TABLE 2 Forces in Truss Members (Fig. 5)
Force
Horizontal Vertical
Member Slope component component kips kN
AJ 0.75 25.3 19.0 -31.7 -141.0
BK 0.75 21.3 16.0 -26.7 -118.8
CL 0.75 21.3 16.0 -26.7 -118.8
DN 0.75 22.7 17.0 -28.3 -125.9
EP 0.75 22.7 17.0 -28.3 -125.9
FQ 0.75 28.0 21.0 -35.0 -155.7
HJ 0.0 21.3 0.0 +21.3 +94.7
GM 0.0 16.0 0.0 +16.0 +71.2
GQ 0.0 28.0 0.0 +28.0 +124.5
JK 0.75 4.0 3.0 -5.0 -22.2
KL oo 0.0 6.0 -6.0 -26.7
LM 2.25 5.3 12.0 +13.1 +58.3
MN 2.25 6.7 15.0 +16.4 +72.9
NP oo 0.0 11.0 -11.0 -48.9
PQ 0.75 5.3 4.0 -6.7 -29.8= O; LMy= 12.0 kips (53.4 kN); LMn= 12.0/2.25 = 5.3 kips (23.6 kN). Substituting in the
first equation gives GM= 16.0 kips (71.2 kN).
Joint 5: ^Fn= 21.3 - 53 + DNH + MNH=Q; ^Fv = -6.0 + 16.0 - 12.0 - 0.75QDN 11 -
2.25MNH = O; DNH = -22.7 kips (-101.0 kN); MNn = 6.7 kips (29.8 kN); DNV = -17.0
kips (-75.6 kN); MNV = 15.0 kips (66.7 kN).
Joint 6: EPn = DNn= 22.7 kips (101.0 kN) of compression; AfP= 11.0 kips (48.9 kN)
of compression.
Joint 7: ^Fn = 22.7- PQn + FQn = O; ^F> - -8.0 - 17.0 - 0.75PQH - 0.75FQH = O;
PQn = -5.3 kips (-23.6 kN); FQ 11 = -28.0 kips (-124.5 kN); PQV = -4.0 kips (-17.8 kN);
FQV = -21.0 kips (-93.4 kN).
Joint 8: ^Fn= 28.0 - GQ = O; GQ = 28.0 kips (124.5 kN);; 2FF= 21.0 - 21.0 = O.
JoiTir P: SF^ =-16.0 - 6.7-5.3+ 28.0 = O; 2FF 15.0-11.0-4.0 = O.
- Complete the computation
Compute the values in the last column of Table 2 and enter them as shown.
TRUSS ANALYSIS BY THE METHOD
OFSECTIONS
Using the method of sections, determine the forces in members BK and LM in Fig. 5a.
Calculation Procedure:
- Draw a free-body diagram of one portion of the truss
Cut the truss at the plane act (Fig. 60), and draw a free-body diagram of the left part of the
truss. Assume that BK is tensile.