Our client is renovating an existing building and adding new mechanical equipment on the roof. To house the new equipment, we are designing a small rigid frame. The frame will attach to very stiff beams, so the frame will be effectively a single story with fixed supports. The members have been

Question

Our client is renovating an existing building and adding new mechanical equipment on the roof. To house the new equipment, we are designing a small rigid frame. The frame will attach to very stiff beams, so the frame will be effectively a single story with fixed supports.

The members have been designed to carry the gravity load. Now the team needs to verify that the members are adequate with regard to lateral load. To help in that process, our role is to analyze the frame subjected to lateral wind load. We need to find the axial force and peak moment generated in each column due to that wind load.

Preliminary member properties:

Beams:	W10 x 22	A = 6.49 in²	[latex]I = 118 in^4[/latex]
Column AD:	HSS5 x 5 x ½	A = 7.88 in²	[latex]I = 26.0 in^4[/latex]
Column BE:	HSSS x 5 x ¼	A = 4.30 in²	[latex]I = 16.0 in^4[/latex]
Steel:	E = 29,000 ksi

Accepted Answer

Estimated Solution:
Estimate an inflection point at midspan of the middle beam; there is no inflection point in the cantilevers. Estimate an inflection point at midheight of each column.(figure 13.44a)

FBD of frame cut at column inflection points:
Since the frame is wider (10 ft) than it is tall (8 ft), use the Portal Method: assume the shear to each column is the same. (figure 13.44b)

[latex]\underrightarrow{+} \sum{F_x} = 0 = 2 k - 2 V [/latex]

[latex]V = 1 k\left(\overset{\leftharpoonup}{\underset{ ightharpoonup}+} ight)[/latex]

FBD of AG:
Because of the Portal Method assumptions, the moments at the top and bottom of each column are the same magnitude and the peak moment in the two columns will be the same. (figure 13.44c)

[latex]+\circlearrowright \sum{M_A} = 0 = 1 k (4 ft) + M_A [/latex]

[latex]M_A = -4 k • ft (+\circlearrowright)[/latex]

FBD of GCH:
Because there are only two columns, equilibrium requires that the axial force in each will be equal and opposite.

[latex]+\circlearrowright \sum{M_H} = 0 = 1 k (4 ft) - N_G(5 ft)[/latex]

[latex]N_G = 0.8 k \left(\overset{\uparrow}{\underset{\downarrow}+} ight)[/latex]

Summary:
Axial force in columns: 0.8 k

Peak moment in columns: 4 k • ft

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

Determine the distribution factors:
Stiffness of each member:

[latex]K_{DC} = 0 ext{(cantilevered member)}[/latex]

[latex]K_{AD} = K_{DA} = \frac{4(29,000 ksi)(26 in^4)}{8 ft} = 3.77 × 10^5\frac{k • in²}{ft}[/latex]

[latex]K_{BE} = K_{EB} = \frac{4(29,000 ksi)(16 in^4)}{8 ft} = 2.32 × 10^5\frac{k • in²}{ft}[/latex]

[latex]K_{DE} = K_{ED} = \frac{4(29,000 ksi)(118 in^4)}{10 ft} = 13.69 × 10^5\frac{k • in²}{ft}[/latex]

[latex]K_{EF} = 0 ext{(cantilevered member )} [/latex]

Distribution factors:

[latex]DF_{AD} = 0 ext{(end of member at fixed support )}[/latex]

[latex]DF_{DC} = 0 ext{(cantilevered member )}[/latex]

[latex]DF_{DA} = \frac{3.77}{0 + 3.77 + 13.69} = 0.216[/latex]

[latex]DF_{DE} = \frac{13.69}{0 + 3.77 + 13.69} = 0.784[/latex]

[latex]DF_{ED} = \frac{13.69}{13.69 + 0 + 2.32} = 0.855[/latex]

[latex]DF_{EB} = \frac{2.32}{13.69 + 0 + 2.32} = 0.145[/latex]

[latex]DF_{EF} = 0 ext{(cantilevered member )}[/latex]

[latex]DF_{BE} = 0 ext{(end of member at fixed support )}[/latex]

Calculate the load effects (no sidesway):
Put a horizontal roller at F to stop Sidesway.

Fixed-end moments:

Load along the axis of the cantilevers causes no moment:
[latex]FEM_{DC} = FEM_{EF} = 0[/latex]

For the columns with no load between the nodes:
[latex]FEM_{ED} = FEM_{EA} = FEM_{BE} = FEM_{EB} = 0[/latex]

For beam DE, no load between the nodes:
[latex]FEM_{DE} =FEM_{ED} = 0[/latex]

Distribute the moments:
There are no moments to distribute, so there are no net moments anywhere in the frame.

Table 13.14

NODE	A	D			E			B
MEMBER	AD	DC	DA	DE	ED	EB	EF	BE
K	3.77	0	3.77	13.69	13.69	2.32	0	2.32
DF	0	0	0.216	0.784	0.855	0.145	0	0	-UM
FEM (k · ft)
Total [latex]M_{NS}[/latex]	0	0	0	0	0	0	0	0

Find the lateral reaction:
FBD of AD:

[latex]+\circlearrowright \sum{M_{D}} = 0 = A_x (8 ft)[/latex]

[latex]A_x = 0[/latex]

FBD of BE:

[latex]+\circlearrowright \sum{M_{E}} = 0 = B_x (8 ft)[/latex]

[latex]B_x = 0[/latex]

FBD of entire frame:

[latex]\underrightarrow{+}\sum{F_x} = 0 = 2 k - F_x[/latex]

[latex]F_x = 2 k (\underleftarrow{+})[/latex]

Observation: If there are no fixed-end moments, the lateral load is carried directly to the horizontal roller.

Calculate the Sidesway effects:
Apply an unknown lateral load at F.

Fixed-end moments:

Loading along the axis of the cantilevers causes no moment:
[latex]FEM_{DC} = FEM_{EF} = 0[/latex]

For the beam, there is no load or differential movement:
[latex]FEM_{DE} = FEM_{ED} = 0[/latex]

For the columns, use formulas from inside the back cover:
[latex]FEM = -\frac{6EIΔ^′}{l²}[/latex]

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

[latex]FEM_{BE} = FEM_{BE} = -\frac{6EI_{BE}Δ^′}{l²_{BE}} ≡ - 100 k • ft[/latex]

Since the columns have different I values but the same Δ′:

[latex]Δ^′ = \frac{100 k • ft (l^2_{BE}}{6EI_{BE}} = \frac{100 k • ft (8 ft)^2}{6E (16.0 in^4)}[/latex]

[latex]FEM_{AD} = FEM_{DA} = -\frac{6EI_{AD}Δ^′}{l^2_{AD}} = -[\frac{6E(26.0 in^4)}{(8 ft)^2}][\frac{100 k • ft (8 ft)^2}{6E(16.0 in^4)}][/latex]

[latex]= -162.5 k • ft[/latex]

Distribute the moments:
Stop when the peak UM is no more than 1 k•ft. There is no need to release node A or B because they are fixed supports.

Table 13.15

NODE	A	D			E			B
MEMBER	AD	DC	DA	DE	ED	EB	EF	BE
K	3.77	0	3.77	13.69	13.69	2.32	0	2.32
DF	0	0	0.216	0.784	0.855	0.145	0	0	-UM
FEM (k · ft)	-162.5		-162.5			-100.0		-100.0
Release D		0.0	35.1	127.4					162.5
Carryover	17.5				63.7
Release E					31.0	5.3	0.0		36.3
Carryover				15.5				2.6
Release D		0.0	-3.4	-12.2					-15.5
Carryover	-1.7				-6.1
Release D					5.2	0.9	0.0		6.1
Carryover				2.6				0.4
Release D		0.0	-0.6	-2.0					-2.6
Carryover	-0.3				-1.0

Total [latex]M^′_S[/latex]	-147.0	0.0	-131.4	131.3	92.8	-93.8	0.0	-97.0

Unbalanced moments:

[latex]UM_D = 0.0 k • ft[/latex]

[latex]UM_E = -1.0 k • ft[/latex]

All are at or below our stopping threshold of 1 k•ft.

∴ Okay to stop

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

[latex]ΣM_B = -131.4 k • ft + 131.3 k • ft = -0.1 k • ft ≈ 0[/latex] ✓

[latex]ΣM_C = 92.8 k • ft - 93.8 k • ft = -1.0 k • ft ≈ 0[/latex] ✓

Find the lateral force:
FBD of AD:

[latex]+\circlearrowright \sum{M_D} = 0 = -147.0 k • ft - A^′_x (8 ft) + (-131.4 k • ft)[/latex]

[latex]A^′_x = -34.8 k (\underrightarrow{+})[/latex]

FBD of BE:

[latex]+\circlearrowright \sum{M_E} = 0 = -97.0 k • ft - B_x(8 ft) + (-93.8 k • ft)[/latex]

[latex]B^′_x = -23.8 k (\underrightarrow{+})[/latex]

FBD of entire frame:

[latex]\underrightarrow{+} ΣF_x = 0 = -34.8 k - 23.8 k + F^′_x[/latex]

[latex]F^′_x = 58.6 k (\underrightarrow{+})[/latex]

Scale results:
Scale factor:

[latex]SF = \frac{2 k}{58.6 k} = 0.0341[/latex]

Scale moments:

[latex]M_{AD} = 0 + 0.0341 (-147.0 k • ft) = -5.01 k • ft (+\circlearrowright)[/latex]

[latex]M_{DA} = 0 + 0.0341 (-131.4 k • ft) = -4.48 k • ft[/latex]

[latex]M_{AD} = 0 + 0.0341 (-93.8 k • ft) = -3.20 k • ft[/latex]

[latex]M_{AD} = 0 + 0.0341 (-97.0 k • ft) = -3.31 k • ft (+\circlearrowright)[/latex]

Column axial forces:
We found the peak column moments by scaling the moment distribution results. Let's use equilibrium to find the column axial forces.

FBD of CF for the original frame:
The moments are the opposite of values from the column FBDs (scaled results).

[latex](+\circlearrowright)ΣM_E = 0 = -4.48 k • ft + N_{DA}(10 ft) + (-3.20 k • ft)[/latex]

[latex]N_{DA} = 0.77 k \left(\overset{\uparrow}{\underset{\downarrow}+} ight)[/latex]

[latex]+\downarrow ΣF_y = 0 = 0.77 k + N_EB[/latex]

[latex]N_{EB} = -0.77 k \left(\overset{\uparrow}{\underset{\downarrow}+} ight)[/latex]

Summary:

Axial force in columns: 0.77 k

Peak moment in AD: 5.01 k • ft

Peak moment in BE: 3.31 k • ft

Evaluation of Results:
Approximation Predicted Outcomes?
Our approximate prediction of the column axial force is about 4% larger than our detailed result, and our approximate prediction of the peak moment in each column is within 21 % of our detailed results. ✓

Conclusion:
These observations suggest we have reasonable results.

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

Therefore, the effect of length change was negligible.

Reaction	Moment Distribution	Computer Aided	Difference
[latex]A_x[/latex]	-1.19 k	-1.19 k	0%
[latex]A_y[/latex]	-0.77 k	-0.77 k	0%
[latex]M_A[/latex]	-5.01 k • ft	-5.03 k • ft	4%
[latex]B_x[/latex]	-0.81 kN	-0.81 kN	0%
[latex]B_y[/latex]	0.77 k	0.77 k	0%
[latex]M_B[/latex]	-3.31 k • ft	-3.30k • ft	0.3%

Question 13.4: Our client is renovating an existing building and adding new...

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