Determine the reactions and the force in each member of the truss shown in Fig. 17.11(a) by the matrix stiffness method.

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Degrees of Freedom. From the analytical model of the truss shown in Fig. 17.11(b), we observe that only joint 3 is free to translate. Thus the truss has two degrees of freedom, [latex]d_1[/latex] and [latex]d_2[/latex], which are the unknown translations of joint 3 in the X and Y directions, respectively.

Structure Stiffness Matrix.

Member 1 As shown in Fig. 17.11(b), joint 1 has been selected as the beginning joint and joint 3 as the end joint for member 1. By applying Eqs. (17.13), we determine

[latex]\cos heta = \frac{X_e - X_b}{L} = \frac{X_e - X_b}{\sqrt{(X_e -X_b)^2 + (Y_e - Y_b)^2}} [/latex] (17.13a)
[latex]\sin heta = \frac{Y_e - Y_b}{L} = \frac{Y_e - Y_b}{\sqrt{(X_e -X_b)^2 + (Y_e - Y_b)^2}} [/latex] (17.13b)

[latex]L=\sqrt{\left(X_{3}-X_{1} ight)^{2}+\left(Y_{3}- Y_{1} ight)^{2}}=\sqrt{(15-0)^{2}+(20-0)^{2}}=25 ft[/latex]

[latex]\cos heta=\frac{X_{3}-X_{1}}{L}=\frac{15}{25}=0.6[/latex]

[latex]\sin heta=\frac{Y_{3}-Y_{1}}{L}=\frac{20}{25}=0.8[/latex]

The member sti¤ness matrix in global coordinates can now be evaluated by using Eq. (17.29)

[latex] K =\frac{EA}{L} \begin{bmatrix} \cos^2 heta & \cos heta \sin heta & -\cos^2 heta & -\cos heta \sin heta \ \cos heta \sin heta & \sin^2 heta & -\cos heta \sin heta & -\sin^2 heta \ -\cos^2 heta & -\cos heta \sin heta & \cos^2 heta & \cos heta \sin heta \ -\cos heta \sin heta & -\sin^2 heta & \cos heta \sin heta & \sin^2 heta \end{bmatrix} [/latex] (17.29)

[latex]K _{1}=\frac{(29,000)(9)}{(25)(12)}\left[\begin{array}{rrrr} 0.36 & 0.48 & -0.36 & -0.48 \ 0.48 & 0.64 & -0.48 & -0.64 \ -0.36 & -0.48 & 0.36 & 0.48 \ -0.48 & -0.64 & 0.48 & 0.64 \end{array} ight][/latex]

or

(1)

From Fig. 17.11(b), we observe that the displacements of the beginning joint 1 for the member are zero, whereas the displacements of the end joint 3 are [latex]d_1[/latex] and [latex]d_2[/latex]. Thus the structure degree of freedom numbers for this member are 0, 0, 1, 2. These numbers are written on the right side and at the top of [latex]K_1[/latex] (see Eq. (1)) to indicate the rows and columns, respectively, of the structure sti¤ness matrix S, where the elements of [latex]K_1[/latex] must be stored. Note that the elements of [latex]K_1[/latex], which correspond to the zero structure degree of freedom number, are simply disregarded. Thus, the element in row 3 and column 3 of [latex]K_1[/latex] is stored in row 1 and column 1 of S, as shown in Fig. 17.11(c). Similarly, the element in row 3 and column 4 of [latex]K_1[/latex] is stored in row 1 and column 2 of S. The remaining elements of [latex]K_1[/latex] are stored in S in a similar manner (Fig. 17.11(c)).

Member 2 From Fig. 17.11(b), we can see that joint 2 is the beginning joint and joint 3 is the end joint for member 2. By applying Eqs. (17.13), we obtain

[latex]\cos heta=\frac{X_{3}-X_{2}}{L}=\frac{15-15}{20}=0[/latex]

[latex]\sin heta=\frac{Y_{3}-Y_{2}}{L}=\frac{20-0}{20}=1[/latex]

Thus, by using Eq. (17.29)

From Fig. 17.11(b), we can see that the structure degree of freedom numbers for this member are 0, 0, 1, 2. These numbers are used to store the pertinent elements of [latex]K_2[/latex] in their proper positions in the structure sti¤ness matrix S, as shown in Fig. 17.11(c).

Member 3 [latex]\cos heta=1 \quad \sin heta=0[/latex]

By using Eq. (17.29),

The structure degree of freedom numbers for this member are 0, 0,1, 2. By using these numbers, the elements of [latex]K_3[/latex] are
stored in S, as shown in Fig. 17.11(c).

Note that the structure sti¤ness matrix S (Fig. 17.11(c)), obtained by assembling the sti¤ness coe‰cients of the three members, is symmetric.

Joint Load Vector. By comparing Fig. 17.11(a) and (b), we realize that

[latex]P_{1}=100 \cos 60^{\circ}=50 k \quad P_{2}=-100 \sin 60^{\circ}=-86.6 k[/latex]

Thus the joint load vector is

[latex]P =\left[\begin{array}{c}50 \ -86.6 \end{array} ight][/latex] (2)

Joint Displacements. The sti¤ness relations for the entire truss can be expressed as (Eq. (17.41)

[latex]P - P _{f}= S d[/latex] (17.41)

with [latex]P_f =0[/latex])

P = Sd (3)

By substituting P from Eq. (2) and S from Fig. 17.11(c), we write Eq. (3) in expanded form as

[latex]\left[\begin{array}{c}50 \-86.6\end{array} ight]=\left[\begin{array}{rr}1,763.2 & 417.6 \417.6 & 1,644.3\end{array} ight]\left[\begin{array}{l}d_{1} \d_{2}\end{array} ight][/latex]

By solving these equations simultaneously, we determine the joint displacements to be

[latex]d_{1}=0.0434 ext { in. } \quad d_{2}=-0.0637 ext { in }[/latex]

or

[latex]d =\left[\begin{array}{r}0.0434 \-0.0637\end{array} ight] ext { in. }[/latex]

Member End Displacements and End Forces.

Member 1 The member end displacements in global coordinates, v, can be obtained by simply comparing the member’s global degree of freedom numbers with the structure degree of freedom numbers for the member, as follows:

[latex]v _{1}=\left[\begin{array}{l}v_{1} \v_{2} \v_{3} \v_{4}\end{array} ight] \begin{array}{l}0 \0 \1\2\end{array}=\left[\begin{array}{c}0 \0 \d_{1} \d_{2}\end{array} ight]=\left[\begin{array}{c}0 \0 \0.0434 \-0.0637\end{array} ight]in[/latex] (4)

Note that the structure degree of freedom numbers for the member (0, 0, 1, 2) are written on the right side of v, as shown in Eq. (4). Since the structure degree of freedom numbers corresponding to [latex]v_1 ext{ and } v_2 [/latex] are zero, this indicates that [latex]v_1 = v_2 = 0[/latex]. Similarly, the numbers 1 and 2 corresponding to [latex]v_3[/latex] and [latex]v_4[/latex], respectively, indicate that [latex]v_3 = d_1[/latex] and [latex]v_4 = d_2[/latex]. It should be realized that these compatibility equations could have been established alternatively simply by a visual inspection of the line diagram of the structure (Fig. 17.11(b)). However, the use of the structure degree of freedom numbers enables us conveniently to program this procedure on a computer.

The member end displacements in local coordinates can now be determined by using the relationship [latex]u = T_v [/latex] (Eq. (17.14)), with T as defined in Eq. (17.22):

[latex] T = \begin{bmatrix} \cos heta & \sin heta &0&0 \ 0 & 0& \cos heta & \sin heta \end{bmatrix} [/latex] (17.22)

[latex]u _{1}=\left[\begin{array}{l}u_{1} \u_{2}\end{array} ight]=\left[\begin{array}{llll}0.6 & 0.8 & 0 & 0 \0 & 0 & 0.6 &0.8\end{array} ight]\left[\begin{array}{l}0 \0 \0.0434 \-0.0637\end{array} ight]=\left[\begin{array}{c}0 \-0.0249\end{array} ight] ext { in. }[/latex]

By using Eq. (17.7), we compute member end forces in local coordinates as

Q = ku

[latex]Q _{1}=\left[\begin{array}{l}Q_{1} \Q_{2}\end{array} ight]=870\left[\begin{array}{rr}1 & -1 \-1 & 1\end{array} ight]\left[\begin{array}{c}0 \-0.0249\end{array} ight]=\left[\begin{array}{r}21.66 \-21.66\end{array} ight] k[/latex]

Thus, as shown in Fig. 17.11(d), the axial force in member 1 is

21.66 k (C) Ans.

By applying Eq. (17.17), we can determine member end forces in global coordinates as

[latex]F = T ^{T} Q[/latex]

[latex]F _{1}=\left[\begin{array}{l}F_{1} \F_{2} \F_{3} \F_{4}\end{array} ight]=\left[\begin{array}{ll}0.6 & 0 \0.8 & 0 \0 & 0.6 \0 & 0.8\end{array} ight]\left[\begin{array}{r}21.66 \-21.66\end{array} ight]=\left[\begin{array}{c}13 \17.33 \-13 \-17.33\end{array} ight] k[/latex]

Member 2 The member end displacements in global coordinates are given by

[latex] v_2 =\begin{bmatrix} v_1 \ v_2 \ v_3 \v_4 \end{bmatrix} \begin{matrix} 0 \ 0 \ 1 \2 \end{matrix} = \begin{bmatrix} 0\0\d_1\d_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0.0434 \ -0.0637 \end{bmatrix} in. [/latex]

By using the relationship u = Tv, we determine the member end displacements in local coordinates to be

[latex]u _{2}=\left[\begin{array}{l}u_{1} \u_{2}\end{array} ight]=\left[\begin{array}{llll}0 & 1 & 0 & 0 \0 & 0 & 0 & 1\end{array} ight]\left[\begin{array}{c}0 \0 \0.0434 \-0.0637\end{array} ight]=\left[\begin{array}{c}0 \-0.0637\end{array} ight] ext { in }[/latex]

Next, the member end forces in local coordinates are computed by using the relationship Q = ku:

[latex]Q _{2}=\left[\begin{array}{l}Q_{1} \Q_{2}\end{array} ight]=1,087.5\left[\begin{array}{rr}1 & -1 \-1 & 1\end{array} ight]\left[\begin{array}{c}0 \-0.0637\end{array} ight]=\left[\begin{array}{r}69.27 \-69.27\end{array} ight] k[/latex]

Thus, as shown in Fig. 17.11(d), the axial force in member 2 is

69.27 k (C)

By using the relationship [latex]F = T ^{T} Q[/latex] , we calculate the member end forces in global coordinates to be

[latex]F _{2}=\left[\begin{array}{l}F_{1} \F_{2} \F_{3} \F_{4}\end{array} ight]=\left[\begin{array}{ll}0 & 0 \1 & 0 \0 & 0 \0 & 1\end{array} ight]\left[\begin{array}{r}69.27 \-69.27\end{array} ight]=\left[\begin{array}{c}0 \69.27 \0 \-69.27\end{array} ight] k[/latex]

Member 3

[latex] v _{3}=\left[\begin{array}{l}v_{1} \v_{2} \v_{3} \v_{4}\end{array} ight] \begin{array}{l}0 \0 \1 \2\end{array}=\left[\begin{array}{c}0 \0 \d_{1} \d_2\end{array} ight]=\left[\begin{array}{c}0 \0 \0.0434 \-0.0637\end{array} ight] ext { in }[/latex]

u = Tv

[latex]u _{3}=\left[\begin{array}{l}u_{1} \u_{2}\end{array} ight]=\left[\begin{array}{llll}1 & 0 & 0 & 0 \0 & 0 & 1 & 0\end{array} ight]\left[\begin{array}{l}0 \0 \0.0434 \-0.0637\end{array} ight]=\left[\begin{array}{l}0 \0.0434\end{array} ight] in[/latex]

Q = ku

[latex]Q _{3}=\left[\begin{array}{l}Q_{1} \Q_{2}\end{array} ight]=1,450\left[\begin{array}{rr}1 & -1 \-1 & 1\end{array} ight]\left[\begin{array}{l}0 \0.0434\end{array} ight]=\left[\begin{array}{r}-62.93 \62.93\end{array} ight] k[/latex]

Thus, the axial force in member 3 is (Fig. 17.11(d))

62.93 k (T)

[latex]F = T ^{T} Q[/latex]

[latex]F _{3}=\left[\begin{array}{l}F_{1} \F_{2} \F_{3} \F_{4}\end{array} ight]=\left[\begin{array}{ll}1 & 0 \0 & 0 \0 & 1 \0& 0\end{array} ight]\left[\begin{array}{r}-62.93 \62.93\end{array} ight]=\left[\begin{array}{c}-62.93 \0 \62.93 \0\end{array} ight] k[/latex]

Support Reactions. As shown in Fig. 17.11(e), the reactions at the support joints 1, 2, and 4 are equal to the forces in global coordinates at the ends of the members connected to these joints.

Equilibrium Check. Applying the equations of equilibrium to the free body of the entire structure (Fig. 17.11(e)), we obtain

[latex]+ ightarrow \sum F_{X}=0 \quad 13-62.93+100 \cos 60^{\circ}=0.07 \approx 0[/latex] Checks

[latex]+\uparrow \sum F_{Y}=0 \quad 17.33+69.27-100 \sin 60^{\circ}=0[/latex] Checks

[latex]+\circlearrowleft \sum M_{(1)}=0 \quad 69.27(15)+62.93(20)-100 \cos 60^{\circ}(20)-100 \sin 60^{\circ}(15)[/latex]

[latex]=-1.39 \approx 0[/latex] Checks

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