CShape Example

See link below for the current CShapes from the AISC v16 steel database.

CShapes

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 cs

Now 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, 11200.0 ksi, 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_f

3.4 inch

The weight of the CShape:

c.weight

33.9 plf

Compression Capacity of CShape:

See link below for the available functions relating to compression for the WTShape member:

Compression API for CShapes

We can calculate the compressive capacity of the WT5X6 shape we just constructed:

L_cx = L_cy = L_cz = 12ft
ϕ_c = 0.9
P_n = cs.Compression.calc_Pn(c, L_cx, L_cy, L_cz)
ϕP_n = ϕ_c * P_n

88.44319924263088 kip

Lets see what the calc_Pn function did:

using Handcalcs
set_handcalcs(precision=2) # sets number of decimals displayed

@handcalcs P_n = cs.Compression.calc_Pn(c, L_cx, L_cy, L_cz)

\[\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_{r} &= 0.56 \cdot \sqrt{\frac{E}{F_{y}}} = 0.56 \cdot \sqrt{\frac{29000\;\mathrm{ksi}}{50\;\mathrm{ksi}}} = 13.49 \\[10pt] \text{Since: }\lambda \leq \lambda_{r} &= 5.23 \leq 13.49 = true \\[10pt] \lambda_{class} &= nonslender \\[10pt] h &= h = 12.36\;\mathrm{inch} \\[10pt] t_{w} &= t_{w} = 0.4\;\mathrm{inch} \\[10pt] \lambda &= \frac{h}{t_{w}} = \frac{12.36\;\mathrm{inch}}{0.4\;\mathrm{inch}} = 30.9 \\[10pt] \lambda_{r} &= 1.49 \cdot \sqrt{\frac{E}{F_{y}}} = 1.49 \cdot \sqrt{\frac{29000\;\mathrm{ksi}}{50\;\mathrm{ksi}}} = 35.88 \\[10pt] \text{Since: }\lambda \leq \lambda_{r} &= 30.9 \leq 35.88 = true \\[10pt] \lambda_{class} &= nonslender \\[10pt] F_{ex} &= \frac{\pi^{2} \cdot E}{\left( \frac{L_{cx}}{r_{x}} \right)^{2}} = \frac{3.14^{2} \cdot 29000\;\mathrm{ksi}}{\left( \frac{12\;\mathrm{ft}}{5.61\;\mathrm{inch}} \right)^{2}} = 434.41\;\mathrm{ksi} \\[10pt] F_{ey} &= \frac{\pi^{2} \cdot E}{\left( \frac{L_{cy}}{r_{y}} \right)^{2}} = \frac{3.14^{2} \cdot 29000\;\mathrm{ksi}}{\left( \frac{12\;\mathrm{ft}}{0.9\;\mathrm{inch}} \right)^{2}} = 11.21\;\mathrm{ksi} \\[10pt] F_{ez} &= \frac{\pi^{2} \cdot E \cdot C_{w}}{L_{cz}^{2}} \cdot \frac{1}{A_{g} \cdot \bar{r}_{0}^{2}} + G \cdot J \cdot \frac{1}{A_{g} \cdot \bar{r}_{0}^{2}} = \frac{3.14^{2} \cdot 29000\;\mathrm{ksi} \cdot 358\;\mathrm{inch}^{6}}{\left( 12\;\mathrm{ft} \right)^{2}} \cdot \frac{1}{10\;\mathrm{inch}^{2} \cdot \left( 5.94\;\mathrm{inch} \right)^{2}} + 11200\;\mathrm{ksi} \cdot 1.01\;\mathrm{inch}^{4} \cdot \frac{1}{10\;\mathrm{inch}^{2} \cdot \left( 5.94\;\mathrm{inch} \right)^{2}} = 46.07\;\mathrm{ksi} \\[10pt] F_{e} &= \frac{F_{ey} + F_{ez}}{2 \cdot H} \cdot \left( 1 - \sqrt{1 - \frac{4 \cdot F_{ey} \cdot F_{ez} \cdot H}{\left( F_{ey} + F_{ez} \right)^{2}}} \right) = \frac{434.41\;\mathrm{ksi} + 46.07\;\mathrm{ksi}}{2 \cdot 0.92} \cdot \left( 1 - \sqrt{1 - \frac{4 \cdot 434.41\;\mathrm{ksi} \cdot 46.07\;\mathrm{ksi} \cdot 0.92}{\left( 434.41\;\mathrm{ksi} + 46.07\;\mathrm{ksi} \right)^{2}}} \right) = 45.64\;\mathrm{ksi} \\[10pt] F_{e} &= \mathrm{min}\left( F_{ex}, F_{ey}, F_{e} \right) = \mathrm{min}\left( 434.41\;\mathrm{ksi}, 11.21\;\mathrm{ksi}, 45.64\;\mathrm{ksi} \right) = 11.21\;\mathrm{ksi} \\[10pt] \text{Since: }\frac{F_{y}}{F_{e}} \leq 2.25 &= \frac{50\;\mathrm{ksi}}{11.21\;\mathrm{ksi}} \leq 2.25 = false \\[10pt] F_{n} &= 0.88 \cdot F_{e} = 0.88 \cdot 11.21\;\mathrm{ksi} = 9.83\;\mathrm{ksi} \\[10pt] \text{Since: }\lambda_{wclass} &= nonslender = true \\[10pt] \text{Since: }\lambda_{fclass} &= nonslender = true \\[10pt] P_{n} &= F_{n} \cdot A_{g} = 9.83\;\mathrm{ksi} \cdot 10\;\mathrm{inch}^{2} = 98.27\;\mathrm{kip} \end{aligned}\]

Flexure Capacity of CShape:

See link below for the available functions relating to flexure for the CShape member:

Flexure API for CShapes

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_nx

172.3233007424337 ft kip

Lets see what the calc_Mnx function did:

@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_ny

18.540000000000003 ft kip

Lets see what the calc_Mny function did:

@handcalcs M_ny = cs.Flexure.calc_Mny(c)