Analysis Results

{% if outputs.type is defined %}

The analysis has converged!

Beam Properties

<table>

<tr><th>Property</th><th>Value</th></tr>

{% for key, value in outputs.beam %}

<tr><td>{{key | katex}}</td><td>{{value}}</td></tr>

{% endfor %}

</table>

{% if outputs.type == '1' %}

Class

For Class 2 classification, the following conditions must be met:

$\frac{b_{el}}{t}$ $\leq$ $\frac{170}{\sqrt{F_y}}$ for flanges and $\frac{h}{t}$ $\leq$ $\frac{1700}{\sqrt{F_y}}$ for the web.

The class for this section is {{outputs.CT}}

Plastic Moment

For class 1 and 2, the plastic moment can then be calculated using the formula: $M_p = Z_xF_y$

For class 3, the plastic moment can then be calculated using the formula: $M_y = S_xF_y$

The plastic moment is calculated as {{outputs.M}} $kNm$

Moment Gradient Factor

The moment gradient is calculated as $\omega_2 = \frac{4M_{max}}{\sqrt{M^{2}_{max} + 4M^2_a + 7M^2_b + 4M^2_c}}$

For a uniformly distributed beam $M_a, M_b, M_c$ can be calculated using:

$M_a = \frac{3wl^2}{32}$

$M_b = M_{max} = \frac{wl^2}{8}$

$M_c = \frac{3\omega l^2}{32}$

The moment gradient factor is {{outputs.w2}}

Critical Elastic Moment

The critical elastic moment can then be calculated using $M_u = \frac{\omega_2\pi}{L}\sqrt{EI_yGJ + (\frac{\pi E}{L})^2I_yC_w}$

The critical elastic moment is {{outputs.Mu}} $kNm$

Ratio

$\frac{M_u}{M_p}$ determines how $M_{rx}$ is calculated for non-laterally supported beams that are class 1 or 2.

$\frac{M_u}{M_y}$ determines how $M_{rx}$ is calculated for non-laterally supported beams that are class 3.

The ratio for this beam is {{outputs.MuM}}.

Max Factor Load

If class 1 or 2:

If $\frac{M_u}{M_p}$ $\geq$ $2.15$ , then: $M_{rx} = \phi M_{p}$

If $2.15$ $\geq$ $\frac{M_u}{M_p}$ $\geq$ $0.67$ , then: $M_{rx} = \phi 1.15M_{p}(1 - 0.28\frac{M_u}{M_p})$

If $\frac{M_u}{M_p}$ $\leq$ $0.67$ , then: $M_{rx} = \phi M_{u}$

If class 3:

If $\frac{M_u}{M_y}$ $\geq$ $2.15$ , then: $M_{rx} = \phi M_{y}$

If $2.15$ $\geq$ $\frac{M_u}{M_y}$ $\geq$ $0.67$ , then: $M_{rx} = \phi 1.15M_{y}(1 - 0.28\frac{M_u}{M_y})$

If $\frac{M_u}{M_y}$ $\leq$ $0.67$ , then: $M_{rx} = \phi M_{u}$

The max factor load is {{outputs.Mrx}} $kNm$

{%if outputs.SR != None %}

Shear Resistance

Factored Shear resistance $V_r = \phi A_wF_s$ , where:

$A_w = d * w$

$F_s = 0.66F_y$

The shear resistance for web is {{outputs.SR}} $kN$

{%endif%}

{% elif outputs.type == '2' %}

Class

For Class 2 classification, the following conditions must be met:

$\frac{b_{el}}{t}$ $\leq$ $\frac{170}{\sqrt{F_y}}$ for flanges and $\frac{h}{t}$ $\leq$ $\frac{1700}{\sqrt{F_y}}$ for the web.

The class is {{outputs.CT}}

Max Factor Load

For laterally supported beams

If class is 1 or 2: $M_{rx} = \phi Z_xF_y$

If class is 3: $M_{rx} = \phi S_xF_y$

The max factor load is {{outputs.Mrx}} $kNm$

{% if outputs.SR != None %}

Shear Resistance

Factored Shear resistance $V_r = \phi A_wF_s$ , where:

$A_w = d * w$

$F_s = 0.66F_y$

The shear resistance for web is {{outputs.SR}} $kN$

{%endif%}

{% endif %}

{%else%}

The analysis has not converged, please contact us with the configuration of your problem

{%endif%}

CSA S16-1