Our team is designing the awning for a new office building. The team leader wants to explore a couple different structural schemes. One of them is a rigid frame. Our task is to analyze the rigid frame option carrying snow load.

In order to compare the different structural schemes, the team leader wants to know the peak moment in each member.

Question

Our team is designing the awning for a new office building. The team leader wants to explore a couple different structural schemes. One of them is a rigid frame. Our task is to analyze the rigid frame option carrying snow load.

In order to compare the different structural schemes, the team leader wants to know the peak moment in each member.

Accepted Answer

Estimated Solution:
Approximate member BC as a simply supported beam. This should provide an upper bound to the peak moment in the beam.

[latex]M_{\max } = \frac{w l^2}{8} = \frac{5 \mathrm{kN} / \mathrm{m}(10 \mathrm{~m})^2}{8} = 62.5 \mathrm{kN} \cdot \mathrm{m}[/latex]

Detailed Analysis:
Determine the distribution factors:
Stiffness of each member:

[latex]K_{B A} = 0 \quad ext { (cantilevered member) }[/latex]

[latex]K_{B C} = K_{C B} = \frac{4 E I}{10 \mathrm{~m}}[/latex]

[latex]K_{B D} = K_{D B} = \frac{4 E I}{4.5 \mathrm{~m}}[/latex]

Distribution factors:

[latex]D F_{B A} = 0 ( cantilevered member )[/latex]

[latex]D F_{B C} = \frac{4 E I / 10 \mathrm{~m}}{0 + 4 E I / 10 \mathrm{~m} + 4 E I / 4.5 \mathrm{~m}} = 0.310[/latex]

[latex]D F_{B D} = \frac{4 E I / 4.5 \mathrm{~m}}{0 + 4 E I / 10 \mathrm{~m} + 4 E I / 4.5 \mathrm{~m}} = 0.690[/latex]

[latex]D F_{C B} = 1 \quad ext { (single member at end with roller support) }[/latex]

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

Calculate the fixed-end moments:

For the cantilever, use statics:

[latex]F E M_{B A} = 5 \mathrm{kN} / \mathrm{m}(2.5 \mathrm{~m})(2.5 \mathrm{~m} / 2) = 15.6 \mathrm{kN} \cdot \mathrm{m}[/latex]

For member BC, use formulas from inside the back cover:

[latex]F E M_{B C} = -\frac{5 \mathrm{kN} / \mathrm{m}(10 \mathrm{~m})^2}{12} = -41.7 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]F E M_{C B} = +\frac{5 \mathrm{kN} / \mathrm{m}(10 \mathrm{~m})^2}{12} = +41.7 \mathrm{kN} \cdot \mathrm{m}[/latex]

For the column with no load between the nodes:

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

Distribute the moments:
Stop when the peak UM is no more than 1.6 kN•m. There is no need to release node D because it is a fixed support.

Table 13.11

NODE	B			C	D
MEMBER	BA	BC	BD	CB	DB
K	0	4EI/10	4EI/4.5	4EI/10	4EI/4.5
DF	0	0.310	0.690	1	0	-UM
FEM (kN•m)	15.6	-41.7		41.7
Release C				-41.7		-41.7
Carryover		-20.9
Release B	0.0	14.6	32.4			47.0
Carryover				7.3	16.2
Release C				-7.3		-7.3
Carryover		-3.6
Release B	0.0	1.1	2.5			3.6
Carryover				0.6	1.3

Total M	15.6	-50.5	34.9	0.6	17. 5

Unbalanced moments:

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

[latex]U M_C = 0.6 \mathrm{kN} \cdot \mathrm{m}[/latex]

All are at or below our stopping threshold of 1.6 kN•m.

∴ Okay to stop

Evaluation of Results:

Satisfied Fundamental Principles?
Check equilibrium of the net moments:

[latex]\Sigma M_B = 15.6 \mathrm{kN} \cdot \mathrm{m} - 50.5 \mathrm{kN} \cdot \mathrm{m} + 34.9 \mathrm{kN} \cdot \mathrm{m} = 0[/latex] ✓

[latex]\Sigma M_C=0.6 \mathrm{kN} \cdot \mathrm{m} \approx 0[/latex] ✓

Find the reactions to evaluate the results:

FBD of BC:

[latex]+\circlearrowright \Sigma M_B = 0 = -50.5 \mathrm{kN} \cdot \mathrm{m} + 5 \mathrm{kN} / \mathrm{m}(10 \mathrm{~m})(10 \mathrm{~m} / 2) - C_y(10 \mathrm{~m})[/latex]

[latex]C_y = 20.0 \mathrm{kN} \quad(+\uparrow)[/latex]

[latex]+\uparrow \Sigma F_y = 0 = V_{B C} - 5 \mathrm{kN} / \mathrm{m}(10 \mathrm{~m}) + 20.0 \mathrm{kN}[/latex]

[latex]V_{B C} = 30.0 \mathrm{kN}[/latex]

FBD of BD:

[latex]+\circlearrowright \Sigma M_B = 0 = 34.9 \mathrm{kN} \cdot \mathrm{m} - D_x(4.5 \mathrm{~m}) + 17.5 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]D_x = 11.6 \mathrm{kN} \quad(\stackrel{+}{ ightarrow})[/latex]

[latex]\stackrel{+}{ ightarrow} \Sigma F_x = 0 = V_{B D} + 11.6 \mathrm{kN}[/latex]

[latex]V_{B D} = -11.6 \mathrm{kN}[/latex]

[latex]+\uparrow \Sigma F_y = 0 = D_y + N_{B D}[/latex]

[latex]D_y = -N_{B D} \quad(+\uparrow)[/latex]

FBD of AC:
The moment is in the opposite direction of the internal force from column BD, not a moment obtained from distribution.

[latex]+\uparrow \Sigma F_y = 0 = -5 \mathrm{kN} / \mathrm{m}(12.5 \mathrm{~m}) - N_{B D} + 20.0 \mathrm{kN}[/latex]

[latex]N_{B D} = -42.5 \mathrm{kN}[/latex]

[latex] herefore D_y = 42.5 \mathrm{kN} \quad(+\uparrow)[/latex]

[latex]\stackrel{+}{ ightarrow} \Sigma F_x = 0 = -(-11.6 \mathrm{kN}) + C_x[/latex]

[latex]C_x = -11.6 \mathrm{kN} \quad(\stackrel{+}{ ightarrow})[/latex]

Summary of reactions:

Evaluation of Results:
Satisfied Fundamental Principles?
Check equilibrium:

[latex]\stackrel{+}{ ightarrow} \Sigma F_x = 11.6 \mathrm{kN} + (-11.6 \mathrm{kN}) = 0[/latex] ✓

[latex]+\uparrow \Sigma F_y = -5 \mathrm{kN} / \mathrm{m}(12.5 \mathrm{~m}) + 42.5 \mathrm{kN} + 20 \mathrm{kN} = 0[/latex] ✓

[latex]\begin{aligned}+\circlearrowright\Sigma M_C =&-5 \mathrm{kN} / \mathrm{m}(12.5 \mathrm{~m})(12.5 \mathrm{~m} / 2) + 17.5 \mathrm{kN} \cdot \mathrm{m} \& \quad + 42.5 \mathrm{kN}(10 \mathrm{~m}) - 11.6 \mathrm{kN}(4.5 \mathrm{~m}) \=&-0.3 \mathrm{kN} \cdot \mathrm{m} \approx 0\end{aligned}[/latex] ✓

Find the peak moment in each member:
Cantilever AB:
Draw the moment as positive for the diagram (not clockwise).

Beam BC:

Column DB:
Draw the internal moment as positive for the diagram (not clockwise).

Summary of peak moments:

[latex]M_{A B \max } = -15.6 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M_{B C ext { max }} = -50.5 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M_{D B \max }=-34.9 \mathrm{kN} \cdot \mathrm{m}[/latex]

Evaluation of Results:

Approximation Predicted Outcomes?

Our approximate prediction of the peak moment in beam BC is larger than the actual, as expected. Our actual peak moment is 19% smaller than the approximate result. ✓

Conclusion:
Our reactions satisfy equilibrium.
Our moment diagrams close to zero or within roundoff error of zero.
Our detailed analysis result for beam BC is below the upper bound.
This evidence suggests that we have reasonable results.

Comparison with computer aided analysis results:
Most of the reactions match the results of the the computer aided analysis to within 1 %, which is within roundoff error. ✓

Reaction	Moment Distribution	Computer Aided	Difference
[latex]D_x[/latex]	11.6 kN	11.5 kN	1%
[latex]D_y[/latex]	42.5 kN	42.5 kN	0%
[latex]M_D[/latex]	17.5 kN•m	17.0 kN•m	3%
[latex]C_x[/latex]	-11.6 kN	-11.5 kN	1%
[latex]C_y[/latex]	20.0 kN	20.0 kN	0%

One reaction, [latex]M_D[/latex], is not small and has a 3%difference; this difference is not all due to roundoff error and is explained in Section 13.4: Sidesway Frames.

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