Analysis of Truss: Lecture 02

Example 02: Calculate the force of member AB, AF, BF, BC, FC and FG of the truss shown in figure 01. Indicate whether they are in tension or compression?

Figure 01

Solution: Calculation of Reactions (Figure 02)

∑M_A = 0

= > 3X 4 + 6X8+ 9X12-R_EX16 =0

= > R_E= 10.5KN

∑F_Y = 0

= > R_A+ R_E -3- 6- 9 =0

= > R_A= 7.5 kN

Figure 02

Considering Joint A:

Here, ɵ = tan^-1 (¾) = 36.87⁰

∑F_Y = 0

= > 7.5 + AF Sin 36.87 =0

= > AF = -12.5 KN

            = 12.5 KN (C)

∑F_X = 0

= > AB – AF Cos 36.87 =0

= > AB – 12.5 Cos 36.87 =0

= > AB = 10KN (T)

Figure 03

Consider Joint B:

∑FY = 0

= > 3 - BF =0

= > BF = 3 KN (T)

∑FX = 0

= > BC- AB =0

= > AB = BC = 10KN (T)

Figure 04

Consider Section (1) ~ (1)

Figure 05

Here, β = tan-1 (2/4) = 26.560

∑MC = 0

= > -3X4 + 7.5 X8 + FG Cos 26.56 X 3 + FG Sin 26.56 X4 =0

= > -12 + 60 + 2.68 FG + 1.786 FG =0

= > FG = -10.76 KN

            = 10.76 KN (T)

∑MF = 0

= > 7.5 X 4 – BC X3 = 0

= > BC = 10KN (T)

∑FX = 0

= > 10 + FC Sin 53.13 – FG Cos 26.56 = 0

= > FC = -0.469

            = 0.50 KN (C)