Tapered Tee Beam

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The tapered tee-beam is a T-section where the depth of the stem or the width of the flange varies along its length. This design provides material efficiency by adapting the section's capacity to the varying stress demands along the member. It's often used for cantilevered brackets, supporting elements with non-uniform loads, or as stiffeners where the force transitions.

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Calculations

Description

Symbol Name

Value

Unit

Comment

Tapered Tee Beam

Depth

d_{t}

100.00

mm

Breadth

b_{t}

50.00

mm

Flange thickness

s_{t}

10.00

mm

Major web thickness

T_{t}

10.00

mm

Minor 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} + \frac{a_{t} \cdot (T_{t} + t_{t})}{2}

=

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

Perimeter

P_{t}

300.00

mm

P_{t}

=

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

=

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

Distance to centroid (x-axis)

C_{x}

25.00

mm

C_{x}

=

\frac{b_{t}}{2}

=

\frac{50.00}{2}

Flange area

A_{f}

500.00

{mm}^{2}

A_{f}

=

b_{t} \cdot s_{t}

=

50.00 \cdot 10.00

Flange centroid

y_{f}

95.00

mm

y_{f}

=

a_{t} + \frac{s_{t}}{2}

=

90.00 + \frac{10.00}{2}

Web area

A_{w}

900.00

{mm}^{2}

A_{w}

=

\frac{(T_{t} + t_{t})}{2} \cdot a_{t}

=

\frac{(10.00 + 10.00)}{2} \cdot 90.00

Web centroid

y_{w}

45.00

mm

y_{w}

=

\frac{a_{t}}{3} \cdot (\frac{(2 \cdot T_{t} + t_{t})}{(T_{t} + t_{t})})

=

\frac{90.00}{3} \cdot (\frac{(2 \cdot 10.00 + 10.00)}{(10.00 + 10.00)})

Distance to centroid (y-axis)

C_{y}

62.86

mm

C_{y}

=

\frac{(A_{f} \cdot y_{f} + A_{w} \cdot y_{w})}{A_{t}}

=

\frac{(500.00 \cdot 95.00 + 900.00 \cdot 45.00)}{1400.00}

Flange term

I_{x,f}

520748

{mm}^{4}

I_{x,f}

=

(\frac{b_{t} \cdot {s_{t}}^{3}}{12} + A_{f} \cdot {(y_{f} - C_{y})}^{2})

=

(\frac{50.00 \cdot {10.00}^{3}}{12} + 500.00 \cdot {(95.00 - 62.86)}^{2})

Web term

I_{x,w}

894490

{mm}^{4}

I_{x,w}

=

(\frac{{a_{t}}^{3} \cdot (t_{t} + 3 \cdot T_{t})}{12} - A_{w} \cdot {y_{w}}^{2}) + A_{w} \cdot {(y_{w} - C_{y})}^{2}

=

(\frac{{90.00}^{3} \cdot (10.00 + 3 \cdot 10.00)}{12} - 900.00 \cdot {45.00}^{2}) + 900.00 \cdot {(45.00 - 62.86)}^{2}

Second moment of area about x-axis

I_{x}

(1.415e+6)

{mm}^{4}

I_{x}

=

I_{x,f} + I_{x,w}

=

520748 + 894490

Second moment of area about x1-axis

I_{x_{1}}

(6.947e+6)

{mm}^{4}

I_{x_{1}}

=

\frac{b_{t} \cdot {s_{t}}^{3}}{12} + b_{t} \cdot s_{t} \cdot {(a_{t} + \frac{s_{t}}{2})}^{2} + \frac{{a_{t}}^{3}}{12} \cdot (t_{t} + 3 \cdot T_{t})

=

\frac{50.00 \cdot {10.00}^{3}}{12} + 50.00 \cdot 10.00 \cdot {(90.00 + \frac{10.00}{2})}^{2} + \frac{{90.00}^{3}}{12} \cdot (10.00 + 3 \cdot 10.00)

Radius or 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 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}}

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}

Calculations for plastic section moduli

Taper

m_{t}

m_{t}

=

\frac{(t_{t} - T_{t})}{a_{t}}

=

\frac{(10.00 - 10.00)}{90.00}

Plastic neutral axis about x-axis

y_{PNA}

30.00

mm

y_{PNA}

=

{\begin{cases} \frac{A_{t}}{(2 \cdot b_{t})} & if \frac{A_{t}}{2} \leq A_{f} \\ {\begin{cases} \frac{(\frac{A_{t}}{2} - A_{f})}{T_{t}} + s_{t} & if m_{t} = 0 \\ s_{t} + (\frac{(\sqrt{{T_{t}}^{2} - m_{t} \cdot (A_{f} - A_{w})} - T_{t})}{m_{t}}) & otherwise. \end{cases} & otherwise. \end{cases}

=

{\begin{cases} \frac{1400.00}{(2 \cdot 50.00)} & if \frac{1400.00}{2} \leq 500.00 \\ {\begin{cases} \frac{(\frac{1400.00}{2} - 500.00)}{10.00} + 10.00 & if 0 = 0 \\ 10.00 + (\frac{(\sqrt{{10.00}^{2} - 0 \cdot (500.00 - 900.00)} - 10.00)}{0}) & otherwise. \end{cases} & otherwise. \end{cases}

Parameters for the PNA in the web case

z_{PNA}

20.00

mm

z_{PNA}

=

y_{PNA} - s_{t}

=

30.00 - 10.00

w_{PNA}

10.00

mm

w_{PNA}

=

T_{t} + m_{t} \cdot z_{PNA}

=

10.00 + 0 \cdot 20.00

A_{w,t}

200.00

{mm}^{2}

A_{w,t}

=

\frac{(z_{PNA} \cdot (T_{t} + w_{PNA}))}{2}

=

\frac{(20.00 \cdot (10.00 + 10.00))}{2}

y_{c,t}

10.00

mm

y_{c,t}

=

z_{PNA} - \frac{(z_{PNA} \cdot (2 \cdot T_{t} + w_{PNA}))}{(3 \cdot (T_{t} + w_{PNA}))}

=

20.00 - \frac{(20.00 \cdot (2 \cdot 10.00 + 10.00))}{(3 \cdot (10.00 + 10.00))}

Z_{w,t}

2000.00

{mm}^{3}

Z_{w,t}

=

A_{w,t} \cdot y_{c,t}

=

200.00 \cdot 10.00

A_{w,b}

700.00

{mm}^{2}

A_{w,b}

=

\frac{(d_{t} - y_{PNA}) \cdot (w_{PNA} + t_{t})}{2}

=

\frac{(100.00 - 30.00) \cdot (10.00 + 10.00)}{2}

y_{c,b}

35.00

mm

y_{c,b}

=

\frac{((d_{t} - y_{PNA}) \cdot (2 \cdot w_{PNA} + t_{t}))}{(3 \cdot (w_{PNA} + t_{t}))}

=

\frac{((100.00 - 30.00) \cdot (2 \cdot 10.00 + 10.00))}{(3 \cdot (10.00 + 10.00))}

PNA in the web case

Z_{w,b}

24500.0

{mm}^{3}

Z_{w,b}

=

A_{w,b} \cdot y_{c,b}

=

700.00 \cdot 35.00

Final Plastic Section Modulus X-axis

Plastic section modulus abouth x-axis

Z_{x}

39000.0

{mm}^{3}

Z_{x}

=

{\begin{cases} \frac{b_{t} \cdot {y_{PNA}}^{2}}{2} + \frac{b_{t} \cdot {(s_{t} - y_{PNA})}^{2}}{2} + A_{w} \cdot (y_{w} - y_{PNA}) & if y_{PNA} \leq s_{t} \\ A_{f} \cdot (y_{PNA} - \frac{s_{t}}{2}) + Z_{w,t} + Z_{w,b} & otherwise. \end{cases}

=

{\begin{cases} \frac{50.00 \cdot {30.00}^{2}}{2} + \frac{50.00 \cdot {(10.00 - 30.00)}^{2}}{2} + 900.00 \cdot (45.00 - 30.00) & if 30.00 \leq 10.00 \\ 500.00 \cdot (30.00 - \frac{10.00}{2}) + 2000.00 + 24500.0 & otherwise. \end{cases}

Plastic Section Modulus Y-axis

Plastic neutral axis about y-axis

x_{PNA}

25.00

mm

x_{PNA}

=

C_{x}

=

25.00

Plastic section modulus about y-axis

Z_{y}

8500.00

{mm}^{3}

Z_{y}

=

\frac{s_{t} \cdot {b_{t}}^{2}}{4} + \frac{a_{t}}{4} \cdot (T_{t} \cdot t_{t} + \frac{{(t_{t} - T_{t})}^{2}}{3})

=

\frac{10.00 \cdot {50.00}^{2}}{4} + \frac{90.00}{4} \cdot (10.00 \cdot 10.00 + \frac{{(10.00 - 10.00)}^{2}}{3})