Handbook of Civil Engineering Calculations

TABLE 1 Forces in Truss Members (Fig. 4)

Force

Member kips kN

AJ +6.1 +29.8

BK +9.5 +42.2

CTV +19.8 +88.0

DE +30.2 -134.2

HJ -6.0 -26.7

GM -13.0 -57.8

FP -20.0 -88.9

JK -4.5 20.0

KM +8.1 +36.0

MN -8.6 -38.2

NP +10.4 +46.2

PE -12.6 56.0

closes that they are tensile forces. Force NM is directed toward the joint; therefore, it is

compressive.

The validity of this procedure lies in the drawing of the vectors representing external

forces while proceeding around the truss in a clockwise direction. Tensile forces are

shown with a positive sign in Table 1; compressive forces are shown with a negative sign.

Related Calculations: Use this general method for any type of truss.

TRUSS ANALYSIS BY THE METHOD

OFJOINTS

Applying the method of joints, determine the forces in the truss in Fig. 5a. The load at
joint 4 has a horizontal component of 4 kips (17.8 kN) and a vertical component of 3 kips
(13.3 kN).

Calculation Procedure:

1. Compute the reactions at the supports
   Using the usual analysis techniques, we find RLV =19 kips (84.5 kN); RLH = 4 kips (17.8
   kN); RR = 21 kips (93.4 kN).


2. List each truss member and its slope
   Table 2 shows each truss member and its slope.


3. Determine the forces at a principal joint
   Draw a free-body diagram, Fig. 5b, of the pin at joint 1. For the free-body diagram, as-
   sume that the unknown internal forces AJ and HJ are tensile. Apply the equations of equi-
   librium to evaluate these forces, using the subscripts H and V, respectively, to identify the
   horizontal and vertical components. Thus ^FH = 4.0 + AJn + HJ = O and ^Fv = 19.0 +