Back

Square Tee Beam

Open in Enji.io

The square tee-beam (T-section), potentially with flange width and stem depth being comparable, consists of a horizontal flange and a vertical web (stem) forming a 'T' shape. It is efficient when the flange is in compression, as in floor systems where the T-beam is part of a slab-beam construction, or as lintels. It's an open section with limited torsional resistance. Applications include built-up members, stiffeners, hangers, and elements in truss structures.

Note: To use these calculations, open the page on your desktop to access the app link.

Calculations

Description

Symbol Name

Value

Unit

Comment

Squared Tee Beam

figure wim4408ab57b0

Section depth

d_{t}

100.00

mm

Section breadth

b_{t}

50.00

mm

Flange thickness

s_{t}

10.00

mm

Web thickness

t_{t}

10.00

mm

Web depth

a_{t}

90.00

mm

a_{t}

=

d_{t} - s_{t}

=

100.00 - 10.00

Area

A_{t}

1400.00

{mm}^{2}

A_{t}

=

b_{t} \cdot s_{t} + a_{t} \cdot t_{t}

=

50.00 \cdot 10.00 + 90.00 \cdot 10.00

Perimeter

P_{t}

300.00

mm

P_{t}

=

2 \cdot (b_{t} + a_{t} + s_{t})

=

2 \cdot (50.00 + 90.00 + 10.00)

Distance to centroid (x-axis)

C_{x}

0

mm

Distance to centrois (y-axis)

C_{y}

62.86

mm

C_{y}

=

d_{t} - \frac{({d_{t}}^{2} \cdot t_{t} + {s_{t}}^{2} \cdot (b_{t} - t_{t}))}{(2 \cdot (b_{t} \cdot s_{t} + a_{t} \cdot t_{t}))}

=

100.00 - \frac{({100.00}^{2} \cdot 10.00 + {10.00}^{2} \cdot (50.00 - 10.00))}{(2 \cdot (50.00 \cdot 10.00 + 90.00 \cdot 10.00))}

Second moment of area about x-axis

I_{x}

(1.415e+6)

{mm}^{4}

I_{x}

=

\frac{(t_{t} \cdot {C_{y}}^{3} + b_{t} \cdot {(d_{t} - C_{y})}^{3} - (b_{t} - t_{t}) \cdot {(d_{t} - C_{y} - s_{t})}^{3})}{3}

=

\frac{(10.00 \cdot {62.86}^{3} + 50.00 \cdot {(100.00 - 62.86)}^{3} - (50.00 - 10.00) \cdot {(100.00 - 62.86 - 10.00)}^{3})}{3}

Second moment of area about y-axis

I_{y}

111667

{mm}^{4}

I_{y}

=

\frac{a_{t} \cdot {t_{t}}^{3}}{12} + \frac{{b_{t}}^{3} \cdot s_{t}}{12}

=

\frac{90.00 \cdot {10.00}^{3}}{12} + \frac{{50.00}^{3} \cdot 10.00}{12}

Second moment of area about x1-axis

I_{x_{1}}

(6.947e+6)

{mm}^{4}

I_{x_{1}}

=

I_{x} + A_{t} \cdot {C_{y}}^{2}

=

(1.415e+6) + 1400.00 \cdot {62.86}^{2}

Second moment of area about y1-axis

I_{y_{1}}

111667

{mm}^{4}

I_{y_{1}}

=

I_{y} + A_{t} \cdot {C_{x}}^{2}

=

111667 + 1400.00 \cdot 0^{2}

Polar moment of inertia about centre

J_{z}

(1.527e+6)

{mm}^{4}

J_{z}

=

I_{x} + I_{y}

=

(1.415e+6) + 111667

Polar moment of inertia about vertex

J_{z_{1}}

(7.058e+6)

{mm}^{4}

J_{z_{1}}

=

I_{x_{1}} + I_{y_{1}}

=

(6.947e+6) + 111667

Radius of gyration about x-axis

K_{x}

31.79

mm

K_{x}

=

\sqrt{\frac{I_{x}}{A_{t}}}

=

\sqrt{\frac{(1.415e+6)}{1400.00}}

Radius of gyration about y-axis

K_{y}

8.93

mm

K_{y}

=

\sqrt{\frac{I_{y}}{A_{t}}}

=

\sqrt{\frac{111667}{1400.00}}

Radius of gyration about z-axis

K_{z}

33.02

mm

K_{z}

=

\sqrt{{K_{x}}^{2} + {K_{y}}^{2}}

=

\sqrt{{31.79}^{2} + {8.93}^{2}}

Radius of gyration about x1-axis

K_{x_{1}}

70.44

mm

K_{x_{1}}

=

\sqrt{\frac{I_{x_{1}}}{A_{t}}}

=

\sqrt{\frac{(6.947e+6)}{1400.00}}

Radius of gyration about y1-axis

K_{y_{1}}

8.93

mm

K_{y_{1}}

=

\sqrt{\frac{I_{y_{1}}}{A_{t}}}

=

\sqrt{\frac{111667}{1400.00}}

Radius of gyration about z1-axis

K_{z_{1}}

71.00

mm

K_{z_{1}}

=

\sqrt{{K_{x_{1}}}^{2} + {K_{y_{1}}}^{2}}

=

\sqrt{{70.44}^{2} + {8.93}^{2}}

Elastic section modulus about x-axis

S_{x}

22515.2

{mm}^{3}

S_{x}

=

\frac{I_{x}}{C_{y}}

=

\frac{(1.415e+6)}{62.86}

Elastic section modulus about y-axis

S_{y}

4466.67

{mm}^{3}

S_{y}

=

\frac{I_{y}}{(\frac{b_{t}}{2})}

=

\frac{111667}{(\frac{50.00}{2})}

Z_{x,1}

39000.0

{mm}^{2}

Z_{x,1}

=

\frac{t_{t} \cdot {(d_{t} - s_{t})}^{2}}{4} + \frac{b_{t} \cdot d_{t} \cdot s_{t}}{2} - \frac{{b_{t}}^{2} \cdot {s_{t}}^{2}}{(4 \cdot t_{t})}

=

\frac{10.00 \cdot {(100.00 - 10.00)}^{2}}{4} + \frac{50.00 \cdot 100.00 \cdot 10.00}{2} - \frac{{50.00}^{2} \cdot {10.00}^{2}}{(4 \cdot 10.00)}

Z_{x,2}

42200.0

{mm}^{2}

Z_{x,2}

=

\frac{t_{t} \cdot {d_{t}}^{2}}{2} + \frac{b_{t} \cdot {s_{t}}^{2}}{4} - \frac{d_{t} \cdot s_{t} \cdot t_{t}}{2} - \frac{{(d_{t} - s_{t})}^{2} \cdot {s_{t}}^{2}}{(4 \cdot b_{t})}

=

\frac{10.00 \cdot {100.00}^{2}}{2} + \frac{50.00 \cdot {10.00}^{2}}{4} - \frac{100.00 \cdot 10.00 \cdot 10.00}{2} - \frac{{(100.00 - 10.00)}^{2} \cdot {10.00}^{2}}{(4 \cdot 50.00)}

Plastic modulus arounx the x-axis

Z_{x}

39000.0

{mm}^{3}

Z_{x}

=

{\begin{cases} Z_{x,2} & if s_{t} > \frac{A_{t}}{(2 \cdot b_{t})} \\ Z_{x,1} & otherwise. \end{cases}

=

{\begin{cases} 42200.0 & if 10.00 > \frac{1400.00}{(2 \cdot 50.00)} \\ 39000.0 & otherwise. \end{cases}

Plastic modulus arounx the y-axis

Z_{y}

8500.00

{mm}^{3}

Z_{y}

=

\frac{s_{t} \cdot {b_{t}}^{2}}{4} + \frac{(d_{t} - s_{t}) \cdot {t_{t}}^{2}}{4}

=

\frac{10.00 \cdot {50.00}^{2}}{4} + \frac{(100.00 - 10.00) \cdot {10.00}^{2}}{4}