**Method of sections for Truss member force calculation**

**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

**Calculation of member forces by method of
sections**

Method of sections is useful when we have to calculate the forces in some of the members, not all. It is expalined in this example. We cut the truss into two parts through section (1) - (1) passing through GF, GC and BC. We consider the equilibrium of the left side part of the truss as shown in figure 3-2(c).

**Figure 3-2(c)**

While considering equilibrium of this part we look only for the support reactions, the external forces acting on the truss, and the member forces which are cut by the section.

Let's assume the unknown forces in members GF, GC and BC as tensile as shown in the figure.

Σ F_{x} = 0; => F_{BC} + F_{GF}
+ F_{GC} cos45 = 0;
(i)

Σ F_{y} = 0; => 50 - 25 - 20 -10
- F_{GC} sin45 = 0; (ii)

F_{GC} sin45 = - 5; **F _{GC}**
= - 7.072 kN;

Σ M = 0 about point G

F_{BC} x 2 - 50 x 2 + 25 x 2 = 0;
(iii)

**F _{BC}** = 25 kN;

Substituting the value of F_{BC} and F_{GC
}in eq. (i) we get

**F _{GF}** = - 20 kN

negative value indicates that the force is opposite to the assumed direction.

Result of Member
forces calculations |
||

Member |
Force (k2N) |
Nature of force |

GC | 7.07 | Compressive |

BC | 25 | Tensile |

GF | 20 | Compressive |

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