RoundHSS_Shape Example

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

Round HSS Shapes

Constructing a RoundHSS_Shape:

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.RoundHSS_Shapes as rhss

Now that the package has been imported, lets construct a HSS20.000x0.500.

rhss_shape = rhss.RoundHSS_Shape("HSS20.000x0.500", F_y=46ksi)

AISCSteel.Shapes.RoundHSS_Shapes.RoundHSS_Shape("HSS20.000X0.500", 104.0 plf, 28.5 inch^2, 20.0 inch, 0.5 inch, 0.465 inch, 1360.0 inch^4, 177.0 inch^3, 136.0 inch^3, 6.91 inch, 1360.0 inch^4, 177.0 inch^3, 136.0 inch^3, 6.91 inch, 2720.0 inch^4, 272.0 inch^3, 29000.0 ksi, 46.0 ksi)

The following went and searched through the AISC v16 steel database and pulled the relevant info to construct a RoundHSS_Shape. You can now access information in the struct like so:

The overall diameter of the shape:

rhss_shape.OD

20.0 inch

The weight of the RoundHSS_Shape:

rhss_shape.weight

104.0 plf

Flexure Capacity of RoundHSS_Shape:

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

Flexure API for Round HSS Shapes

Compact Shape

We can calculate the flexural capacity of the HSS20.000x0.500 shape we just constructed:

ϕ_b = 0.9
M_n = rhss.Flexure.calc_Mn(rhss_shape)
ϕM_nx = ϕ_b * M_n

610.6500000000001 ft kip

Lets see what the calc_Mn function did:

using Handcalcs
@handcalcs M_n = rhss.Flexure.calc_Mn(rhss_shape)

\[\begin{aligned} \lambda &= \frac{D}{t} = \frac{20\;\mathrm{inch}}{0.465\;\mathrm{inch}} = 43.01075268817204 \\[10pt] \lambda_{p} &= \frac{0.07 \cdot E}{F_{y}} = \frac{0.07 \cdot 2.9 \cdot 10^{4}\;\mathrm{ksi}}{46\;\mathrm{ksi}} = 44.1304347826087 \\[10pt] \lambda_{r} &= \frac{0.31 \cdot E}{F_{y}} = \frac{0.31 \cdot 2.9 \cdot 10^{4}\;\mathrm{ksi}}{46\;\mathrm{ksi}} = 195.43478260869566 \\[10pt] \text{Since: }\lambda \leq \lambda_{p} &= 43.01075268817204 \leq 44.1304347826087 = true \\[10pt] class &= compact \\[10pt] M_{p} &= F_{y} \cdot Z = 46\;\mathrm{ksi} \cdot 177\;\mathrm{inch}^{3} = 678.5\;\mathrm{ft}\,\mathrm{kip} \\[10pt] M_{nFY} &= M_{p} = 678.5\;\mathrm{ft}\,\mathrm{kip} \\[10pt] \text{Since: }\lambda_{class} &= compact = true \\[10pt] M_{nLB} &= M_{p} = 678.5\;\mathrm{ft}\,\mathrm{kip} \\[10pt] M_{n} &= \mathrm{min}\left( M_{nY}, M_{nLB} \right) = \mathrm{min}\left( 678.5\;\mathrm{ft}\,\mathrm{kip}, 678.5\;\mathrm{ft}\,\mathrm{kip} \right) = 678.5\;\mathrm{ft}\,\mathrm{kip} \end{aligned}\]

Non-Compact Shape

We can calculate the flexural capacity of the HSS20.000x0.375 shape:

rhss_shape = rhss.RoundHSS_Shape("HSS20.000x0.375", F_y=46ksi)
ϕ_b = 0.9
M_n = rhss.Flexure.calc_Mn(rhss_shape)
ϕM_nx = ϕ_b * M_n

441.69099000000006 ft kip

Lets see what the calc_Mn function did:

using Handcalcs
@handcalcs M_n = rhss.Flexure.calc_Mn(rhss_shape)

\[\begin{aligned} \lambda &= \frac{D}{t} = \frac{20\;\mathrm{inch}}{0.349\;\mathrm{inch}} = 57.30659025787966 \\[10pt] \lambda_{p} &= \frac{0.07 \cdot E}{F_{y}} = \frac{0.07 \cdot 2.9 \cdot 10^{4}\;\mathrm{ksi}}{46\;\mathrm{ksi}} = 44.1304347826087 \\[10pt] \lambda_{r} &= \frac{0.31 \cdot E}{F_{y}} = \frac{0.31 \cdot 2.9 \cdot 10^{4}\;\mathrm{ksi}}{46\;\mathrm{ksi}} = 195.43478260869566 \\[10pt] \text{Since: }\lambda_{p} < \lambda \leq \lambda_{r} &= 44.1304347826087 < 57.30659025787966 \leq 195.43478260869566 = true \\[10pt] class &= noncompact \\[10pt] M_{p} &= F_{y} \cdot Z = 46\;\mathrm{ksi} \cdot 135\;\mathrm{inch}^{3} = 517.5\;\mathrm{ft}\,\mathrm{kip} \\[10pt] M_{nFY} &= M_{p} = 517.5\;\mathrm{ft}\,\mathrm{kip} \\[10pt] \text{Since: }\lambda_{class} &= noncompact = true \\[10pt] M_{nLB} &= \left( \frac{0.021 \cdot E}{\frac{D}{t}} + F_{y} \right) \cdot S = \left( \frac{0.021 \cdot 2.9 \cdot 10^{4}\;\mathrm{ksi}}{\frac{20\;\mathrm{inch}}{0.349\;\mathrm{inch}}} + 46\;\mathrm{ksi} \right) \cdot 104\;\mathrm{inch}^{3} = 490.8\;\mathrm{ft}\,\mathrm{kip} \\[10pt] M_{n} &= \mathrm{min}\left( M_{nY}, M_{nLB} \right) = \mathrm{min}\left( 517.5\;\mathrm{ft}\,\mathrm{kip}, 490.8\;\mathrm{ft}\,\mathrm{kip} \right) = 490.8\;\mathrm{ft}\,\mathrm{kip} \end{aligned}\]