AJV= O; /. AJV= 19.0 kips (-84.5 kN); AJn = - 19.0/0.75 = -25.3 kips (-112.5 kN). Sub-
stituting in the first equation gives HJ= 21.3 kips (94.7 kN).
The algebraic signs disclose that AJ is compressive and HJ is tensile. Record these re-
sults in Table 2, showing the tensile forces as positive and compressive forces as nega-
tive.
- Determine the forces at another joint
Draw a free-body diagram of the pin at joint 2 (Fig. 5c). Show the known force AJ as
compressive, and assume that the unknown forces BK and JK are tensile. Apply the equa-
tions of equilibrium, expressing the vertical components of BK and JK in terms of their
horizontal components. ThusSF^ =25.3 + BKH + JKH=Q\ 2FF=-6.0+ 19.0 + 0.15BKH
-Q.15JKff=0.
Solve these simultaneous equations, to obtain BKn = -21.3 kips (-94.7 kN); JKH =
-4.0 kips (-17.8 kN); BKV = -16.0 kips (-71.2 kN); JKV = -3.0 kips (-13.3 kN). Record
these results in Table 2.
5. Continue the analysis at the next joint
Proceed to joint 3. Since there are no external horizontal forces at this joint, CLn = BKn =
21.3 kips (94.7 kN) of compression. Also, KL = 6 kips (26.7 kN) of compression.
6. Proceed to the remaining joints in their numbered order
Thus Jor joint 4:2FH = ^.0-213 +4.0+ LMH+GM= 0;2FV = -3.0-3.0-6.0+ LMV
(b) Free-body diogrom (c) Free-body diagram
of joint I of joint 2FIGURE 5(a) Truss diagram6 panels