Respuesta :
[tex]\phi M n=263 \mathrm{ft} \cdot \mathrm{kip}[/tex] > M OK [tex]\mathrm{Mu}=263.1 \mathrm{ft}.\mathrm{kip}[/tex]
[tex]A s=4.0 \text { in } 2[/tex]
Explanation:
Given properties:
[tex]f_{y}[/tex] = 600000 psi
[tex]f_{c}[/tex] = 4000 psi
I: = 24 ft
[tex]W_{I}[/tex]: = 2000 lbf/ft
[tex]W_{c}[/tex]: = 150 lbf / [tex]ft^{3}[/tex]
For weight determination, the beam dimension is estimated.
[tex]b:=21-in[/tex]
[tex]d:=18-in[/tex]
[tex]h:=12-i n[/tex]
[tex]\mathrm{wo}=\mathrm{b} \cdot \mathrm{h} \cdot \mathrm{wc}[/tex]
[tex]\text { wu }:=1.2-\mathrm{w} \ o+1.6 \mathrm{wI}[/tex]
[tex]w_{u}=3.515 \frac{\mathrm{kip}}{\mathrm{ft}}[/tex]
[tex]M_{u}:=w_{u} \cdot \frac{I^{2}}{8}[/tex]
[tex]\mathrm{Mu}=253.1 \mathrm{kip} \cdot \mathrm{ft}[/tex]
[tex]\beta _{1}[/tex] = [tex]0.85-\frac{0.05\left(\mathrm{f}_{\mathrm{c}}-4000 \mathrm{psi}\right)}{1000 \mathrm{psi}}[/tex]
[tex]\beta _{1}[/tex] = 0.85
[tex]\rho_{\max }:=0.85 \cdot \beta_{1} \cdot \frac{\mathrm{f}_{\mathrm{c}}}{\mathrm{f}_{\mathrm{y}}} \cdot \frac{0.003}{0.003+0.004}[/tex]
= 0.0206
a) The section property is found to carry Mu.
For ρ = 0.6 max.
ρ = 0.0124
Resistance = [tex]\rho \mathrm{f}_{\mathrm{y}}\left(1-0.588 \frac{\rho \mathrm{f}_{\mathrm{y}}}{\mathrm{f}_{\mathrm{c}}}\right)[/tex]
Resistance ρ<ρ0.005 then Φ = 0.90
[tex]b:=\sqrt[3]{\frac{M_{u}}{2.25 \cdot \phi \cdot R e s i s t a n c e}}[/tex] = 13.134 in
d = 1.5 b
d = 19.7 in
[tex]A S:=\rho \cdot b \cdot d b[/tex] = [tex]A s=3.78 \text { in } 2[/tex]
The solution is used for 4-#8
b = 14" d = 20.5" h = 22"
[tex]A s=4-0.79 i n 2[/tex] b = 14.in d = 20.5.in h = 22.in
Design capacity is
[tex]\mathbf{a}:=\frac{\mathbf{A}_{\mathbf{s}} \cdot \mathbf{f}_{\mathbf{y}}}{0.85 \cdot \mathbf{f}_{\mathbf{c}} \cdot \mathbf{b}}[/tex] = 3.98 in
[tex]\phi M_{n}:=\phi \cdot A_{s} \cdot f_{y}\left(d-\frac{a}{2}\right)[/tex]
[tex]\phi M n=263 \mathrm{ft} \cdot \mathrm{kip}[/tex] > M OK [tex]\mathrm{Mu}=263.1 \mathrm{ft}.\mathrm{kip}[/tex]
b)
[tex]\rho_{005}:=0.85 \cdot \beta_{1} \cdot \frac{\mathrm{f}_{\mathrm{c}}}{\mathrm{f}_{\mathrm{y}}} \cdot \frac{0.003}{0.003+0.005}[/tex]
[tex]\rho:=\rho 005[/tex] [tex]\rho=\rho 0.0181[/tex]
Resistance = [tex]\rho \mathrm{f}_{\mathrm{y}}\left(1-0.588 \frac{\rho \mathrm{f}_{\mathrm{y}}}{\mathrm{f}_{\mathrm{c}}}\right)[/tex]
Resistance = 911 psi
[tex]b:=\sqrt[3]{\frac{M_{u}}{1.5^{2} \cdot \phi \cdot R e s i s t a n c e}}[/tex]
= 11.8 in
d = 1.5 b
d = 17.7 in
[tex]A S:=\rho \cdot b \cdot d[/tex] [tex]A s=3.78 \text { in } 2[/tex]
4-#9 is satisfied [tex]A s=4.0 \text { in } 2[/tex]
Dimensions of beam b = 12" d = 18" h = 21"
Less area of steel increases the depth.
