**Problem
3-2**
**
Using method of sections determine the
forces in the members BC, GC and GF of the pin-jointed plane truss shown in
figure 3-2(a)**
**Figure 3-2(a)**
**
Solution:**
This
truss is
determinate as it satisfies the condition of determinacy of plane
truss: m = 2j - 3
**Reactions at the support:**
Support A is on the roller, therefore
it will have only vertical reaction and no horizontal
reaction. Support D being hinged will experience both horizontal and
vertical reactions. (refer to figure 3-2(b)).
**Figure 3-2 (b)**
(i) There is no horizontal force acting on the truss
therefore the horizontal reaction at D will be zero to
satisfy
Σ F_{x} = 0;
(ii) Applying the condition of
equilibrium Σ F_{y} = 0; we get
A_{y}+ D_{y}
- 25 -10 - 20 - 15 = 0;
A_{y}+ D_{y} = 70;
(iii) Σ M_{z} = 0; Considering
z-axis perpendicular to the plane and passing through joint A. Take
moment of all the forces about z-axis (taking clock-wise negative and anticlock-wise positive);
A_{y }x 0 +
D_{y } x 6 - 20 x 2 - 15 x 4 - 10 x 2 + 25 x 0 =0;
we get
**D**_{y }= 120/6 = 20 kN;
Therefore **A**_{y} = 70 - 20 = 50 kN |