cold formed x-brace shear wall - example (2)

8/19/2019 Cold Formed X-Brace Shear Wall - Example (2)

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X-Brace Shear Wall (2) SteelSmart® System 7.0 SP2

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Table of Contents

1. Objective ................................................................................................................................. 3

2. Modeling using SteelSmart® System ....................................................................................... 5

3. Analysis .................................................................................................................................... 8

3.1 General ................................................................................................................................. 8

3.2 Axial Forces in Members due to Seismic Loads (N0i) ........................................................... 9

3.3 Axial Forces in Members due to Lateral Unit Load (N1i) ................................................... 10

3.4 Lateral Deflection and Inter-Storey Drift Calculation ....................................................... 11

4. Verification of SteelSmart® System ....................................................................................... 13


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1.

Objective

The objective of this verification sample is to check the inter-storey drift calculation of a 2-storey

x-brace shear wall subjected to a combination of dead and seismic loads.

This check should be based on the following parameters:

Design Code: 2012 IBC w/ AISI S100-07/ S2-10

Design Method: ASD

Columns and strap plates will be designed for (D + 0.7E) as “Max Compression” load

combination

Lateral deflection will be calculated based on the full seismic loads

All connections will be designed for (0.6Duplift + 0.7E) as “Uplift Dead Load” load combination

Average floor depth = 6”

Building Risk Category: III

Seismic Design Category (SDC): A

Response Modification Coefficient (R) = 3

System Overstrength Factor (Ω0) = 2

Distortional Buckling Rotational Stiffness (k) = 0

Maximum Bridging spacing = 5 ft

Lateral bracing: @ bridging spacing

Torsional bracing: @ bridging spacing

Maximum inter-storey drift from seismic = 0.02h, where h is the storey height

Column layout: single in 1st storey, back-to-back in 2nd storey with fastener spacing = 12”

Columns are designed for an exact depth of 6”

Uplift dead load will be auto calculated

Infill stud spacing = 16”

Tributary width for uplift dead load calculation = 16”

Effect of the standard punch-out is considered

Strength increase due to cold-work of forming is considered

The layout of the shear wall with numbering of joints and members is shown in Figure 1, and the

applied dead and seismic loads are shown in Figure 2.


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12 ft

10 ft

10 ft

2 3

5

6 7

9

1

4

85 6

3 4

1 2

Figure 1 Layout of the Shear Wall

2 3

5

6 7

9

1

4

856

34

1 2

6 kips

4 kips

Seismic Loads

2 3

5

6 7

9

1

4

85 6

3 4

1 2

0.5 kip

Dead Loads

0.5 kip

0.5 kip 0.5 kip

Figure 2 Loads on the Shear Wall


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2.

Modeling using SteelSmart® System

The shear wall is assumed consisted of a group of link members that carry axial forces only,

therefore; the bracing members in compression are omitted from the model (drawn as dotted

lines in Figures 1 and 2) since they are assumed of no contribution in the stiffness of the shear

wall due to their high slenderness. Each horizontal member in the shear wall model represents a

rigid concrete element of squared shape (100” X 100”).


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3. Analysis

3.1

General

The lateral deflection of the shear wall at a specified level () can be calculated using the Virtual

Work method according to the following relation:

External Work = Internal Work

* (Unit Lateral Load) =

9

1

10

i ii

iii

A E

L N N

where,

i = member ID

N0i = axial force induced in the ith

member due to seismic loads

N1i = axial force induced in the ith

member due to unit lateral load placed at the same joint

that we want to calculate its deflection

Li = geometric length of the ith member

Ei = modulus of elasticity of the ith

member

Ai = cross-sectional area of the ith member

Table 1 displays the values of (L, E, and A) for each member in accordance with the design

output of SteelSmart® System 8.0.


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Table 1 Properties of Members

Member Cross-Section L (in.) E (ksi) A (in.2)

1 100” X 100” 120 3500 10,000

2 (2) 600C/STW250-43 144 29500 1.1058

3 (2) 600C/STW250-43 144 29500 1.1058

4 100” X 100” 120 3500 10,000

5 (2) 4” X 54 mils 187.45 29500 0.4528

6 600C/STW250-43 120 29500 0.5529

7 600C/STW250-43 120 29500 0.5529

8 100” X 100” 120 3500 10,000

9 (2) 4” X 54 mils 169.71 29500 0.4528

3.2

Axial Forces in Members due to Seismic Loads (N 0i )

The axial force in each member of the shear wall due to seismic loads (N0i) can be calculated by

the solver of SteelSmart® System 8.0 for the (E) load case. Values of N0i are displayed in Table 2.

Negative signs denote compression forces, while no sign denotes tension forces.

Table 2 Axial Forces in Shear Wall Members due to Full Seismic Loads

Member N0i (kips)

1 0.000

2 6.000

3 -18.000

4 -10.000

5 15.620

6 0.000

7 -6.000

8 -6.000

9 8.485


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3.3

Axial Forces in Members due to Lateral Unit Load (N 1i )

To calculate the lateral deflection at a specified joint, the axial force in each member of the

shear wall due to lateral unit load (N1i) can be calculated manually by placing 1 kip laterally at

this joint (Figure 3), then constructing the corresponding axial force diagram. Alternatively, the

values of N1i for the shear wall members could be obtained using the solver of SteelSmart®

System 8.0 regardless to the designed sections since it is a statically determinate structure.

Lateral Unit Load @ Joint 5

2 3

5

6 7

9

1

4

856

3 4

1 2

1 kip

Lateral Unit Load @ Joint 3

2 3

5

6 7

9

1

4

856

34

1 2

1 kip

Figure 3 Position of Unit Lateral Load for Calculation of Lateral Deflection at Joints 3, 5

Values of N1i are displayed in Table 3 for lateral unit load placed @ joint 5, while they are

displayed in Table 4 for lateral unit load placed @ joint 3.

Table 3 Axial Forces in Shear Wall Members due to Unit Lateral Load Placed at Joint 5

Member N1i (kips)

1 0.000

2 1.000

3 -2.200

4 -1.000

5 1.562

6 0.000

7 -1.000

8 -1.000

9 1.414


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Table 4 Axial Forces in Shear Wall Members due to Unit Lateral Load Placed at Joint 3

Member N1i (kips)

1 0.000

2 0.000

3 -1.200

4 -1.000

5 1.562

6 0.000

7 0.000

8 0.000

9 0.000

3.4

Lateral Deflection and Inter-Storey Drift Calculation

As discussed in Section 3.1, the lateral deflection at joint 3 or joint 5 (3 or 5, respectively) shall

be calculated as follows:

= Load Lateral Unit

A E

L N N

i ii

iii

9

1

10

According to Figure 4, the internal work for unit lateral load placed at joint 5 = 0.740 kip-in.

N0 N1 Length "L" Modulus ofElasticity "E"

Area "A" (N0N1L)/(EA)

(kips) (kips) (in.) (ksi) (in.2) (kips-in.)

1 0.000 0.000 120 3500 10,000 0.000

2 6.000 1.000 144 29500 1.1058 0.026

3 -18.000 -2.200 144 29500 1.1058 0.175

4 -10.000 -1.000 120 3500 10,000 0.000

5 15.620 1.562 187.45 29500 0.4528 0.342

6 0.000 0.000 120 29500 0.5529 0.000

7 -6.000 -1.000 120 29500 0.5529 0.044

8 -6.000 -1.000 120 3500 10,000 0.000

9 8.485 1.414 169.71 29500 0.4528 0.152

0.740Internal Work (kips-in.)

Member Number

Internal Work for Unit Load Placed @ Joint 5

Figure 4 Calculation of the Internal Work for Unit Lateral Load Placed at Joint 5

5 =1

740.0 = 0.740 in.


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According to Figure 5, the internal work for unit lateral load placed at joint 3 = 0.438 kip-in.

N0 N1 Length "L" Modulus of

Elasticity "E" Area "A" (N0N1L)/(EA)

(kips) (kips) (in.) (ksi) (in.2) (kips-in.)

1 0.000 0.000 120 3500 10,000 0.0002 6.000 0.000 144 29500 1.1058 0.000

3 -18.000 -1.200 144 29500 1.1058 0.095

4 -10.000 -1.000 120 3500 10,000 0.000

5 15.620 1.562 187.45 29500 0.4528 0.342

6 0.000 0.000 120 29500 0.5529 0.000

7 -6.000 0.000 120 29500 0.5529 0.000

8 -6.000 0.000 120 3500 10,000 0.000

9 8.485 0.000 169.71 29500 0.4528 0.000

0.438Internal Work (kips-in.)

Internal Work for Unit Load Placed @ Joint 3

Member Number

Figure 5 Calculation of the Internal Work for Unit Lateral Load Placed at Joint 3

3 =1

438.0

= 0.438 in.

The inter-storey drift should be calculated in accordance with Section 12.8.6 ASCE 7-10 as

follows:

= inter-storey drift

=e

ed

I

C (Equation 12.8-15 ASCE 7-10)

where,

Cd = deflection amplification factor in Table 12.2-1 ASCE 7-10

= 3.5 for light-framed (cold-formed steel) wall systems using flat strap bracing

e = inter-storey drift determined by an elastic analysis

Ie = importance factor determined in accordance with Section 11.5.1 ASCE 7-10

= 1.25 (for Risk Category III)

2-1 = inter-storey drift regarding stories 2 and 1

=(

e

d

I

C 35

=(

25.1

438.0740.0*5.3 = 0.846 in. < (0.02h2 = 0.02*10*12 = 2.4 in.)

1 = drift of storey 1

=e

d

I

C 3

=25.1

438.0*5.3 = 1.226 in. < (0.02h1 = 0.02*12*12 = 2.88 in.)


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4.

Verification of SteelSmart® System


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cold formed x-brace shear wall - example (2)

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