13.46 and 13.47 Determine the reactions and the force in each member of the trusses shown in Figs. P13.46 and P13.47 using the method of consistent deformations. FIG. P13.46

Calculate the support reactions and the member forces of the truss using method of consistent deformation.

Given information:

The structure is given in the Figure.

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).

Condition for zero force members:

1. 1. If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.
2. 2. If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the degree of indeterminacy of the structure using the equation:

i=m+r−2j

Here, m is the number of members, r is the reaction of the truss, and j is the number of joints.

Substitute 15 for m, 2 for r, and 8 for j.

i=15+2−2(8)=17−16=1

Therefore, the degree of indeterminacy of the truss is i=1.

Select the member DG as redundant.

The truss and the loading are symmetric. Hence, only analyze the right half CEGH of the truss.

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

Refer Figure 1,

Find the vertical reactions.

The loading of the truss is symmetric.

Ay=Ey=60+60+602Ay=Ey=90 kN

Remove the redundant member DG.

At C, the half of downward vertical load consider to the right half.

Sketch the free body diagram of the portion CEGH as shown in Figure 2.

Refer Figure 2,

Consider the forces in the member due to external load as FO.

Find the length of member EH.

LEH=62+82=10 ft

Find the member forces using method of joints.

Consider joint E.

Summation of forces along y-direction is equal to 0.

ΣFy=0

FEHO(610)+90=00.6FEHO=−90FEHO=−150 kFEHO=150 k(C)

Summation of forces along x-direction is equal to 0.

ΣFx=0

FEHO(810)+FEDO=00.8(−150)=−FEDOFEDO=−120 kFEDO=120 k(T)

Consider joint D.

Summation of forces along y-direction is equal to 0.

ΣFy=0

FDHO−60=0FDHO=60 k (T)

Summation of forces along x-direction is equal to 0.

ΣFx=0

−FDCO+FDEO=0FDCO=120 k (T)

Consider joint H.

Summation of forces along y-direction is equal to 0.

ΣFy=0

−FHCO(610)−FHEO(610)−FHDO=0−0.6FHCO−(−150)(0.6)−60=00.6FHCO=90−60FHCO=50 k (T)

Summation of forces along x-direction is equal to 0.

ΣFx=0

−FHGO−FHCO(810)+FHEO(810)=0−FHGO−0