A Stringer beam is a structural member that

Stringer beams when utilized in staircase construction can be designed either with two edge beams (simply supported) or with a central beam (double cantilever) *See Figure 1*.

**Figure 1: Types of stringer beam**

In the design of staircases with two edge beams, the staircase section is idealized as simply supported between the two edge beams. Therefore, the beams are designed as Inverted L beams while the stair itself as simply supported slab. For the double cantilever, the staircase section is idealized as a T- section therefore the stringer beam is designed as a flanged T beam and the waste of the staircase as spanning transversely (like a cantilever). In each case a primary beam/support will be required to support the stringer beam. These supports are normally designed as carrying point loads from the stringer beam in addition to any other load they may be carrying.

### Worked Example

*Figure 2.0 *shows the plan of the second flight of a staircase spanning 7.0m between supports. Assuming we design a staircase spanning transversely and supported by beams at the supports, this would result in an uneconomical staircase section due to the thickness of the waste that would ensue and quantity of steel that will be required to control deflection. A Stringer beam is therefore provided in order to make take care of the foregoing reasons. In this case a central stringer beam (double cantilever) type is designed using the design data in

*Table 1*.

*Figure 2: Worked example-stair plan*

Concrete | C20/25 |

Reinforcement | 410Mpa |

Stair width | 1500mm |

Risers | 150mm |

Threads | 300mm |

Imposed Loading (Mall) | 4kN/m^{2} |

*Table 1: Design Data*#### Actions

##### Permanent Actions** **

Permanent action includes the weight of the stair waste, weight of steps, load from finishes and balustrades. Because the waste of the stair is inclined, the permanent actions due to the slab waste, finishes and any other applied loads except the step must be corrected by a slope factor.

a.\quad 150mm\quad concrete\quad waste\quad \\ \quad =0.15\times 25\quad =3.75kN/{ m }^{ 2}

b.\quad finishes =1.5kN/{ m }^{ 2 }

c.\quad balustrade =0.5kN/{ m }^{ 2 }

d.\quad steps=\frac { 0.15 }{ 2 } \times 25=1.88kN/{ m }^{ 2 }

g_{k}=(3.75+1.5+0.5)\times1.22+1.88 \\ =8.32kN/m^{2}

##### Variable Actions

b.\quad Imposed\quad load =4kN/{ m }^{ 2 }

q_{k} =4kN/m^{2}

##### Design Value of Actions

Using Eqn 6.10 of BS EN 1990:

n_{s}=\quad 1.35{ g }_{ k }+{ 1.5q }_{ k }

=(1.35\times8.32)+(1.5\times4)\\ \qquad \qquad \qquad = 17.2kN/{ m }^{ 2 }

##### Actions on Stringer Beam

a.\quad stair =17.2\times1.5=26.6kN/m^{2}

b.\quad beam=(0.225\times0.3)\times 25\times1.35 =2.3kN/m^{2}

w=26.6+2.3=28.9kN/m^{2}

#### Structural Analysis

Since the beam is simply supported, the equation for obtaining the bending moment and shear force is pretty straightforward.

##### Bending Moment

M_{Ed}=\frac { w l^{2} }{ 8 } = \frac { 28.9 \times7^{2} }{ 8 }=177kN.m

##### Shear Force

V_{Ed}=\frac { w l }{ 2 } = \frac { 28.9 \times7 }{ 2}=101.2kN

#### Flexural Design

M_{Ed}=177kN.m

Assuming cover to reinforcement of 30mm, 20mm tensile bars in two layers, 16mm compression bars & 8mm links

{ d }^{ ' }=\left( { c }_{ nom }+links+\phi /2 \right) \\=50+10+16/2=68mm

{ d }=h-\left( { c }_{ nom }+links+\phi /2+\phi \right)\\=450-\left( 25+8+20/2+20 \right) =387mm

b={ b }_{ eff }=1500mm

k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 177\times { 10 }^{ 6 } }{ 1500\times { 387 }^{ 2 }\times 20 } =0.039

z=0.95d=0.95\times387=367.7mm

{ A }_{ s,req }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 177\times { 10 }^{ 6 } }{ 0.87\times 410\times 367.7 } \\=1349.5{ mm }^{ 2 }

**Use 5Y20mm bars Bottom. (Tw****o layers)**** (As, prov = 1570mm2)**

#### Shear Design

{ V }_{ Ed }=\quad 101.2kN

By inspection, the maximum shear force is less than max. shear resistance of the concrete section, therefore the section is adequate however, nominal shear reinforcement must be provided:

\theta =0.5{ sin }^{ -1 }\left( \frac { { 5.56V }_{ Ed } }{ { b }_{ w }d\left( 1-{ f }_{ ck }/250 \right) { f }_{ ck } } \right)

=0.5{ sin }^{ -1 }\left( \frac { 5.56\times 101.2\times { 10 }^{ 3 } }{ 225\times 387\left( 1-20/250 \right) 20 } \right) \\=10.3

cot\theta =5.5>\quad 2.5\quad ;\quad take\quad cot\theta =2.5

Therefore, the area of shear reinforcement required:

\frac { { A }_{ sv } }{ { s }_{ v } } \ge \frac { { V }_{ Ed } }{ zcot\theta { f }_{ wd } }

=\frac { 101.2\times { 10 }^{ 3 } }{ \left( 0.9\times 387\times 2.5 \right) \times 0.87\times410 } =0.33

max\quad spacing\quad =0.75d=290.25mm

**Use Y8mm – 200mm cente****rs (0.51)**

**Deflection Verification **

Deflection of the Stringer beam is carried out by censuring that the actual span – depth ratio is less than the limiting value. In this case, this beam has been sized conservatively based on the preliminary sizing rules, hence the check shouldn’t be critical.

##### Limiting Span-Depth Ratio

{ \left[ \frac { l }{ d } \right] }_{ limit }=N\cdot K\cdot F1\cdot F2\cdot F3

\rho =\frac { { A }_{ s,req } }{ { A }_{ c } }

=\frac { 1349.5 }{ \left( 1500\times 387 \right) } =0.0023

{ \rho }_{ o }={ 10 }^{ -3 }\sqrt { { f }_{ ck } } ={ 10 }^{ -3 }\sqrt { 30 } =0.0055

N=11+\frac { 1.5\sqrt { { f }_{ ck } } { \rho }_{ o } }{ \rho } +3.2\sqrt { { f }_{ ck } } { \left( \frac { { \rho }_{ o } }{ \rho } \right) }^{ 3/2 } \\ =79.9

\frac { { b }_{ eff } }{ b } =\frac { 1500 }{ 225 } =6.7>3\quad ;\quad F1\quad =0.8

F2\quad =\frac { 7 }{ { l }_{ eff } } =\frac { 7 }{ 7 } =1\quad \\ K=1.3\quad (end\quad spans)

F3=\frac { 310 }{ { \sigma }_{ s } } \le 1.5

{ \sigma }_{ s }=\frac { { f }_{ yk } }{ { \gamma }_{ s } } \left[ \frac { { g }_{ k }+{ \varphi }_{ 2 }{ q }_{ k } }{ { w }_{ d } } \right] \left( \frac { { A }_{ s,req } }{ { A }_{ s,pro } } \right)

=\frac { 460 }{ 1.15 } \left[ \frac { 8.32+0.6(4) }{ 17.2 } \right] \frac { 1349.5 }{ 1570 } \\=214.3Mpa

F3=\frac { 310 }{ 214.3 } =1.45\le 1.5

\left[ \frac { l }{ d } \right] _{ limit }=79.9\times 1.3\times 0.8\times 1\times 1.45\\ =120.5

##### Actual Span – Depth Ratio

{ \left[ \frac { l }{ d } \right] }_{ Actual }=\frac { span }{ effective\quad depth }

= \frac { 7000 }{ 387 } =18.09

The limiting span-effective depth ratio is greater than actual; thus, deflection is deemed satisfactory.

See: Preliminary Sizing of Structural Elements

**Stair Waste **

The stair waste itself require steel bars at the top in-order to accommodate the imposed tensile stresses on the cantilever. This can be designed for the stair design load, as a cantilever slab, however, for most staircases nominal reinforcement steel based on minimum area of steel should be sufficient to satisfy this requirement.

**Use Y10mm – 200mm centers both**** ways (Top) (****393mm2/m)**

figure 3 shown below, shows the typical section through the stair flight.

**Figure 3: Typical section through stair flight**

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