# Forces in Statically Determinate Trusses

A convenient method for determining the member forces in a truss is to isolate a portion of the truss. A section should be chosen such that it is possible to determine the forces in the cut members with the equations of equilibrium [Eq. (3. 11) or (3.12)]. Compressive forces act toward the panel point, and tensile forces act away from the panel point.

## Method of Sections

To calculate the force in member a of the truss in Fig. 3.67a, the portion of the truss in Fig. 3.67b is isolated by passing section xx through members a, b, and c. Equilibrium of this part of the truss is maintained by the 10-kip loads at panel points U1 and U2, the 25-kip reaction, and the forces Sa, Sb, and Sc in members a, b, and c, respectively. Sa can be determined by equating to zero the sum of the moments of all the external forces about panel point L3, because the other unknown forces Sb and Sc pass through L3 and their moments therefore equal zero. The corresponding equilibrium equation is

– 9Sa +  36 x 25 - 24 x 10 - 12 x 10 = 0

a Solution of this equation yields Sa  60 kips. Similarly, Sb can be calculated by equating to zero the sum of the moments of all external forces about panel point U2:

-9Sb + 24 x 25 - 12 x 10 = 0

for which Sb = 53.3 kips. Since members a and b are horizontal, they do not have a vertical component. Hence diagonal c must carry the entire vertical shear on section
xx: 25 - 10 - 10 = 5 kips.
With 5 kips as its vertical component and a length of 15 ft on a rise of 9 ft,

Sc = 15â„9 x 5 = 8.3 kips

When the chords are not horizontal, the vertical component of the diagonal may be found by subtracting from the shear in the section the vertical components of force in the chords.

## Method of Joints

A special case of the method of sections is choice of sections that isolate the joints. With the forces in the cut members considered as external forces, the sum of the horizontal components
and the sum of the vertical components of the external forces acting at each joint must equal zero.
Since only two equilibrium equations are available for each joint, the procedure is to start with a joint that has two or fewer unknowns (usually a support). When these unknowns have been found, the procedure is repeated at successive joints with no more than two unknowns.
For example, for the truss in Fig. 3.68a, at joint 1 there are three forces: the reaction of 12 kips, force Sa in member a, and force Sc in member c. Since c is horizontal, equilibrium of vertical forces requires that the vertical component of force in member a be 12 kips.
From the geometry of the truss,

Sa = 12 - 15â„9 =  20 kips.

The horizontal component of

Sa is 20 x 12â„15 = 16 kips.

Since the sum of the horizontal components of all forces acting at joint 1 must equal zero, Sc = 16 kips.
At joint 2, the force in member e is zero because no vertical forces are present there.
Hence, the force in member d equals the previously calculated 16-kip force in member c.
Forces in the other members would be determined in the same way (see Fig. 3.68d, e, and Æ’). 