- Determine the magnitude and character of the first force
Take moments with respect to joint 4. Since each halt of the truss forms a 3-4-5 right tri-
angle, d = 20(3/5) = 12 ft (3.7 m), SM 4 = 19(20) - 6(10) + UBK = O, andBK=-26.7 kips
(-118.SkN).
The negative result signifies that the assumed direction of BK is incorrect; the force is,
therefore, compressive.
- Use an alternative solution
Alternatively, resolve BK (again assumed tensile) into its horizontal and vertical compo-
nents at joint 1. Take moments with respect to joint 4. (A force may be resolved into its
components at any point on its action line.) Then, 2M 4 = 19(20) + 2OBK 7 = -16.0 kips
(-71.2 kN); BK = -16.0(5/3) = -26.7 kips (-118.8 kN).
- Draw a second free-body diagram of the truss
Cut the truss at plane bb (Fig. 6b\ and draw a free-body diagram of the left part. Assume
LM is tensile.
- Determine the magnitude and character of the second force
Resolve LM into its horizontal and vertical components at joint 4. Take moments with
respect to joint 1: SM 1 = 6(10 + 20) + 3(20) - 20LMV = O; LMV = 12.0 kips (53.4 kN);
LMH= 12.0/2.25 = 3.3 kips (23.6 kN); LM= 13.1 kips (58.3 kN).
REACTIONS OFA THREE-HINGEDARCH
The parabolic arch in Fig 7 is hinged at A, B, and C. Determine the magnitude and direc-
tion of the reactions at the supports
Calculation Procedure:
- Consider the entire arch as a free body and take moments
Since a moment cannot be transmitted across a hinge, the bending moments at A, B, and C
FIGURE 6
