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.

The connection to the concrete support at C will actually deform as the truss tries to lift up due to the live load. An engineer who is experienced in connection design predicts that the lift under the full live load will be 2 mm. Our job is to determine the effect of that lift on the vertical displacement of A. 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 lift at C. We can superimpose the results.

Accepted Answer

Estimated Solution:
Because the reactions are determinate, the truss will not change shape as the support at C rises. Therefore, we can use similar triangles to predict the vertical displacement at A.

[latex]\frac{\Delta_{A y}}{3.5 \mathrm{~m}} = \frac{2 \mathrm{~mm}}{3 \mathrm{~m}}[/latex]

[latex]\Delta_{A y} = 2.33 \mathrm{~mm} \quad(+\downarrow)[/latex]

Since the truss does not change shape, the elements do not get longer or shorter. That means there are no internal forces.

Since the reactions are determinate and there is no applied load, the reactions are all zero.

Direct Stiffness Method:
Assemble [latex]Q_f[/latex] and [latex]D_S[/latex]:
Since we are considering only the support deformation, there are no applied loads:

[latex][Q_f] = \begin{matrix} 1 \ 2 \ 3 \ 7 \ 8 \end{matrix} \left[\begin{array}{l}0 \0 \0 \0 \0\end{array} ight][/latex]

The only support deformation is at DOF 6:

[latex][D_S] = \begin{matrix} 4 \ 5 \ 6 \end{matrix}\left[\begin{array}{c}0 \0 \2 \mathrm{~mm}\end{array} ight][/latex]

Find the deformations:

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

[latex]\begin{matrix} 1 \ 2 \ 3 \ 7 \ 8 \end{matrix}\left[\begin{array}{l}0 \0 \0 \0 \0\end{array} ight] = \begin{matrix} 1 \ 2 \ 3 \ 7 \ 8 \end{matrix}\begin{bmatrix} \overset{1}{5780} \frac{kN}{cm} & \overset{2}{806.1} \frac{kN}{cm} & \overset{3}{-4840} \frac{kN}{cm} & \overset{7}{-940.9} \frac{kN}{cm} & \overset{8}{-806.1} \frac{kN}{cm} \ 806.1 \frac{kN}{cm} & 690.9 \frac{kN}{cm} & 0 & -806.1 \frac{kN}{cm} & -690.9 \frac{kN}{cm} \ -4840 \frac{kN}{cm} & 0 & 10487 \frac{kN}{cm} & 0 & 0 \ -940.9 \frac{kN}{cm} & -806.1 \frac{kN}{cm} & 0 & 1826.5 \frac{kN}{cm} & -80.0 \frac{kN}{cm} \ -806.1 \frac{kN}{cm} & -690.9 \frac{kN}{cm} & 0 & -80.0 \frac{kN}{cm} & 7224 \frac{kN}{cm} \end{bmatrix}\left[\begin{array}{c}\Delta_{A x} \\Delta_{A y} \\Delta_{B x} \\Delta_{D x} \\Delta_{D y}\end{array} ight] + \begin{gathered}1 \2 \3 \7 \8\end{gathered}\left[\begin{array}{ccc}\overset{4}{0} & \overset{5}{0} & \overset{6}{0} \0 & 0 & 0 \0 & -5647 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 \0 & -886.1 \frac{\mathrm{kN}}{\mathrm{cm}} & 886.1 \frac{\mathrm{kN}}{\mathrm{cm}} \-5647 \frac{\mathrm{kN}}{\mathrm{cm}} & 886.1 \frac{\mathrm{kN}}{\mathrm{cm}} & -886.1 \frac{\mathrm{kN}}{\mathrm{cm}}\end{array} ight]\left[\begin{array}{c}0 \0 \0.2 \mathrm{~cm}\end{array} ight] \begin{aligned}&4 \&5 \&6\end{aligned}[/latex]

[latex]\left[\mathbf{D}_f ight] = \left[\mathbf{K}_{f f} ight]^{-1}\left(\left[\mathbf{Q}_f ight] - \left[\mathbf{K}_{f s} ight]\left[\mathbf{D}_s 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]=\begin{array}{r}1 \2 \ 3 \7 \8\end{array}\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 & 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}0 \0 \0 \177.22 \mathrm{kN} \-177.22 \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}{c}0 \mathrm{~cm} \-0.2333 \mathrm{~cm} \0 \mathrm{~cm} \-0.2000 \mathrm{~cm} \0 \mathrm{~cm}\end{array} ight]=\left[\begin{array}{c}0 \mathrm{~mm} \-2.3 \mathrm{~mm} \0 \mathrm{~mm} \-2.0 \mathrm{~mm} \0 \mathrm{~mm}\end{array} ight] \begin{aligned}&1 \&2 \&7 \&8\end{aligned}[/latex]

Find the reactions:
Evaluation of Results:
Satisfied Fundamental Principles?
We expect all the reactions to be zero, so this serves as a check.

[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}} \6 & 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} \-0.2333 \mathrm{~cm} \0 \mathrm{~cm} \-0.2000 \mathrm{~cm} \0 \mathrm{~cm}\end{array} ight]\begin{matrix} 4 \ 5 \ 6 \end{matrix} \left[\begin{array}{ccc}\overset{4}{5647} \frac{\mathrm{kN}}{\mathrm{cm}} & \overset{5}{0} & \overset{6}{0} \0 & 6533 \frac{\mathrm{kN}}{\mathrm{cm}} & -886.1 \frac{\mathrm{kN}}{\mathrm{cm}} \0 & -886.1 \frac{\mathrm{kN}}{\mathrm{cm}} & 886.1 \frac{\mathrm{kN}}{\mathrm{cm}}\end{array} ight]\left[\begin{array}{c}0 \0 \0.2 \mathrm{~cm}\end{array} ight][/latex]

[latex]\left[\begin{array}{l}B_y \C_x \C_y\end{array} ight]=\left[\begin{array}{c}0 \mathrm{kN} \177.22 \mathrm{kN} \-177.22 \mathrm{kN}\end{array} ight]+\left[\begin{array}{c}0 \mathrm{kN} \-177.22 \mathrm{kN} \177.22 \mathrm{kN}\end{array} ight]=\left[\begin{array}{l}0 \0 \0\end{array} ight][/latex] ✓

Find the internal force:
Evaluation of Results:
Satisfied Fundamental Principles?
We expect all the internal forces to be zero, so this too serves as a check.
Let's check just two members.
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 \-0.2333 \mathrm{~cm} \-0.2000 \mathrm{~cm} \0\end{array} ight]\&=1631.2 \frac{\mathrm{kN}}{\mathrm{cm}}(0+0.15183 \mathrm{~cm}-0.15186 \mathrm{~cm}+0)\&=-0.05 \mathrm{kN} \approx 0\end{aligned}[/latex] ✓

Since there are no applied loads, it is not obvious what a sufficiently small number is. However, we generated force magnitudes when we calculated the reactions: 177 kN. When we compare with those magnitudes, 0.05 kN is essentially zero.

Member CD:
Note that we include the support displacement.

[latex]\begin{aligned}q^{\prime C D}_F &=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 \0.2000 \mathrm{~cm} \-0.2000 \mathrm{~cm} \0\end{array} ight] \&=1772.3 \mathrm{kN} / \mathrm{cm}(0-0.14142 \mathrm{~cm}+0.14142 \mathrm{~cm}+0) \&=0\end{aligned}[/latex] ✓

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

Similar triangles: [latex]\Delta_{A y}=-2.33 \mathrm{~mm}(+↑ )[/latex]

Direct Stiffness Method: [latex]\Delta_{A y} = -2.33 \mathrm{~mm} \quad(+\uparrow)[/latex]

The predicted vertical displacements at A are the same to at least three significant
figures. ✓

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

Combine results:
Since we assume the structure remains linear elastic, we can superimpose the results for live load and support movement:

[latex]\left[\begin{array}{c}\Delta_{A x} \\Delta_{A y} \\Delta_{B x} \\Delta_{D x} \\Delta_{D y}\end{array} ight]=\left[\begin{array}{c}0.1343 \mathrm{~cm} \-1.3011 \mathrm{~cm} \0.0620 \mathrm{~cm} \-0.5101 \mathrm{~cm} \-0.1151 \mathrm{~cm}\end{array} ight]+\left[\begin{array}{c}0 \mathrm{~cm} \-0.2333 \mathrm{~cm} \0 \mathrm{~cm} \-0.2000 \mathrm{~cm} \0 \mathrm{~cm}\end{array} ight]=\left[\begin{array}{c}0.1343 \mathrm{~cm} \-1.5344 \mathrm{~cm} \0.0620 \mathrm{~cm} \-0.7101 \mathrm{~cm} \-0.1151 \mathrm{~cm}\end{array} ight]\begin{aligned}&1 \&2 \&3 \&7 \&8\end{aligned}[/latex]

[latex]\left[\begin{array}{l}B_y \C_x \C_y\end{array} ight]=\left[\begin{array}{c}650 \mathrm{kN} \-0.1 \mathrm{kN} \-350 \mathrm{kN}\end{array} ight]+\left[\begin{array}{l}0 \0 \0\end{array} ight]=\left[\begin{array}{c}650 \mathrm{kN} \-0.1 \mathrm{kN} \-350 \mathrm{kN}\end{array} ight] \begin{gathered}4 \5 \6\end{gathered}[/latex]

Notice that when we consider the full live load and the lift at the support, the net vertical displacement at A is larger than the allowable displacement:

[latex]\Delta_{A y}^{ ext {Total }} = 15.3 \mathrm{~mm}>\Delta_{A y}^{ ext {Allowable }} = 15 \mathrm{~mm}[/latex]

Therefore, our team will need to redesign the truss to reduce the anticipated displacement. The Virtual Work Method we discussed in Section 5.3 is a great tool to help in that process.

Direct Stiffness Method with Combined Loading:
Our supervisor has challenged us to verify that superposition does produce the same results as considering both the live load and support lift simultaneously.

Find the deformations:

[latex]\begin{aligned}&1 \&2 \&3 \&7 \&8\end{aligned}\left[\begin{array}{c}0 \-300 k N \0 \0 \0\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 & 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]\left[\begin{array}{l}\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}{ccc}\overset{4}{0} & \overset{5}{0} & \overset{6}{0} \0 & 0 & 0 \0 & -5647 \frac{\mathrm{kN}}{\mathrm{cm}} & 0 \0&-886.1 \frac{\mathrm{kN}}{\mathrm{cm}}&886.1 \frac{\mathrm{kN}}{\mathrm{cm}}\-5647 \frac{\mathrm{kN}}{\mathrm{cm}} & 886.1 \frac{\mathrm{kN}}{\mathrm{cm}} & -886.1 \frac{\mathrm{kN}}{\mathrm{cm}}\end{array} ight]\left[\begin{array}{c}0 \0 \0.2 \mathrm{~cm}\end{array} ight]\begin{aligned}&4 \&5 \&6\end{aligned}[/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}}{\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{c}} & 0 & -806.1 \frac{\mathrm{kN}}{\mathrm{c}} & -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 \-300 \mathrm{kN} \0 \0 \0\end{array} ight]-\left[\begin{array}{c}0 \0 \0 \177.22 \mathrm{kN} \-177.22 \mathrm{kN}\end{array} 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]=\left[\begin{array}{r}0.1342 \mathrm{~cm} \-1.5345 \mathrm{~cm} \0.0619 \mathrm{~cm} \-0.7102 \mathrm{~cm} \-0.1151 \mathrm{~cm}\end{array} ight]=\left[\begin{array}{r}1.3 \mathrm{~mm} \-15.3 \mathrm{~mm} \0.6 \mathrm{~mm} \-7.1 \mathrm{~mm} \-1.2 \mathrm{~mm}\end{array} ight]\begin{aligned}&1 \&2 \&3 \&7 \&8\end{aligned}[/latex]

Find the reactions:

[latex]\left[\begin{array}{l}B_y \C_x \C_y\end{array} ight] = \begin{aligned}&4 \&5 \&6\end{aligned}\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}0.1342 \mathrm{~cm} \-1.5345 \mathrm{~cm} \0.0619 \mathrm{~cm} \-0.7102 \mathrm{~cm} \-0.1151 \mathrm{~cm}\end{array} ight] + \begin{aligned}&4 \&5 \&6\end{aligned}\left[\begin{array}{ccc}\overset{4}{5647} \frac{\mathrm{kN}}{\mathrm{cm}} & \overset{5}{0} & \overset{6}{0} \0 & 6533 \frac{\mathrm{kN}}{\mathrm{cm}} & -886.1 \frac{\mathrm{kN}}{\mathrm{cm}} \0 & -886.1 \frac{\mathrm{kN}}{\mathrm{cm}} & 886.1 \frac{\mathrm{kN}}{\mathrm{cm}}\end{array} ight]\left[\begin{array}{c}0 \0 \0.2 \mathrm{~cm}\end{array} ight][/latex]

[latex]\left[\begin{array}{l}B_y \C_x \C_y\end{array} ight]=\left[\begin{array}{c}650.0 \mathrm{kN} \177.77 \mathrm{kN} \-527.3 \mathrm{kN}\end{array} ight]+\left[\begin{array}{c}0 \mathrm{kN} \-177.22 \mathrm{kN} \177.22 \mathrm{kN}\end{array} ight]=\left[\begin{array}{c}650.0 \mathrm{kN} \0.55 \mathrm{kN} \-350.1 \mathrm{kN}\end{array} ight] \begin{aligned}&4 \&5 \&6\end{aligned}[/latex]

Evaluation of Results:
Satisfied Fundamental Principles?
Indeed, the results from superposition match the results from considering both loading cases simultaneously. ✓

Question 14.4: Our firm has designed a large pipeline along the edge of a c...

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