## Analysis and Design of Building Components

Subject: Architecture | Topics:

Design of the Building Components:

The basic function of a building is to provide structurally sound and environmentally controlled spaces to house and protect occupants and contents. If this basic function is not achieved, it is because some aspect of the building has failed. Exponent’s architects, engineers, and scientists have a broad range of expertise with failures in the built environment, and providing clients with in-depth investigations of individual building components, as well as the interdependence of components with each other and the outside environment. Failures of basic building functions can range from defects in single components such as windows, to extensive deficiencies in an entire exterior wall system, to the inability of HVAC systems to properly condition the air. The source of these deficiencies can include inadequate design, improper execution of the work, defective materials, or simply normal and expected aging perhaps coupled with lack of maintenance.

Beam Design:

Option I Building:

Design of Beam 1:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M = 3.390  k-ft

-M = 6.780  k-ft

‘d’ Check:

Mmax= 6.780k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  3.424″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 158.60 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x 9.5 = 0.31 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 50.08 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 3.92 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 2:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M = 6.007  k-ft

-M = 12.014  k-ft

‘d’ Check:

Mmax= 12.014 k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  4.56″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 177.50 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x 9.5 = 0.31 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 88.75 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 4.71 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 3:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M = 4.078  k-ft

-M = 22.287  k-ft

‘d’ Check:

Mmax= 22.287 k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  6.21″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 329.26 psi

= 0.006

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.006x 10 x 9.5 = 0.57 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 60.25 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 5.36 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 4:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M =5.174  k-ft

-M = 10.348  k-ft

‘d’ Check:

Mmax= 10.348  k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  4.22″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 152.88 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.003x 10 x 9.5 = 0.31 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 76.44 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 5.386 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 5:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M =7.132  k-ft

-M = 14.263  k-ft

‘d’ Check:

Mmax= 14.263  k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  4.97″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 210.72psi

= 0.0036

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.0036x 10 x 9.5 = 0.34 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 105.37 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 4.89 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 6:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M =5.254  k-ft

-M = 22.646  k-ft

‘d’ Check:

Mmax= 22.646  k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  6.26″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 334.57 psi

= 0.006

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.006x 10 x 9.5 = 0.57 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 77.62psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 3.76 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 7:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M =5.099  k-ft

-M = 10.199 k-ft

‘d’ Check:

Mmax= 10.199  k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  4.20″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 150.68 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.003x 10 x 9.5 = 0.31 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 75.33 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 6.273 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 8:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M = 6.377  k-ft

-M = 12.754  k-ft

‘d’ Check:

Mmax= 12.754 k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  4.7″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 188.42 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.003x 10 x 9.5 = 0.31 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 94.21 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 4.635 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 9:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M =3.088  k-ft

-M = 6.177  k-ft

‘d’ Check:

Mmax= 6.177  k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  3.27″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 91.26 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.003x 10 x 9.5 = 0.31 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 45.62 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 4.635 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 10:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M = 3.295  k-ft

-M = 6.59  k-ft

‘d’ Check:

Mmax= 6.59k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  3.37″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 97.36 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.003x 10 x 9.5 = 0.31 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 48.68 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 3.938 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 11:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M = 4.328  k-ft

-M = 8.656  k-ft

‘d’ Check:

Mmax= 8.656 k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  3.87″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 127.88 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.003x 10 x 9.5 = 0.31 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 63.94 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 4.881 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

Design of Beam 12:

Here,

Beam size = 10″ x 12″

Effective depth, d = 12″ – 2.5″ = 9.5″

From Etabs-

+M = 1.752 k-ft

-M = 3.503  k-ft

‘d’ Check:

Mmax= 3.503 k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  2.46″ < 9.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 45.1 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.003x 10 x 9.5 = 0.31 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 25.88 psi < 200 psi

min  = 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x9.5 = 0.31 in

Use 2 # 5 Straight bar’s

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 2.669 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 10.41k

Vc = 0.75 x 10.41 = 7.81k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 4.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 4.5″ c/c

S = 48 x  = 18″ c / c

S = Minimum width of column = 10″ c / c

We use # 3 bar @ 10″ c / c

Others beam are same as above beam calculation.

Option II Building:

Design of Beam 25:

Here,

Beam size = 10″ x20″

Effective depth, d = 20″ – 2.5″ = 17.5″

From Etabs-

+M = 97.467  k-ft

-M = 14.803  k-ft

‘d’ Check:

Mmax= 97.467  k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  12.98″ < 17.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 64.45 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x17.5 = 0.0.58 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 424.35 psi

= 0.0068

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0068 x 10 x17.5 = 1.19 in

Use 2 # 5 Straight bars  and  2#5 Ext. bottom bars.

Ap = 1.24 in

Stirrup design:-

From Etabs-

Vu  = 5.804 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 19.17 k

Vc = 0.75 x 19.17 = 14.38 k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 8.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 8.5″ c/c

Design of Beam 26:

Here,

Beam size = 10″ x 20″

Effective depth, d = 20″ – 2.5″ = 17.5″

From Etabs-

+M = 75.896  k-ft

-M = 9.106  k-ft

‘d’ Check:

Mmax= 75.896  k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  11.45″ < 17.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 39.65 psi < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x 17.5 = 0.58 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 330.43 psi

= 0.0058

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0058 x 10 x17.5 = 1.015 in

Use 2 # 5 Straight bars  and  1#6 Ext. bottom bars.

Ap = 1.06 in

Stirrup design:-

From Etabs-

Vu  = 5.79 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 19.17 k

Vc = 0.75 x 19.17 = 14.38 k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 8.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 8.5″ c/c

Design of Beam 27:

Here,

Beam size = 10″ x 20″

Effective depth, d = 20″ – 2.5″ = 17.5″

From Etabs-

+M = 72.632  k-ft

-M = 8.439  k-ft

‘d’ Check:

Mmax= 72.632  k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  11.21″ < 17.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 36.74 psi  < 200 psi

= 0.003

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x 17.5 = 0.58 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 316.22 psi

= 0.0056

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0056 x 10 x17.5 = 0.98 in

Use 2 # 5 Straight bars  and  1#6 Ext. bottom bars.

Ap = 1.06 in

Stirrup design:-

From Etabs-

Vu  = 7.509 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 19.17 k

Vc = 0.75 x 19.17 = 14.38 k

Vu < Vc,  So, Stirrup is no required.

Use minimum shear reinforcement-

Smax = =  = 8.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 8.5″ c/c

Design of Beam 28:

Here,

Beam size = 10″ x 20″

Effective depth, d = 20″ – 2.5″ = 17.5″

From Etabs-

+M = 26.048  k-ft

-M = 52.096  k-ft

‘d’ Check:

Mmax= 52.096  k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  9.49″ < 17.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 226.81 psi

= 0.0038

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.0038 x 10 x 17.5 = 0.67 in

Use 2 # 5 Straight bars  and  1#5 Ext. top bars.

Ap = 0.93 in

Steel for mid span:-

R= =

= 113.41 psi  < 200 psi

= 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x17.5 = 0.58 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 19.870 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 19.17 k

Vc = 0.75 x 19.17 = 14.38 k

Vu > Vc,  So, Stirrup is required.

Vs = (Vu – Vc) = 19.87  – 14.38 = 5.49 kip

4bwd = 4 x 0.75 x 10 x  = 28.76 >Vs

Smax =   = 8.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 8.5″ c/c  Throught the beam length.

Design of Beam 29:

Here,

Beam size = 10″ x 20″

Effective depth, d = 20″ – 2.5″ = 17.5″

From Etabs-

+M = 17.468  k-ft

-M = 34.936  k-ft

‘d’ Check:

Mmax= 34.936  k-ft

b    = 0.85 x ßl   x

b   =  0.85 x 0.85 x x

= 0.021

max = 0.75 b = 0.75 x 0.021= 0.0158

Mmax      = bd fy (1- 0.59  x

Or, d =

=  7.77″ < 17.5″

So, ok

Reinforcement Calculation:

Steel for Ext. Support:-

R==

= 152.10 psi  < 200 psi

= 0.0033

min   =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x 17.5 = 0.58 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Steel for mid span:-

R= =

= 76.77 psi  < 200 psi

= 0.003

min  =  = 0.0033

min =  = 0.00274

As = bd = 0.0033 x 10 x17.5 = 0.58 in

Use 2 # 5 Straight bars

Ap = 0.62 in

Stirrup design:-

From Etabs-

Vu  = 14.507 k

Concrete shear strength,

Vc  = 2bwd = 2 x 10 x = 19.17 k

Vc = 0.75 x 19.17 = 14.38 k

Vu > Vc,  So, Stirrup is required.

Vs = (Vu – Vc) = 14.507  – 14.38 = 0.127  kip

4bwd = 4 x 0.75 x 10 x  = 28.76 >Vs

Smax =   = 8.75″

Smax = 24″

Use, 2 legs Stirrup #3 @ 8.5″ c/c  Throught the beam length.

Others beam are same as option I beam.

Column Design:

Option I Building:

Design Data:–

Colomn height = 10′  c / c

Column type     = Tied

Clear cover        = 1.5″

For tied column-

= 0.80

= 0.65

g = 0.02

Materials:

Fy = 60 ksi

f′c = 3 ksi

wc = 150 psf

Design of column C1:

Here, Pu = 49.001 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 49.001 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 25.47 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 49.001 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C2:

Here, Pu = 98.846 k

Ag (Provided)  = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 98.846 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 51.2 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 98.846 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C3:

Here, Pu = 97.368 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 97.368= 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 50.62 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 97.368= 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C4:

Here, Pu = 60.253 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 60.253 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 31.33 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 60.253 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C5:

Here, Pu = 62.259 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 62.259 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 32.37 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 62.259 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C6:

Here, Pu = 126.655 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 126.655 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 65.847 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 126.655 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C7:

Here, Pu = 170.054 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 170.054 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 88.41 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 170.054 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C8:

Here, Pu = 110.947 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 110.947 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 57.68 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 110.947 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C9:

Here, Pu = 49.019 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 49.019 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 25.48 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 49.019 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C10:

Here, Pu = 99.013 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 99.013 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 51.48 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 99.013 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C11:

Here, Pu = 108.54 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 108.54 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 56.43inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 108.54= 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C12:

Here, Pu = 60.848 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 60.848 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 31.634 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 60.848 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C13:

Here, Pu = 35.069k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 35.069 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 18.23 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 35.069 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C14:

Here, Pu = 69.45 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 69.45 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 36.106 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 69.45 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C15:

Here, Pu = 62.943 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 62.943 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 32.724 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 62.943 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C16:

Here, Pu = 40.405 k

Ag (Provided) = 10 x 15 = 150 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 40.405 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 21 inch< 150 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 40.405 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Option II Building:

Design of column C17:

Here, Pu = 126.557 k

Ag (Provided) = 10 x 18 = 180 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 126.557 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 65.80 inch< 180 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 126.557 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C18:

Here, Pu = 126.557 k

Ag (Provided) = 10 x 18 = 180 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 301.559  = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 156.778  inch< 180 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 301.559 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = 3.44 inch

Use 8 # 6 bars, Ap = 3.52 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 12″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C19:

Here, Pu = 125.959 k

Ag (Provided) = 10 x 18 = 180 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 125.959 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 64.75 inch< 180 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 125.959 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C20:

Here, Pu = 123.383 k

Ag (Provided) = 10 x 18 = 180 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 123.383 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 64.15 inch< 180 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 123.383 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C21:

Here, Pu = 115.408 k

Ag (Provided) = 10 x 18 = 180 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 115.408 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 59.00 inch< 180 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 115.408 = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c / c

Design of column C22:

Here, Pu = 118.007  k

Ag (Provided) = 10 x 18 = 180 inch

Check for size:

Pu  = Ag [ .85f′c (1- g) + g fy]

Or, 118.007  = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 60)]

Ag = 61.35 inch< 180 inch, ok

Calculation of steel:-

Pu =  [0.85 x f′c (Ag – Ast) + Ast fy]

Or, 118.007  = 0.80 x .65 [0.85 x 3 x (150- Ast) +60 Ast]

Ast = ( – )

So, 1% reinforcement should be provided.

As = 0.01 x 150 = 1.5 inch

Use 6 # 5 bars, Ap = 1.86 inch

Tie bar Calculation / Spacing :

Use # 3 bar as Tie:-

i) S = 16 x   = 10″  c/c

ii) S = 48 x   = 18″ c/c

iii) S = Minimum thickness = 10″  c/c

So we use # 3 bar @ 10″ c/c

Others column are same as option I column.

Bracing  Design:

Design of Bracing BR1:

Design for A36 steel.

Here,  Mmax  = 20.498 k-ft = 245.976 k-in

Assume compact section,

The allowable bending stress, Fb = 0.66 Fy

= 0.66 x 36 = 24 ksi

Hence, Required section modulus, Sx =

=

= 10.249 in

Try, W 8 x 15 section

Sx = 11.8 in, A= 4.44 in, d = 8.11 in,  tw = 0.245 in

b= 4.015 in, t= 0.315 in

Checking width thickness ratio for compact section:

=  = 6.373 < = =10.83  (ok)

=   = 33.102 < = = 106.67 (ok)

So, the section is a compact section.

Checking bending stress of the beam:

Bending stress, f=  = = 20.85 ksi < Allowable stress = 24 ksi (ok)

So, the section W8 x 15 is acceptable.

Design of Bracing BR2:

Design for A36 steel.

Here,  Mmax  = 18.866 k-ft = 226.392 k-in

Assume compact section,

The allowable bending stress, Fb = 0.66 Fy

= 0.66 x 36 = 24 ksi

Hence, Required section modulus, Sx =

=

= 9.433 in

Try, W 8 x 15 section

Sx = 11.8 in, A= 4.44 in, d = 8.11 in, tw = 0.245 in

b= 4.015 in, t= 0.315 in

Checking width thickness ratio for compact section:

=  = 6.373 < = =10.83 (ok)

=   = 33.102 < = = 106.67 (ok)

So, the section is a compact section.

Checking bending stress of the beam:

Bending stress, f=  = = 19.19 ksi < Allowable stress = 24 ksi (ok)

So, the section W8 x 15 is acceptable.

Design of Bracing BR3:

Design for A36 steel.

Here,  Mmax  = 20.349 k-ft = 244.188 k-in

Assume compact section,

The allowable bending stress, Fb = 0.66 Fy

= 0.66 x 36 = 24 ksi

Hence, Required section modulus, Sx =

=

= 10.175  in

Try, W 8 x 15 section

Sx = 11.8 in, A= 4.44 in, d = 8.11 in,  tw = 0.245 in

b= 4.015 in, t= 0.315 in

Checking width thickness ratio for compact section:

=  = 6.373 < = =10.83  (ok)

=   = 33.102 < = = 106.67 (ok)

So, the section is a compact section.

Checking bending stress of the beam:

Bending stress, f=  = = 20.71 ksi < Allowable stress = 24 ksi (ok)

So, the section W8 x 15 is acceptable.

Design of Bracing BR4:

Design for A36 steel.

Here,  Mmax  = 15.630 k-ft = 187.56 k-in

Assume compact section,

The allowable bending stress, Fb = 0.66 Fy

= 0.66 x 36 = 24 ksi

Hence, Required section modulus, Sx =

=

= 7.815 in

Try, W 8 x 15 section

Sx = 11.8 in, A= 4.44 in, d = 8.11 in,  tw = 0.245 in

b= 4.015 in, t= 0.315 in

Checking width thickness ratio for compact section:

=  = 6.373 < = =10.83  (ok)

=   = 33.102 < = = 106.67 (ok)

So, the section is a compact section.

Checking bending stress of the beam:

Bending stress, f=  = = 15.89 ksi < Allowable stress = 24 ksi (ok)

So, the section W8 x 15 is acceptable.

Design of Bracing BR5:

Design for A36 steel.

Here,  Mmax  = 20.608 k-ft = 247.296 k-in

Assume compact section,

The allowable bending stress, Fb = 0.66 Fy

= 0.66 x 36 = 24 ksi

Hence, Required section modulus, Sx =

=

= 10.304 in

Try, W 8 x 15 section

Sx = 11.8 in, A= 4.44 in, d = 8.11 in,  tw = 0.245 in

b= 4.015 in, t= 0.315 in

Checking width thickness ratio for compact section:

=  = 6.373 < = =10.83  (ok)

=   = 33.102 < = = 106.67 (ok)

So, the section is a compact section.

Checking bending stress of the beam:

Bending stress, f=  = = 20.96 ksi < Allowable stress = 24 ksi (ok)

So, the section W8 x 15 is acceptable.

Design of Bracing BR6:

Design for A36 steel.

Here,  Mmax  = 16.204 k-ft = 194.448 k-in

Assume compact section,

The allowable bending stress, Fb = 0.66 Fy

= 0.66 x 36 = 24 ksi

Hence, Required section modulus, Sx =

=

= 8.102 in

Try, W 8 x 15 section

Sx = 11.8 in, A= 4.44 in, d = 8.11 in,  tw = 0.245 in

b= 4.015 in, t= 0.315 in

Checking width thickness ratio for compact section:

=  = 6.373 < = =10.83  (ok)

=   = 33.102 < = = 106.67 (ok)

So, the section is a compact section.

Checking bending stress of the beam:

Bending stress, f=  = = 16.48 ksi < Allowable stress = 24 ksi (ok)

So, the section W8 x 15 is acceptable.

Some are parts:

Analysis and Design of Building Components (Part 1)

Analysis and Design of Building Components (Part 2)

Analysis and Design of Building Components (Part 3)

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