Question

Hi I only need question 6.67.

2 solutions can be found online but the sectioning/cutting is not clear for me.

Through how many joints can we cut through maximum?

A small explanation for every step for and after cutting the truss structure is appreciated thank you

Counters Β ΔΟΛ F 6.67 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the force in member DE and in the counters that are acting under the given loading. AC EG 6.68 Solve Prob. 6.67 assuming that the 9-kip load has been removed. 6 kips 9 kips 12 kips |8f***8 ft 8 ft**8 ft Fig. P6.67 6.69 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.)

Answer 1

Coun ters BE, CD, DG, EF are Counters u Jokips lo rakips 12kips a & 8or & ft & 8 ft * ett * There is no limit for the number of members to becut through in section method. We can cut through a numb of members until and unless the sectioned truss is in equilibrium. 0ܝܚܝ ܚܝܝܝ ܝܝܝܝܝ ܝܝܝܝ SOLO ank Steps to be followed while solving a truss with counters: 1 1. We have to determine the reaction forces of supports 2. Remember when a truss is loaded, one of the pair of counters becomes taunt (tension) while the other becomes slack, and there is not pretension in any counter. - Out of a pair of counters, one will have magnitude I of a force in tension and other to counter has zero 1 magnitude of force. | 3. So cut a section in such a way that the counter I which is in tension is exposed. 14. Solve for the force equilibrium in yo direction with I a one counter. If the value is positive (ine in tension this is the solution - If not, solve for force equili- CSbriumed with the other counter

the truss, % Let us solve Reqctions: Awy ET v skips K eft x t 8ft 19 kips x 8ft 112 kips oft > EFxzo +) EMA =O → [ Ax =0 → Hux 32 -(648) - (9X16) - (12%24)=0 + [ Hy = 15 kips (7) I ti E Ey=0 . AL Hy-lo 12 +9+ 6) + Ay =0 → Ay = 27 - Hy = 27-15 Ay = 12 kips] (1). Now, let us cut through the members BD, BE, CD and CE. As mentioned earlier, Let us (considerleen the member 1 BE is in tension Ekips and co 12 kips is a zero.force member Cslack). 1. TFcp =0 From the load diagram, Oz tan! ( ) = 36.8694 FCE Sumo Vakips 12kips I skips > FBD Scanned with CamScanner 12 kips Ekips

our in tension, and ue is positive X - B). For the above FBD) +1 EFy=O 12-6 - FBE Sin O zo 12-6 – FBE Sim (36.8690) =0 | Fee = 10 kips (tension) Since the above value į positive and in assumption is correct. Now, for other pair of counters, Let us cut through members, DF, EF, DG and EG. Again, let us assume, Am member EF is in tension 12 kips Grips akips | 12 kips and DG is a zero force member as we did earlier. :FDG = 0 From the truss diagram F öch oz tani' ( ) = 36.8694 IG E Tis is kips isti Efy=0 Forek - FGE 15 kips ~ 12 kips ☆ -12+15 - FFE Sin O =0 = Fre sin (36.869) = 3 FFE = 5 kips (tension) Since the above value is positive and is ourannassumption is right. in tension, CamScanner

Blow by method of joints, I at joint D, As we know FOX For FDG = Foc to Foc 1 FDG Now +18Fy =o + FDE = 0 FDE 1. Feozo FBE = 10 kips (T) Fog = 0 FEF = 5 kips (T) FDE=0 Ics Scanned with CamScanner