\rm{M}}_{\rm{z}}} = 0{\rm{\;}}\)

So, m – 3 = m_{1} + m_{2}

∴ m = m_{1} + m_{2} + 3