Calculation of Truss member forces by method of joints

Problem 3-1

Use method of joints to determine the forces in all the members of pin-jointed plane truss shown in figure 3-1(a).

Figure 3-1(a)

Solution:

In the given truss the support at A is roller and C is hinged. First we will find whether this truss is determinate or indeterminate.

Condition of determinacy of plane truss: m = 2j - 3

In this truss  j = 6, which requires 2×6 - 3 members for the truss to be determinate. It is confirmed from the figure that there are 9 members in this truss.

Therefore the the given truss is statically determinate.

Reactions at the support:

Support A is on the roller, therefore it will have only vertical reaction and no  horizontal reaction. Support C being hinged will experience both horizontal and vertical reactions. (refer to figure 3-1(b).

Considering horizontal reaction at C to be in the +ve x direction and. Applying the conditions of static equilibrium, we get;

(i) Σ F_x = 0; therefore C_x - 15 = 0;               eq (1)

Solving the equation we get C_x =15 kN. The positive sign of this value indicates that our assumption in the direction of C_x was correct.

(ii) Σ F_y = 0; yields A_y+ C_y - 25 -10 - 20 = 0;        

                            A_y+ C_y = 55;                      eq. (2)

(iii) Σ M_z = 0; Considering z-axis perpendicular to the plane and passing through joint A. Take moment of all the forces about z-axis (taking clock-wise negative and anticlock-wise positive);

A_y x 0 + C_y x 4 -  C_x x 0  - 20 x 2 + 15 x 2  - 10 x 2 + 25 x 0 =0;

we get  C_y = 7.5 kN;

Therefore  A_y = 47.5 kN;

Figure 3-1(b)

Calculation of member forces

We use method of joints to find all the forces in the members of the given truss.

First of all look for the joint which does not have more than 2 unknown forces. In this truss we find that joint D and F have only two unknown forces.

Let's start with joint D; In the beginning assume all the unknown forces as tensile. Tensile forces are shown with an outward arrow whereas compressive forces are shown with  an inward arrow at the joint. The forces acting in the +ve directions of axes are taken as +ve whereas those acting in the -ve directions of axes are taken as -ve.

Equilibrium of joint D

Σ F_x = 0;  

- F_DE  - 15 kN = 0;

F_DE = - 15 kN The -ve sign of F_DE indicates that the assumed direction (tensile) was wrong, therefore the actual nature of force F_DE will be compressive.

Σ F_y = 0; => - F_DC = 0; therefore F_DC = 0

Equilibrium of joint F

Σ F_x = 0;   => F_FE = 0

Σ F_y = 0; => - 25 - F_FA = 0;

F_FA = - 25 kN

The -ve sign of F_FA indicates that the assumed direction of (tension) for this forces is not correct, therefore the actual direction of force in member FA is to be compressive.

Now the joints A and C have only two unknown forces whereas joint  B and E have three unknown forces. So we come to joint A.

Equilibrium of joint A

The force in member AE is making an angle of 45 degree with +x-axis, therefore it has to be resolved into its rectangular components along  x-axis  (F_AEcos45) and along y-axis ( F_AE sin45).

Σ F_x = 0;   => F_AE cos45 + F_AB = 0                  (i)

Σ F_y = 0; => F_AE sin45 + A_y - F_AF = 0              (ii)

substituting the magnitudes  of A_y and F_AF  (equal to F_FA)  into eq.(i) we get

F_AE sin45 + 47.5 - 25 = 0;

Therefore F_AE = - 22.5/sin45=  - 31.82 kN; negative sign of F_AE indicates that this force should be in the opposite sense, hence it should be a compressive force of 31.82 kN.

Substituting this value of F_AE in eq. (i) yields.

F_AB = 22.5 kN;

Equilibrium of joint C

      Σ F_y = 0;

     F_CE sin45 + C_y + F_CD = 0      (iii)

        F_CE sin45 + 7.5 + 0 = 0; Therefore F_CE = - 7.5/sin45 = - 10.61 kN; again the negative sign of this force indicates that it should be in the reverse direction, hence compressive 10.61 kN.

Σ F_x = 0;   => - F_CE cos45 - F_CB + C_x = 0       (iv)

Substituting the value of F_CE  and C_x in eq. (iv) we get;

  F_CB  = + 7.5 + 15 = + 22.5 kN

Equilibrium of joint B

Σ F_y = 0; => F_BE = 20 kN

From the previous calculations about joint A and joint C, it is also evident that Σ F_x = 0;  because F_BA = - F_AB and F_BC = - F_CB ; This confirms our calculations.

Result of Member forces calculations
Member	Force (kN)	Nature of force
AB	22.5	Tensile
AF	25	Compressive
AE	31.82	Compressive
BC	22.5	Tensile
BE	20	Tensile
CD	0
CE	10.61	Compressive
DE	15	Compressive
EF	0

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