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
bf = 4.015 in, tf = 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, fb = = = 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
bf = 4.015 in, tf = 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, fb = = = 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
bf = 4.015 in, tf = 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, fb = = = 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
bf = 4.015 in, tf = 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, fb = = = 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
bf = 4.015 in, tf = 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, fb = = = 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
bf = 4.015 in, tf = 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, fb = = = 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)
