4.6 through 4.28 Determine the force in each member of the truss shown by the method of Joints. FIG. P4. 19

Find the forces in the members of the truss by the method of joints.

Given information:

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero (∑Fx=0), consider the forces acting towards right side as positive (→+) and the forces acting towards left side as negative (←−).
• For summation of forces along y-direction is equal to zero (∑Fy=0), consider the upward force as positive (↑+) and the downward force as negative (↓−).
• For summation of moment about a point is equal to zero (∑Mat a point=0), consider the clockwise moment as negative and the counter clockwise moment as positive.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates Tension (T).

Calculation:

Show the free body diagram of the truss as shown in Figure 1.

Refer Figure 1.

Consider the horizontal and vertical reaction at A are denoted by Ax and Ay.

Consider the vertical reaction at G is denoted by Gy.

Find the value of the angle θ as follows:

tanθ=2448θ=tan−1(2448)θ=26.565°

Consider the triangles ABH, ACI, ADJ.

BHAB=CIAC=DJADBH16=CI32=2448

Solve the above relation.

BH=2448×16BH=8 ft

CI=2448×32CI=16 ft

Calculate the value of the angle α as follows:

tanα=CICDtanα=1616α=tan−1(1)α=45°

Consider the triangle GDJ.

GJ=GD2+DJ2=482+242=53.66 ft

Consider the triangles GFL, GEK, and GDJ.

GLGF=GKGE=GJGDGL16=GK32=53.6648

Solve the above relation.

GL=53.6648×16GL=17.88 ft

GK=53.6648×32GK=35.77 ft

Take the sum of the forces in the horizontal direction as zero.

∑Fx=0Ax−(10+20+20)sin26.565°=0Ax=22.36 k

Take the sum of the forces in the vertical direction as zero.

∑Fy=0Ay+Gy−10−(10+20+20)cos26.565°=0Ay+Gy=54.721        (1)

Take the sum of the moment about G as zero.

∑MG=0−10×80−10×GJ−20×GK−20×GL+Ay×96=0−10×80−10×53.66−20×35.77−20×17.88+Ay×96=0Ay=2409.696Ay=25.1 k

Substitute 25.1 k for Ay in Equation (1).

25.1+Gy=54.721Gy=29.621 k

Show the joint A as shown in Figure 2.

Figure 2

Find the forces in the members AH as follows:

For Equilibrium of forces,

∑Fy=0Ay+FAHsin26.565°=025.10+FAHsin26.565°=0FAH=−25.10sin26.565°FAH=−56.12 kFAH=56.12 k(C)

Find the forces in the members AB as follows:

For Equilibrium of forces,

∑Fx=0Ax+FAB+FAHcos26.565°=022.36+FAB−56.12cos26.565°=0FAB=−22.36+56.12cos26.565°FAB=27.83 k(T)

Show the joint B as shown in Figure 3.

Refer Figure 3.

Find the forces in the members BC as follows:

For Equilibrium of forces,

∑Fx=0−FAB+FBC=0FBC=FABFBC=27.83 k(T)

Find the forces in the members BH as follows:

For Equilibrium of forces,

∑Fy=0FBH=0

Show the joint H as shown in Figure 4.

Refer Figure 4.

Find the forces in the members CH and IH as follows:

For Equilibrium of forces,

∑Fy=0−10+FIHsin26.565°−FAHsin26.565°−FCHsin26.565°=0FIHsin26.565°−FCHsin26.565°=10−FAHsin26.565°FIH−FCH=10−56.12sin26.565°sin26.565°FIH−FCH=10−FAHsin26.565°sin26.565°FIH−FCH=−33.759        (2)

∑Fx=0−FAHcos26.565°+FCHcos26.565°+FIHcos26.565°=0−FAH+FCH+FIH=0FCH+FIH=FAHFCH+FIH=−56.12        (3)

Solve Equation (2) and (3).

FIH=44.94 k(C)FCH=11.18 k(C)

Show the joint C in Figure 5.

Refer Figure 5.

Find the forces in the members IC as follows:

For Equilibrium of forces,

∑Fy=0FIC+FCHsin26.565°=0FIC=−(−11.18)sin26.565°FIC=5 k(T)

Find the forces in the members CD as follows:

For Equilibrium of forces,

∑Fx=0FCD−FBC−FCHcos26.565°=0FCD=FBC+FCHcos26