Our firm has designed a large pipeline along the edge of a canal in Singapore. Our team has been tasked with designing a structure to support the pipe. Other engineers on the team have designed the truss members for strength. Because we are new to the firm, our supervisor wants to verify our

Question

Our firm has designed a large pipeline along the edge of a canal in Singapore. Our team has been tasked with designing a structure to support the pipe. Other engineers on the team have designed the truss members for strength.

Because we are new to the firm, our supervisor wants to verify our knowledge of the Direct Stiffness Method. A determinate truss is a great way to do that.

Our final test from the supervisor is to analyze the truss to find the reactions due to just the self-weight of the truss. We are not concerned with the vertical displacement at A because the pipe will be attached after the truss displaces under its self-weight.

Member properties:

Diagonals:	Double angle	120 × 120 × 8	[latex]w_o = 0.288 kN/m [/latex]
All others:	Universal columns	250 × 250 × 66.5	[latex]w_o = 0.652 kN/m [/latex]

Accepted Answer

Loading:
We can convert the distributed load to loads at the nodes by taking the total load on each member and applying half to each end.Member AB:

[latex]\begin{aligned}l_{A B} & = 3.50 \mathrm{~m} \W_{A B} & = 3.500 \mathrm{~m}(0.652 \mathrm{kN} / \mathrm{m}) = 2.282 \mathrm{kN} \P_{A y}^{A B} & = 2.282 \mathrm{kN} / 2 = 1.141 \mathrm{kN} \quad(+\downarrow) \P_{B y}^{A B} & = 1.141 \mathrm{kN} \quad(+\downarrow)\end{aligned}[/latex]

Member AD:

[latex]\begin{aligned}&l_{A B} = 4.610 \mathrm{~m} \&W_{A D} = 4.610 \mathrm{~m}(0.288 \mathrm{kN} / \mathrm{m}) = 1.328 \mathrm{kN} \&P_{A y}^{A D} = 1.328 \mathrm{kN} / 2 = 0.664 \mathrm{kN} \quad(+\downarrow) \&P_{D y}^{A D} = 0.664 \mathrm{kN} \quad(+\downarrow)\end{aligned}[/latex]

Member BC:

[latex]\begin{aligned}&l_{B C} = 3.000 \mathrm{~m} \&W_{B C} = 3.000 \mathrm{~m}(0.652 \mathrm{kN} / \mathrm{m}) = 1.956 \mathrm{kN} \&P_{B y}^{B C} = 1.956 \mathrm{kN} / 2 = 0.978 \mathrm{kN}\quad(+\downarrow) \&P_{C y}^{B C} = 0.978 \mathrm{kN} \quad(+\downarrow)\end{aligned}[/latex]

Member BD:

[latex]\begin{aligned}&l_{B D} = 3.000 \mathrm{~m} \&W_{B D} = 3.000 \mathrm{~m}(0.652 \mathrm{kN} / \mathrm{m}) = 1.956 \mathrm{kN} \&P_{B y}^{B D} = 1.956 \mathrm{kN} / 2 = 0.978 \mathrm{kN}\quad(+\downarrow) \&P_{D y}^{B D} = 0.978 \mathrm{kN} \quad(+\downarrow)\end{aligned}[/latex]

Member CD:

[latex]\begin{aligned}&l_{C D} = 4.243 \mathrm{~m} \&W_{C D} = 4.243 \mathrm{~m}(0.288 \mathrm{kN} / \mathrm{m}) = 1.222 \mathrm{kN} \&P_{C y}^{C D} = 1.222 \mathrm{kN} / 2 = 0.611 \mathrm{kN}\quad(+\downarrow) \&P_{D y}^{C D} = 0.611 \mathrm{kN} \quad(+\downarrow)\end{aligned}[/latex]

Net forces on nodes:

[latex]\begin{aligned}&P_{A y} = 1.141 \mathrm{kN} + 0.664 \mathrm{kN} = 1.805 \mathrm{kN} \quad(+\downarrow) \&P_{B y} = 1.141 \mathrm{kN} + 0.978 \mathrm{kN} + 0.978 \mathrm{kN} = 3.097 \mathrm{kN} \quad(+\downarrow) \&P_{C y} = 0.978 \mathrm{kN} + 0.611 \mathrm{kN} = 1.589 \mathrm{kN} \quad(+\downarrow) \&P_{D y} = 0.664 \mathrm{kN} + 0.978 \mathrm{kN} + 0.611 \mathrm{kN} = 2.253 \mathrm{kN} \quad(+\downarrow)\end{aligned}[/latex]

Predicted Solution:

Apply equilibrium:

[latex]+\circlearrowleft \Sigma M_C = 0 = 1.805 \mathrm{kN}(6.5 \mathrm{~m}) + 2.253 \mathrm{kN}(3 \mathrm{~m}) + 3.097 \mathrm{kN}(3 \mathrm{~m}) - B_y(3 \mathrm{~m})[/latex]

[latex]B_y = 9.261 \mathrm{kN} \quad(+\uparrow)[/latex]

[latex]+\uparrow \Sigma F_y = 0 = -1.805 \mathrm{kN} - 2.253 \mathrm{kN} - 3.097 \mathrm{kN} + 9.261 \mathrm{kN} - 1.589 \mathrm{kN} + C_y[/latex]

[latex]C_y = -0.517 \mathrm{kN} \quad(+\uparrow)[/latex]

[latex]\stackrel{+}{ ightarrow} \Sigma F_x = 0 = C_x[/latex]

[latex]C_x = 0[/latex]

Direct Stiffness Method:
Assemble [latex]Q_f[/latex] and [latex]Q_{so}[/latex]:

[latex][Q_f] = \begin{matrix} 1 \ 2 \ 3 \ 7 \ 8 \end{matrix} \left[\begin{array}{c}0 \-1.805 \mathrm{kN} \0 \0 \-2.253 \mathrm{kN}\end{array} ight][/latex]

Since there are loads applied directly to the supported DOFs, we reverse the signs before adding them to [latex][Q_{so}][/latex]:

[latex][Q_{so}] = \begin{matrix} 4 \ 5 \ 6 \end{matrix} \left[\begin{array}{c}+3.097 \mathrm{kN} \0 \+1.589 \mathrm{kN}\end{array} ight][/latex]

Find the deformations:

[latex]\left[\mathbf{Q}_f ight] = \left[\mathbf{K}_{f f} ight]\left[\mathbf{D}_f ight] + \underbrace{ [\mathbf{K}_{f s}] [\mathbf{D}_s]}_{0}[/latex]

[latex]\begin{aligned}&1 \&2 \&3 \&7 \&8\end{aligned}\left[\begin{array}{c}0 \-1.805 \mathrm{kN} \0 \0 \-2.253 \mathrm{kN}\end{array} ight]=\begin{aligned}&1 \&2 \&3 \&7 \&8\end{aligned}\left[\begin{array}{ccccc}\overset{1}{5780 \frac{\mathrm{kN}}{\mathrm{cm}}} & \overset{2}{806.1 \frac{\mathrm{kN}}{\mathrm{cm}}} & \overset{3}{-4840 \frac{\mathrm{kN}}{\mathrm{cm}}} & \overset{7}{-940.9 \frac{\mathrm{kN}}{\mathrm{cm}}} & \overset{8}{-806.1 \frac{\mathrm{kN}}{\mathrm{cm}}} \806.1 \frac{\mathrm{kN}}{\mathrm{cm}} & 690.9 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 & -806.1 \frac{\mathrm{kN}}{\mathrm{cm}} & -690.9 \frac{\mathrm{kN}}{\mathrm{cm}} \-4840 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 & 10847 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 & 0 \-940.9 \frac{\mathrm{kN}}{\mathrm{cm}} & -806.1 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 & 1826.5 \frac{\mathrm{kN}}{\mathrm{cm}} & -80.0 \frac{\mathrm{kN}}{\mathrm{cm}} \-806.1 \frac{\mathrm{kN}}{\mathrm{cm}} & -690.9 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 & -80.0 \frac{\mathrm{kN}}{\mathrm{cm}} & 7224 \frac{\mathrm{kN}}{\mathrm{cm}}\end{array} ight]\left[\begin{array}{c}\Delta_{A x} \\Delta_{A y} \\Delta_{B x} \\Delta_{D x} \\Delta_{D y}\end{array} ight][/latex]

[latex]\left[\mathbf{D}_f ight] = \left[\mathbf{K}_{f f} ight]^{-1}\left(\left[\mathbf{Q}_f ight] ight)[/latex]

[latex]\left[\begin{array}{c}\Delta_{A x} \\Delta_{A y} \\Delta_{B x} \\Delta_{D x} \\Delta_{D y}\end{array} ight] =\begin{aligned}&1 \&2 \&3 \&7 \&8\end{aligned}\left[\begin{array}{ccccc}\overset{1}{5780 \frac{\mathrm{kN}}{cm}} & \overset{2}{806.1 \frac{\mathrm{kN}}{\mathrm{cm}}} & \overset{3}{-4840 \frac{\mathrm{kN}}{\mathrm{cm}}} & \overset{7}{-940.9 \frac{\mathrm{kN}}{\mathrm{cm}}} & \overset{8}{-806.1 \frac{\mathrm{kN}}{\mathrm{cm}}} \806.1 \frac{\mathrm{kN}}{\mathrm{cm}} & 690.9 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 & -806.1 \frac{\mathrm{kN}}{\mathrm{cm}} & -690.9 \frac{\mathrm{kN}}{\mathrm{cm}} \-4840 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 & 10487 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 & 0 \-940.9 \frac{\mathrm{kN}}{\mathrm{cm}} & -806.1 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 & 1826.5 \frac{\mathrm{kN}}{\mathrm{cm}} & -80.0 \frac{\mathrm{kN}}{\mathrm{cm}} \-806.1 \frac{\mathrm{kN}}{\mathrm{cm}} & -690.9 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 & -80.0 \frac{\mathrm{kN}}{\mathrm{cm}} & 7224 \frac{\mathrm{kN}}{\mathrm{cm}}\end{array} ight]^{-1}\left(\left[\begin{array}{c}0 \-1.805 \mathrm{kN} \0 \0 \-2.253 \mathrm{kN}\end{array} ight] ight)[/latex]

[latex]\left[\begin{array}{l}\Delta_{A x} \\Delta_{A y} \\Delta_{B x} \\Delta_{D x} \\Delta_{D y}\end{array} ight] = \left[\begin{array}{r}8.077 imes 10^{-4} \mathrm{~cm} \-8.693 imes 10^{-3} \mathrm{~cm} \3.728 imes 10^{-4} \mathrm{~cm} \-3.468 imes 10^{-3} \mathrm{~cm} \-1.0916 imes 10^{-3} \mathrm{~cm}\end{array} ight] \begin{aligned}&1 \&2 \&3 \&7 \&8\end{aligned}[/latex]

Find the reactions:

[latex]\left[\mathbf{Q}_s ight] = \left[\mathbf{K}_{s f} ight]\left[\mathbf{D}_f ight] + \underbrace{\left[\mathbf{K}_{s s} ight] [\mathbf{D}_s]}_{0} + [\mathbf{Q}_{s o}][/latex]

[latex]\left[\begin{array}{c}B_y \C_x \C_y\end{array} ight]=5\left[\begin{array}{ccccc}\overset{1}{0} & \overset{2}{0} & \overset{3}{0} & \overset{7}{0} & \overset{8}{-5647 \frac{\mathrm{kN}}{\mathrm{cm}}} \0 & 0 & -5647 \frac{\mathrm{kN}}{\mathrm{cm}} & -886.1 \frac{\mathrm{kN}}{\mathrm{cm}} & 886.1 \frac{\mathrm{kN}}{\mathrm{cm}} \0 & 0 & 0 & 886.1 \frac{\mathrm{kN}}{\mathrm{cm}} & -886.1 \frac{\mathrm{kN}}{\mathrm{cm}}\end{array} ight]\left[\begin{array}{r}8.077 imes 10^{-4} \mathrm{~cm} \-8.693 imes 10^{-3} \mathrm{~cm} \3.728 imes 10^{-4} \mathrm{~cm} \-3.468 imes 10^{-3} \mathrm{~cm} \-1.0916 imes 10^{-3} \mathrm{~cm}\end{array} ight]+\left[\begin{array}{c}+3.097 \mathrm{kN} \0 \+1.589 \mathrm{kN}\end{array} ight][/latex]

[latex]=\left[\begin{array}{c}6.164 \mathrm{kN} + 3.097 \mathrm{kN} \0.001 \mathrm{kN} - 0 \-2.106 \mathrm{kN} + 1.589 \mathrm{kN}\end{array} ight] = \left[\begin{array}{r}9.261 \mathrm{kN} \0.001 \mathrm{kN} \-0.517 \mathrm{kN}\end{array} ight] \begin{gathered}4 \5 \6\end{gathered}[/latex]

Evaluation of Results:
Satisfied Fundamental Principles?
Compare the results from equilibrium with the results from the Direct Stiffness Method.

The results are identical to at least three significant figures. ✓
The horizontal reaction at C is within three significant figures compared with the magnitudes of the other reactions.

Conculsion:
This suggests that we have reasonable results.

Question 14.6: Our firm has designed a large pipeline along the edge of a c...

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