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.

In addition to the live load in the pipe, the truss might experience a temperature change in only the diagonal members as the sun heats it up in the late morning. A reasonable temperature increase to plan for is 10°C. To minimize the possibility of leaks in the pipeline, we need to make sure that the truss does not deflect more than 15 mm when the pipeline changes from empty to full and the temperature increases.

We found the vertical displacement due to the live load in Example 14.3.Our supervisor wants us to find the effect of just the temperature change. We can superimpose these results with the live load results to predict the total effect and then evaluate the adequacy of the design. The coefficient of thermal expansion for steel is [latex]11.7 × 10^{-6}/°C[/latex].

Accepted Answer

Estimated Solution.:
Because the truss is determinate, the temperature change will not cause reactions or internal forces.Increasing the temperature of members AD and CD will cause them to get longer. Considering the geometry of the truss and the support conditions, both lengthening members will cause node A to displace down.Direct Stiffness Method:
Nodal equivalent loads:
Member AD -temperature change:
From Example 14.1,

[latex]\cos heta = 0.7593[/latex]

[latex]\sin heta = 0.6508[/latex]

[latex]A = 37.6 \mathrm{~cm}^2[/latex]

Nodal loads in local coordinates:

[latex]\begin{aligned}q_{N_0}^{\prime} &= +\alpha \Delta T A E \&= \left(11.7 imes 10^{-6} /{ }^{\circ} \mathrm{C} ight)\left(+10^{\circ} \mathrm{C} ight)\left(37.6 \mathrm{~cm}^2 ight)\left(20,000 \mathrm{kN} / \mathrm{cm}^2 ight) \&= 87.98 \mathrm{kN}\end{aligned}[/latex]

[latex]q_{F \mathrm{o}}^{\prime} = -87.98 \mathrm{kN}[/latex]

Nodal loads in global coordinates:

[latex]\begin{aligned}{\left[\begin{array}{l}q_{ ext {Nxo }} \q_{ ext {Nyo }} \q_{ ext {Fxo }} \q_{ ext {Fyo }}\end{array} ight]=} & {\left[\begin{array}{cc}\cos heta & 0 \\sin heta & 0 \0 & \cos heta \0 & \sin heta\end{array} ight]\left[\begin{array}{c}q_{N x o}^{\prime} \q_{F x o}^{\prime}\end{array} ight] } \=& {\left[\begin{array}{cc}0.7593 & 0 \0.6508 & 0 \0 & 0.7593 \0 & 0.6508\end{array} ight]\left[\begin{array}{r}87.98 \mathrm{kN} \-87.98 \mathrm{kN}\end{array} ight] } \=& {\left[\begin{array}{c}66.80 \mathrm{kN} \57.26 \mathrm{kN} \-66.80 \mathrm{kN} \-57.26 \mathrm{kN}\end{array} ight] \begin{array}{l}1 \ 2 \7 \ 8\end{array} }\end{aligned}[/latex]

Member CD -temperature change:
From Example 14.1,

[latex]\cos heta = -0.7071[/latex]

[latex]\sin heta = 0.7071[/latex]

[latex]A = 37.6 \mathrm{~cm}^2[/latex]

Nodal loads in local coordinates:

[latex]\begin{aligned}q_{N o}^{\prime} &=\left(11.7 imes 10^{-6} /{ }^{\circ} \mathrm{C} ight)\left(10^{\circ} \mathrm{C} ight)\left(37.6\mathrm{~cm}^2 ight)\left(20,000 \mathrm{kN} / \mathrm{cm}^2 ight) \&=87.98 \mathrm{kN}\end{aligned}[/latex]

[latex]q_{F 0}^{\prime} = -87.98 \mathrm{kN}[/latex]

Nodal loads in global coordinates:

[latex]\begin{aligned}{\left[\begin{array}{l}q_{N x \mathrm{O}} \q_{N y \mathrm{O}} \q_{ ext {Fxo }} \q_{ ext {Fyo }}\end{array} ight]=} & {\left[\begin{array}{cc}-0.7071 & 0 \0.7071 & 0 \0 & -0.7071 \0 & 0.7071\end{array} ight]\left[\begin{array}{r}87.98 \mathrm{kN} \-87.98 \mathrm{kN}\end{array} ight] } \=& {\left[\begin{array}{c}-62.21 \mathrm{kN} \62.21 \mathrm{kN} \62.21 \mathrm{kN} \-62.21 \mathrm{kN}\end{array} ight] \begin{array}{l}5 \ 6 \ 7 \8\end{array} }\end{aligned}[/latex]

Net nodal loads:

[latex]\begin{aligned}&Q_{i o} = \Sigma q_{i o} \&Q_{1o} = 66.80 \mathrm{kN} \&Q_{2o} = 57.26 \mathrm{kN} \&Q_{3o} = 0 \&Q_{4o} = 0 \&Q_{5o} = -62.21 \mathrm{kN} \&Q_{6o} = 62.21 \mathrm{kN} \&Q_{7o} = -66.80 \mathrm{kN} + 62.21 \mathrm{kN} = -4.59 \mathrm{kN} \&Q_{8o} = -57.26 \mathrm{kN} - 62.21 \mathrm{kN} = -119.47 \mathrm{kN}\end{aligned}[/latex]

[latex]\left[\mathbf{Q}_o ight] = \begin{array}{r}1 \2 \3 \4\5 \6 \7 \8\end{array}\left[\begin{array}{r}66.80 \mathrm{kN} \57.26 \mathrm{kN} \0 \0 \-62.21 \mathrm{kN} \62.21 \mathrm{kN} \-4.59 \mathrm{kN} \-119.47 \mathrm{kN}\end{array} ight][/latex]

[latex]\left[\mathbf{Q}_{f o} ight] = \begin{array}{r}1 \2 \ 3 \7 \8\end{array}\left[\begin{array}{c}66.80 \mathrm{kN} \57.26 \mathrm{kN} \0 \-4.59 \mathrm{kN} \-119.47 \mathrm{kN}\end{array} ight][/latex]

[latex][Q_{so}] = \begin{matrix} 4 \ 5 \ 6 \end{matrix} \left[\begin{array}{c}0 \-62.21 \mathrm{kN} \62.21 \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{\left[\mathbf{K}_{f s}] [\mathbf{D}_s ight]}_{0} + \left[\mathbf{Q}_{f o} ight][/latex]

[latex]\begin{aligned}&1 \&2 \&3 \&7 \&8\end{aligned}\left[\begin{array}{l}0 \0 \0 \0 \0\end{array} ight] = \begin{aligned}&1 \&2 \&3 \&7 \&8 \&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 & 1087 \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]+\begin{aligned}&1 \&2 \&3 \&7 \&8\end{aligned}\left[\begin{array}{c}66.80 \mathrm{kN} \57.26 \mathrm{kN} \0 \-4.59 \mathrm{kN} \-119.47 \mathrm{kN}\end{array} ight] [/latex]

[latex]\left[\mathbf{D}_f ight] = \left[\mathbf{K}_{f f} ight]^{-1}\left(\left[\mathbf{Q}_f ight] - \left[\mathbf{Q}_{f 0} 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}{rrrrrr}\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 & 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}{l}0 \0 \0 \0 \0\end{array} ight] - \left[\begin{array}{c}66.80 \mathrm{kN} \57.26 \mathrm{kN} \0 \-4.59 \mathrm{kN} \-119.47 \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}{l}-0.0000 \mathrm{~cm} \-0.16481 \mathrm{~cm} \-0.0000 \mathrm{~cm} \-0.07023 \mathrm{~cm} \-0.0000 \mathrm{~cm}\end{array} ight] = \left[\begin{array}{c}0.00 \mathrm{~mm} \-1.6 \mathrm{~mm} \0.00 \mathrm{~mm} \-0.70 \mathrm{~mm} \0.00 \mathrm{~mm}\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}] [\mathbf{D}_s ight]}_{0} +\left[\mathbf{Q}_{s o} ight][/latex]

[latex]\left[\begin{array}{c}B_y \C_x \C_y\end{array} ight] = \begin{matrix} 4 \ 5 \ 6 \end{matrix}\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}{c}0 \mathrm{~cm} \-0.16481 \mathrm{~cm} \0 \mathrm{~cm} \-0.07023 \mathrm{~cm} \0 \mathrm{~cm}\end{array} ight] + \left[\begin{array}{c}0 \-62.21 \mathrm{kN} \62.21 \mathrm{kN}\end{array} ight] = \left[\begin{array}{c}0 + 0 \62.23 \mathrm{kN} - 62.21 \mathrm{kN} \-62.23 \mathrm{kN} + 62.21 \mathrm{kN}\end{array} ight] = \left[\begin{array}{c}0 \0.02 \mathrm{kN} \-0.02 \mathrm{kN}\end{array} ight]\begin{matrix} 4 \ 5 \ 6 \end{matrix} [/latex]

When compared with the components, these values are essentially zero.

[latex]\left[\begin{array}{l}B_y \C_x \C_y\end{array} ight] = \left[\begin{array}{l}0 \0 \0\end{array} ight] \begin{aligned}&4 \&5 \ &6\end{aligned}[/latex]

Find the internal forces:
Member AB:

[latex]q_F^{\prime A B} = 4840 \frac{\mathrm{kN}}{\mathrm{cm}}\left[\begin{array}{llll}-1 & 0 & 1 & 0\end{array} ight]\left[\begin{array}{c}0 \-0.16481 \mathrm{~cm} \0 \0\end{array} ight] = 0[/latex]

Member AD:

[latex]\begin{aligned}q_F^{\prime A D} = 1631.2 \frac{\mathrm{kN}}{\mathrm{cm}}\left[\begin{array}{llll}-0.7593 & -0.6508 & 0.7593 & 0.6508\end{array} ight]\left[\begin{array}{c}0 \-0.16481 \mathrm{~cm} \-0.07023 \mathrm{~cm} \0\end{array} ight] \+(-87.98 \mathrm{kN}) = 0\end{aligned}[/latex]

Member BC:

[latex]q_F^{\prime B C}=5647 \frac{\mathrm{kN}}{\mathrm{cm}}\left[\begin{array}{llll}-1 & 0 & 1 & 0\end{array} ight]\left[\begin{array}{l}0 \0 \0 \0\end{array} ight] = 0[/latex]

Member BD:

[latex]q_F^{\prime B D} = 5647 \frac{\mathrm{kN}}{\mathrm{cm}}\left[\begin{array}{llll}0 & -1 & 0 & 1\end{array} ight]\left[\begin{array}{c}0 \0 \-0.07023 \mathrm{~cm} \0\end{array} ight] = 0[/latex]

Member CD:

[latex]\begin{aligned}q_F^{\prime C D} = 1772.3 \frac{\mathrm{kN}}{\mathrm{cm}}\left[\begin{array}{llll}0.7071 & -0.7071 & -0.7071 & 0.7071\end{array} ight]\left[\begin{array}{c}0 \0 \-0.07023 \mathrm{~cm} \0\end{array} ight] \+(-87.98 \mathrm{kN}) = 0.03 \mathrm{kN}\end{aligned}[/latex]

Evaluation of Results:
Satisfied Fundamental Principles?
The reactions and internal forces are zero, as expected because the truss
is determinate.
The vertical displacement at A is downward, as expected:

[latex]\Delta_{A y} = -1.6 \mathrm{~mm} \quad(+\uparrow)[/latex]

Convulsion:
These results plus the checks on the reactions and internal forces strongly suggest that we have reasonable results.

Evaluation of Design:
The vertical displacement will not push the live load displacement beyond the limit.

[latex]\Delta_{A y}^{\mathrm{Temp}} = -1.6 \mathrm{~mm} \quad(+\uparrow)[/latex]

[latex]\Delta_{A y}^{ ext {Total }} = -1.6 \mathrm{~mm} - 13.0 \mathrm{~mm} = -14.6 \mathrm{~mm} \quad(+\uparrow)[/latex]

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