Analysis of Truss: Lecture 01

Example 01: Calculate the force of member AB, AH, BH, BC, BG, and HG 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

            = > 2x20+ 2x40+ 2X 60+ 6X40+5X60-R_E =0

            = > R_E = 9.75 kips

           ∑F_y = 2+ 2+ 2+6+5-9.75-R_A=0

           = > R_A= 7.25 kips

 

Figure 02 

Taking the joints A (Figure 03):

Figure 03

∑F_y = 0

= > 7.25 + AB_Y= 0

= > 7.25 + AB Sin30⁰=0

= > AB= -14.5kips

           = 14.5kips (C )

∑F_x = 0

= > AH – AB_X= 0

= > AH= 14.5 Cos 30⁰

            = 12.6 kips (T)

Taking Joint H (Figure 04):

Figure 04

∑F_x = 0

= > AH – GH= 0

= > AH= GH = 12.6 kips

∑F_y = 0

= > BH=0; Such type of member with zero force is considered as zero force member. 

Taking the section (1) ~ (1) of figure 5:

 

Figure 05

[We are in need of calculating the member force of BC, BG and HG. If we can take moment with respect to B, then the unknown load value of BC and BG will be zero and there will only be one load (HG) which will be easily calculated]

 ∑M_B= 0

= > HGx 11.54 – 7.25x 20= 0

= > HG = 12.56 kips (T)

[Now, If we can take moment with respect to G, then the unknown load value of BG and known load of HG will be zero and there will only be one unknown load (BC) which will be easily calculated. Moreover, BC will have two components and needed to be considered separately]

∑M_G= 0

= > 7.25 X 40 – 2X 20 + BC_X X 11.54 + BC_Y X 20 = 0

= > 250 + BC Cos 30⁰ X 11.54 + BC Sin 30⁰ X 20 = 0

= > 250 + 9.99 BC + 10 BC = 0

= > BC= -12.5 kips

           = 12.5 kips (C)

∑F_Y= 0

= > 7.25 – 2 – 12.5 Sin 30⁰ – BG Cos 60⁰ = 0

= > BG = 2 kips (T)