A school installed a lightweight frame to mark the start and finish line at the track. For a special event coming up, the organizers want to hang a single heavy piece of equipment from the frame. They contacted our firm to evaluate the frame to determine whether it is safe to hang the equipment

Question

A school installed a lightweight frame to mark the start and finish line at the track. For a special event coming up, the organizers want to hang a single heavy piece of equipment from the frame. They contacted our firm to evaluate the frame to determine whether it is safe to hang the equipment from it. After inspecting the frame in the field, our team has determined that the weakest parts are the moment connections at the corners and the connections to the foundations.

One of our concerns is a possible seismic event while the equipment is hanging from the frame. Another team member has calculated the static equivalent seismic load.

Our task is to predict the resulting moment at each of the beam-column connections and the reactions at the foundations.

Accepted Answer

Estimated Solution:
Approximate member BC as a fix-fix supported beam. If there was only vertical load, this would provide an upper bound to the moments at the ends. However, since there is also lateral load on the frame, this is only an approximation.

Use the formulas from inside the back cover:

[latex]M_{B C} = -\frac{5 \mathrm{kN}(2 \mathrm{~m})(9 \mathrm{~m})^2}{(11 \mathrm{~m})^2} = -6.69 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M_{C B} = +\frac{5 \mathrm{kN}(9 \mathrm{~m})(2 \mathrm{~m})^2}{(11 \mathrm{~m})^2} = +1.49 \mathrm{kN} \cdot \mathrm{m}[/latex]

Detailed Analysis:

Assumption:
The effect of the length change of members is negligible.

Determine the distribution factors:
Stiffness of each member:

[latex]K_{A B} = K_{B A} = \frac{4 E I}{3 \mathrm{~m}}[/latex]

[latex]K_{B C} = K_{C B} = \frac{4 E I}{11 \mathrm{~m}}[/latex]

[latex]K_{C D} = K_{D C} = \frac{4 E I}{3 \mathrm{~m}}[/latex]

Distribution factors:

[latex]D F_{A B} = 0 ( end of member at fixed support )[/latex]

[latex]D F_{B A} = \frac{4 E I / 3 \mathrm{~m}}{4 E I / 3 \mathrm{~m} + 4 E I / 11 \mathrm{~m}} = 0.786[/latex]

[latex]D F_{B C} = \frac{4 E I / 11 \mathrm{~m}}{4 E I / 3 \mathrm{~m} + 4 E I / 11 \mathrm{~m}} = 0.214[/latex]

[latex]D F_{C B} = \frac{4 E I / 11 \mathrm{~m}}{4 E I / 11 \mathrm{~m} + 4 E I / 3 \mathrm{~m}} = 0.214[/latex]

[latex]D F_{C D} = \frac{4 E I / 3 \mathrm{~m}}{4 E I / 11 \mathrm{~m} + 4 E I / 3 \mathrm{~m}} = 0.786[/latex]

[latex]D F_{D C} = 0 ( end of member at fixed support ) [/latex]

Calculate the load effects (no sidesway):
Put a horizontal roller at C to stop sidesway.
Fixed-end moments:

For the columns with no load between the nodes:

[latex]F E M_{A B} = F E M_{B A} = F E M_{C D} = F E M_{D C} = 0[/latex]

For the beam, we calculated the moments for approximate analysis:

[latex]F E M_{B C} = -6.69 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]F E M_{C B} = +1.49 \mathrm{kN} \cdot \mathrm{m}[/latex]

Distribute the moments:
Stop when the peak UM is no more than 0.015 kN•m. There is no need to release node A or D because they are fixed supports.

Table 13.12

NODE	A	B		C		D
MEMBER	AB	BA	BC	CB	CD	DC
K	4EI/3	4EI/3	4EI/11	4EI/11	4EI/3	4EI/3
DF	0	0.786	0.214	0.214	0.786	0	-UM
FEM (kN•m)			- 6.690	1.490
Release B		5.256	1.434				6.690
Carryover	2.628			0.717
Release C				-0.473	-1.734		-2.207
Carryover			-0.236			-0.867
Release B		0.186	0.051				0.236
Carryover	0.093			0.025
Release C				-0.005	-0.020		-0.025
Carryover			- 0.003			- 0.010

Total [latex]M_{NS}[/latex]	2.721	5.442	-5.444	1.754	-1.754	-0.877

Unbalanced moments:

[latex]U M_B = -0.002 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]U M_C = 0.000 \mathrm{kN} \cdot \mathrm{m}[/latex]

All are at or below our stopping threshold of 0.015 kN•m.

∴ Okay to stop

Evaluation of Results:
Satisfied Fundamental Principles?
Check equilibrium of the net moments:

[latex]\begin{aligned}\Sigma M_B &= 5.442 \mathrm{kN} \cdot \mathrm{m} - 5.444 \mathrm{kN} \cdot \mathrm{m} \&= -0.002 \mathrm{kN} \cdot \mathrm{m} \approx 0\end{aligned}[/latex] ✓

[latex]\Sigma M_C = 1.754 \mathrm{kN} \cdot \mathrm{m} - 1.754 \mathrm{kN} \cdot \mathrm{m} = 0[/latex] ✓

Find the lateral reaction:
FBD of AB:

[latex]\begin{gathered}+\circlearrowright \Sigma M_B = 0 = 2.72 \mathrm{kN} \cdot \mathrm{m} - A_x(3 \mathrm{~m}) \+ 5.44 \mathrm{kN} \cdot \mathrm{m}\end{gathered}[/latex]

[latex]A_x = 2.72 \mathrm{kN}[/latex]

FBD of CD:

[latex]\begin{aligned}& +\circlearrowright \Sigma M_C = 0 = -0.88 \mathrm{kN} \cdot \mathrm{m} - D_x(3 \mathrm{~m})\&- 1.75 \mathrm{kN} \cdot \mathrm{m}\end{aligned}[/latex]

[latex]\left.D_x = -0.88 \mathrm{kN} \quad \stackrel{+}{ ightarrow} ight)[/latex]

FBD of entire frame:

[latex]\stackrel{+}{ ightarrow} \Sigma F_x = 0 = 2.72 \mathrm{kN} + 0.5 \mathrm{kN} - C_x + (-0.88 \mathrm{kN})[/latex]

[latex]C_x = 2.34 \mathrm{kN}[/latex]

Calculate the Sidesway effects:
Apply an unknown lateral load at C.
Fixed-end moments:

For the columns, use formulas from inside the back cover:
[latex]F E M_{A B} = -\frac{6 E I \Delta^{\prime}}{l^2}[/latex]

Since we choose Δ', we can choose it to have a certain fixed-end moment:

[latex]F E M_{A B} \equiv -100 \mathrm{kN} \cdot \mathrm{m}[/latex]

Since both columns have the same E, I, and l:

[latex]F E M_{A B} = F E M_{B A} = F E M_{C D} = F E M_{D C} = -100 \mathrm{kN} \cdot \mathrm{m}[/latex]

For the beam, there is no load or differential movement:

[latex]F E M_{B C} = F E M_{C B} = 0[/latex]

Distribute the moments:
Stop when the peak UM is no more than 1 kN•m. There is still no need to release node A or D because they are fixed supports.

Table 13.13

NODE	A	B		C		D
MEMBER	AB	BA	BC	CB	CD	DC
K	4EI/3	4EI/3	4EI/11	4EI/11	4EI/3	4EI/3
DF	0	0.786	0.214	0.214	0.786	0	-UM
FEM (kN•m)	-100.0	-100.0			-100.0	-100.0
Release B		78.6	21.4				100.0
Carryover	39.3			10.7
Release C				19.1	70.2		89.3
Carryover			9.6			35.1
Release B		-7.5	-2.0				-9.6
Carryover	-3.8			-1.0

Total M^′_{S}	-64.5	-28.9	29.0	28.8	-29.8	-64.9

Unbalanced moments:

[latex]U M_B = 0.0 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]U M_C = -1.0 \mathrm{kN} \cdot \mathrm{m}[/latex]

All are at or below our stopping threshold of 1 kN•m.

∴ Okay to stop

Evaluation of Results:
Satisfied Fundamental Principles?
Check equilibrium of the net moments:

[latex]\Sigma M_B = -28.9 \mathrm{kN} \cdot \mathrm{m} + 29.0 \mathrm{kN} \cdot \mathrm{m} = 0.1 \mathrm{kN} \cdot \mathrm{m} \approx 0[/latex] ✓

[latex]\Sigma M_C = 28.8 \mathrm{kN} \cdot \mathrm{m} - 29.8 \mathrm{kN} \cdot \mathrm{m} = -1.0 \mathrm{kN} \cdot \mathrm{m} \approx 0[/latex] ✓

Find the lateral force:
FBD of AB:

[latex]\begin{aligned}+\circlearrowright \Sigma M_B = 0 =&- 64.5 \mathrm{kN} \cdot \mathrm{m} - A_x^{\prime}(3 \mathrm{~m}) \&- 28.9 \mathrm{kN} \cdot \mathrm{m}\end{aligned}[/latex]

[latex]A_x^{\prime} = -31.1 \mathrm{kN} \quad(\stackrel{+}{ ightarrow})[/latex]

FBD of CD:

[latex]\begin{aligned}&+\circlearrowright\Sigma M_C = 0 = -64.9 \mathrm{kN} \cdot \mathrm{m} - D_x^{\prime}(3 \mathrm{~m})\&- 29.8 \mathrm{kN} \cdot \mathrm{m}\end{aligned}[/latex]

[latex]D_x^{\prime} = -31.6 \mathrm{kN}[/latex]

FBD of entire frame:

[latex]\stackrel{+}{ ightarrow} \Sigma F_x = 0 = (-31.1 \mathrm{kN}) + (-31.6 \mathrm{kN}) + C_x^{\prime}[/latex]

[latex]C_x^{\prime} = 62.7 \mathrm{kN} \quad(\stackrel{+}{ ightarrow})[/latex]

Scale results:
Scale factor:

[latex]S F = \frac{2.34 \mathrm{kN}}{62.7 \mathrm{kN}} = 0.0373[/latex]

Scale moments:

[latex]\begin{aligned}&M_{S A B} = 0.0373(-64.5 \mathrm{kN} \cdot \mathrm{m}) = -2.41 \mathrm{kN} \cdot \mathrm{m} \&M_{S B A} = 0.0373(-28.9 \mathrm{kN} \cdot \mathrm{m}) = -1.08 \mathrm{kN} \cdot \mathrm{m} \&M_{S B C} = 0.0373(+29.0 \mathrm{kN} \cdot \mathrm{m}) = +1.08 \mathrm{kN} \cdot \mathrm{m} \&M_{S C B} = 0.0373(+28.8 \mathrm{kN} \cdot \mathrm{m}) = +1.07 \mathrm{kN} \cdot \mathrm{m} \&M_{S C D} = 0.0373(-29.8 \mathrm{kN} \cdot \mathrm{m}) = -1.11 \mathrm{kN} \cdot \mathrm{m} \&M_{S D C} = 0.0373(-64.9 \mathrm{kN} \cdot \mathrm{m}) = -2.42 \mathrm{kN} \cdot \mathrm{m}\end{aligned}[/latex]

Superimpose the effects:
Superimpose the moments:

[latex]M_{A B} = +2.72 \mathrm{kN} \cdot \mathrm{m} - 2.41 \mathrm{kN} \cdot \mathrm{m} = +0.31 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M_{B A} = +5.44 \mathrm{kN} \cdot \mathrm{m} - 1.08 \mathrm{kN} \cdot \mathrm{m} = +4.36 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M_{B C} = -5.44 \mathrm{kN} \cdot \mathrm{m} + 1.08 \mathrm{kN} \cdot \mathrm{m} = -4.36 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M_{C B} = +1.75 \mathrm{kN} \cdot \mathrm{m} + 1.07 \mathrm{kN} \cdot \mathrm{m} = +2.82 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M_{C D} = -1.75 \mathrm{kN} \cdot \mathrm{m} - 1.11 \mathrm{kN} \cdot \mathrm{m} = -2.86 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M_{D C} = -0.88 \mathrm{kN} \cdot \mathrm{m} - 2.42 \mathrm{kN} \cdot \mathrm{m} = -3.30 \mathrm{kN} \cdot \mathrm{m}[/latex]

Superimpose the horizontal reactions:
We need to scale the Sidesway values:

[latex]A_x = +2.72 \mathrm{kN} + 0.0373(-31.1 \mathrm{kN}) = +1.56 \mathrm{kN} \quad(\stackrel{+}{ ightarrow})[/latex]

[latex]D_x = -0.88 \mathrm{kN} + 0.0373(-31.6 \mathrm{kN}) = -2.06 \mathrm{kN} \quad(\stackrel{+}{ ightarrow})[/latex]

Other reactions:
We found the moment and horizontal reactions by superimposing the moment distribution results. Let's use equilibrium to find the vertical reactions.

FBD of BC for the original frame:
The moments are net values from distribution, not the opposite values from the column FBDs.

[latex]+\circlearrowright \Sigma M_C = 0 = -4.36 \mathrm{kN} \cdot \mathrm{m} - N_{B A}(11 \mathrm{~m}) - 5 \mathrm{kN}(9 \mathrm{~m}) + 2.86 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]N_{B A} = -4.22 \mathrm{kN}[/latex]

[latex]+\uparrow \Sigma F_y = 0 = -(-4.22 \mathrm{kN}) - 5.0 \mathrm{kN} - N_{C D}[/latex]

[latex]N_{C D} = -0.78 \mathrm{kN}[/latex]

FBD of AB for the original frame:
The moment is the net value from distribution, not the opposite value from the beam FBD.

[latex]+\uparrow \Sigma F_y = 0 = A_y + (-4.22 \mathrm{kN})[/latex]

[latex]A_y = 4.22 \mathrm{kN} \quad(+\uparrow)[/latex]

FBD of CD for the original frame:
The moment is the net value from distribution, not the opposite value from the beam FBD.

[latex]+\uparrow \Sigma F_y = 0 = D_y + (-0.78 \mathrm{kN})[/latex]

[latex]D_y = 0.78 \mathrm{kN}[/latex]

Summary:

Evaluation of Results:

Satisfied Fundamental Principles?

Check equilibrium:
Note that the internal moments at the beam-column connections are shown but do not factor into the equilibrium calculations of the whole frame.

[latex]\stackrel{+}{ ightarrow} \Sigma F_x = 1.56 \mathrm{kN} + 0.5 \mathrm{kN} - 2.06 \mathrm{kN} = 0[/latex] ✓

[latex]+\uparrow \Sigma F_y = 4.22 \mathrm{kN} - 5.0 \mathrm{kN} + 0.78 \mathrm{kN} = 0[/latex] ✓

[latex]\begin{aligned}+\circlearrowright \Sigma M_A =& 0.31 \mathrm{kN} \cdot \mathrm{m} + 5.0 \mathrm{kN}(2 \mathrm{~m}) + 0.5 \mathrm{kN}(3 \mathrm{~m}) \&- 3.30 \mathrm{kN} \cdot \mathrm{m} - 0.78 \mathrm{kN}(11 \mathrm{~m}) \=&- 0.07 \mathrm{kN} \cdot \mathrm{m} \approx 0\end{aligned}[/latex] ✓

Approximation Predicted Outcomes?

Our approximate prediction of the moment at BC is about 50% larger than our detailed result, but our approximate prediction of the moment at CB is about 50% of our detailed result. This suggests that our results are the right order of magnitude. ✓

Conclusion:
Our reactions satisfy equilibrium.
Since our reactions are based on the calculation of the internal moments, this suggests we have reasonable results.

Comparison with computer aided analysis results:
If we assume the size of the pipes, we can use their individual areas to calculate the cross-sectional properties for the frame. If we use the following net properties for the frame members, we get the computer aided analysis results in the table:

[latex]A = 20 \mathrm{~cm}^2 \quad I = 6300 \mathrm{~cm}^4 \quad E = 200 \mathrm{GPa} = 20,000 \mathrm{kN} / \mathrm{cm}^2[/latex]

Reaction	Moment Distribution	Computer Aided	Difference
[latex]A_x[/latex]	1.56 kN	1.50 kN	4%
[latex]A_y[/latex]	4.22 kN	4.23 kN	0.2%
[latex]M_A[/latex]	0.31 kN•m	0.18 kN • m	72%
[latex]D_x[/latex]	—2.06 kN	—2.00 kN	3%
[latex]D_y[/latex]	0.78 kN	0.77 kN	1%
[latex]M_D[/latex]	—3.03 kN•m	—3.18 kN•m	5%

Most of the reactions match the results of the computer aided analysis to within 5%. ✓
Roundoff error accounts for about 1 % of the difference; the rest is due to length change of the members.
The reaction with a much larger percentage difference is a relatively small magnitude.

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