CShape Example
See link below for the current CShapes from the AISC v16 steel database.
Constructing a CShape:
Before constructing, import the AISCSteel package. Also import the StructuralUnits package since we will use it later on.
using StructuralUnits
import AISCSteel
import AISCSteel.Shapes.CShapes as csNow that the package has been imported, lets construct a C15x33.9.
c = cs.CShape("C15x33.9")AISCSteel.Shapes.CShapes.CShape("C15X33.9", 33.9 plf, 10.0 inch^2, 15.0 inch, 3.4 inch, 0.4 inch, 0.65 inch, 1.44 inch, 12.36 inch, 0.788 inch, 0.896 inch, 0.332 inch, 315.0 inch^4, 50.8 inch^3, 42.0 inch^3, 5.61 inch, 8.07 inch^4, 6.19 inch^3, 3.09 inch^3, 0.901 inch, 1.01 inch^4, 358.0 inch^6, 15.1 inch^2, 10.4 inch^4, 7.55 inch^6, 3.81 inch^6, 14.0 inch^3, 25.2 inch^3, 5.94 inch, 0.92, 1.13 inch, 14.4 inch, 38.8 inch, 42.2 inch, 33.4 inch, 36.8 inch, 12.125 inch, 2.0 inch, 29000.0 ksi, 50.0 ksi)The following went and searched through the AISC v16 steel database and pulled the relevant info to construct a CShape. You can now access information in the struct like so:
The width of the flange:
c.b_f3.4 inchThe weight of the CShape:
c.weight33.9 plfFlexure Capacity of CShape:
See link below for the available functions relating to flexure for the CShape member:
Major Axis Bending
We can calculate the flexural capacity about the x-axis of the C15x33.9 shape we just constructed:
L_b = 5ft
ϕ_b = 0.9
M_nx = cs.Flexure.calc_Mnx(c, L_b)
ϕM_nx = ϕ_b * M_nx172.3233007424337 ft kipLets see what the calc_Mnx function did:
using Handcalcs
set_handcalcs(precision=2) # sets number of decimals displayed
@handcalcs M_nx = cs.Flexure.calc_Mnx(c, L_b)\[\begin{aligned} c &= \frac{h_{0}}{2} \cdot \sqrt{\frac{I_{y}}{C_{w}}} = \frac{14.4\;\mathrm{inch}}{2} \cdot \sqrt{\frac{8.07\;\mathrm{inch}^{4}}{358\;\mathrm{inch}^{6}}} = 1.08 \\[10pt] M_{p} &= F_{y} \cdot Z_{x} = 50\;\mathrm{ksi} \cdot 50.8\;\mathrm{inch}^{3} = 211.67\;\mathrm{ft}\,\mathrm{kip} \\[10pt] L_{p} &= 1.76 \cdot r_{y} \cdot \sqrt{\frac{E}{F_{y}}} = 1.76 \cdot 0.9\;\mathrm{inch} \cdot \sqrt{\frac{29000\;\mathrm{ksi}}{50\;\mathrm{ksi}}} = 3.18\;\mathrm{ft} \\[10pt] L_{r} &= 1.95 \cdot r_{ts} \cdot \frac{E}{0.7 \cdot F_{y}} \cdot \sqrt{\frac{J \cdot c}{S_{x} \cdot h_{0}} + \sqrt{\left( \frac{J \cdot c}{S_{x} \cdot h_{0}} \right)^{2} + 6.76 \cdot \left( \frac{0.7 \cdot F_{y}}{E} \right)^{2}}} = 1.95 \cdot 1.13\;\mathrm{inch} \cdot \frac{29000\;\mathrm{ksi}}{0.7 \cdot 50\;\mathrm{ksi}} \cdot \sqrt{\frac{1.01\;\mathrm{inch}^{4} \cdot 1.08}{42\;\mathrm{inch}^{3} \cdot 14.4\;\mathrm{inch}} + \sqrt{\left( \frac{1.01\;\mathrm{inch}^{4} \cdot 1.08}{42\;\mathrm{inch}^{3} \cdot 14.4\;\mathrm{inch}} \right)^{2} + 6.76 \cdot \left( \frac{0.7 \cdot 50\;\mathrm{ksi}}{29000\;\mathrm{ksi}} \right)^{2}}} = 11.21\;\mathrm{ft} \\[10pt] F_{cr} &= \frac{C_{b} \cdot \pi^{2} \cdot E}{\left( \frac{L_{b}}{r_{ts}} \right)^{2}} \cdot \sqrt{1 + 0.08 \cdot \frac{J \cdot c}{S_{x} \cdot h_{0}} \cdot \left( \frac{L_{b}}{r_{ts}} \right)^{2}} = \frac{1 \cdot 3.14^{2} \cdot 29000\;\mathrm{ksi}}{\left( \frac{5\;\mathrm{ft}}{1.13\;\mathrm{inch}} \right)^{2}} \cdot \sqrt{1 + 0.08 \cdot \frac{1.01\;\mathrm{inch}^{4} \cdot 1.08}{42\;\mathrm{inch}^{3} \cdot 14.4\;\mathrm{inch}} \cdot \left( \frac{5\;\mathrm{ft}}{1.13\;\mathrm{inch}} \right)^{2}} = 119.99\;\mathrm{ksi} \\[10pt] M_{nFY} &= M_{p} = 211.67\;\mathrm{ft}\,\mathrm{kip} \\[10pt] \text{Since: }L_{p} < L_{b} \leq L_{r} &= 3.18\;\mathrm{ft} < 5\;\mathrm{ft} \leq 11.21\;\mathrm{ft} = true \\[10pt] M_{nLTB} &= C_{b} \cdot \left( M_{p} - \left( M_{p} - 0.7 \cdot F_{y} \cdot S_{x} \right) \cdot \frac{L_{b} - L_{p}}{L_{r} - L_{p}} \right) = 1 \cdot \left( 211.67\;\mathrm{ft}\,\mathrm{kip} - \left( 211.67\;\mathrm{ft}\,\mathrm{kip} - 0.7 \cdot 50\;\mathrm{ksi} \cdot 42\;\mathrm{inch}^{3} \right) \cdot \frac{5\;\mathrm{ft} - 3.18\;\mathrm{ft}}{11.21\;\mathrm{ft} - 3.18\;\mathrm{ft}} \right) = 191.47\;\mathrm{ft}\,\mathrm{kip} \\[10pt] M_{n} &= \mathrm{min}\left( M_{nFY}, M_{nLTB} \right) = \mathrm{min}\left( 211.67\;\mathrm{ft}\,\mathrm{kip}, 191.47\;\mathrm{ft}\,\mathrm{kip} \right) = 191.47\;\mathrm{ft}\,\mathrm{kip} \end{aligned}\]
Minor Axis Bending
We can calculate the flexural capacity about the y-axis of the C15x33.9 shape we just constructed:
M_ny = cs.Flexure.calc_Mny(c)
ϕM_ny = ϕ_b * M_ny18.540000000000003 ft kipLets see what the calc_Mny function did:
@handcalcs M_ny = cs.Flexure.calc_Mny(c)\[\begin{aligned} b &= b_{f} = 3.4\;\mathrm{inch} \\[10pt] t &= t_{f} = 0.65\;\mathrm{inch} \\[10pt] \lambda &= \frac{b}{t} = \frac{3.4\;\mathrm{inch}}{0.65\;\mathrm{inch}} = 5.23 \\[10pt] \lambda_{p} &= 0.38 \cdot \sqrt{\frac{E}{F_{y}}} = 0.38 \cdot \sqrt{\frac{29000\;\mathrm{ksi}}{50\;\mathrm{ksi}}} = 9.15 \\[10pt] \lambda_{r} &= 1 \cdot \sqrt{\frac{E}{F_{y}}} = 1 \cdot \sqrt{\frac{29000\;\mathrm{ksi}}{50\;\mathrm{ksi}}} = 24.08 \\[10pt] \text{Since: }\lambda \leq \lambda_{p} &= 5.23 \leq 9.15 = true \\[10pt] \lambda_{class} &= compact \\[10pt] M_{p} &= \mathrm{min}\left( F_{y} \cdot Z_{y}, 1.6 \cdot F_{y} \cdot S_{y} \right) = \mathrm{min}\left( 50\;\mathrm{ksi} \cdot 6.19\;\mathrm{inch}^{3}, 1.6 \cdot 50\;\mathrm{ksi} \cdot 3.09\;\mathrm{inch}^{3} \right) = 20.6\;\mathrm{ft}\,\mathrm{kip} \\[10pt] F_{cr} &= \frac{0.7 \cdot E}{\left( \frac{b}{t_{f}} \right)^{2}} = \frac{0.7 \cdot 29000\;\mathrm{ksi}}{\left( \frac{3.4\;\mathrm{inch}}{0.65\;\mathrm{inch}} \right)^{2}} = 741.93\;\mathrm{ksi} \\[10pt] M_{nFY} &= M_{p} = 20.6\;\mathrm{ft}\,\mathrm{kip} \\[10pt] \text{Since: }\lambda_{fclass} &= compact = true \\[10pt] M_{nFLB} &= M_{p} = 20.6\;\mathrm{ft}\,\mathrm{kip} \\[10pt] M_{n} &= \mathrm{min}\left( M_{nFY}, M_{nFLB} \right) = \mathrm{min}\left( 20.6\;\mathrm{ft}\,\mathrm{kip}, 20.6\;\mathrm{ft}\,\mathrm{kip} \right) = 20.6\;\mathrm{ft}\,\mathrm{kip} \end{aligned}\]