The document provides design details for stairs including steps, stair beams, and beams at half floor height. Key details include: 1) Steps are designed as z-sections made of checkered plate with dimensions calculated to support the weight and live loads. 2) Stair beams are usually channels with dimensions selected to support beam weight and live loads from steps and railings. Connections to steps are designed using butt welds. 3) Beams at half floor height are usually I-beams with dimensions selected to support reactions from the stair beam and live loads from the stairs. Connections to stair beams are also designed.

1. Stairs
1/7-Stairs

2. Stairs
2/7-Stairs
Design of step:
Steps are made of checkered plate
6/8 as shown
Assume o.w. of step = 20 kg / m
L.L = 300 → 500 kg / m2
F.C. may be used
W =
1000
20
+ (F.C. + L.L.) b
Calculate M =
8
2
wa
Q =
2
wa
Fbcx = 1.4 t/cm2
(lip is small, so neglect L.T.B.)
Check f, q, a
L
L
=
→
≤
300
δ
Width of stairs is about 1→1.5 m
Slope is 1:2 in most cases.
Design of stair beam: (Usually used as channel)
26.5
θ ≈
assume o.w. = 25 kg/m
weight of step, handrail, (all steel structure)
ws = 70 kg/m2
/hz proj
L.L = 300 → 500 kg / m2
F.C. = …….. if any
W =
1000
25
+ (70 + F.C. + L.L.) (a/2)
Draw M, Q, (neglect normal)
M =
8
2
wL
Q =
2
wL
Lu act = zero
b = 30 cm
Checkered plate
5-7
a
(6.00)
a
(4.50)
(3.00)
a
a
Beam at 1/2 floor level
R 1
M
2
M
w = ...... t/m'
1
2

3. Stairs
3/7-Stairs
Fbcx = 1.4 t/cm2
dmin of channel →
Check f, q, δ
Design of connection using butt weld:
h
hw=
h =
weld
h
2
cos
channel
C
1
3
°
2
Calculate Ix
Check y
I
M
x
= …..< 0.7 ft (allowable of good butt weld subjected to tension due
to moment or due to tension). Look ECP page
Design of beam at 1/2 floor height: (Usually IPE)
R = Reaction of stair beam
Calculate M, Q
If channel is flushed & checkered plate is
welded, so Lu act = zero
If we use grating Lu act = a
Design of connection between channel & I beam:
Force = Rt → Rss, Rb as before
3
2
1
0
L70*7
26.5
4
2
h-2t
a a
R RR R

4. Stairs
4/7-Stairs
Solved example:
It is required a complete design of stairs
(steps, beam of stairs with its connection &
beam at 1/2 floor height)
Given that: L.L. = 400 kg/m2
, F.C. =150
kg/m2
Solution:
Design of step:
Assume o.w. of step = 20 kg/m
Take shape of step as Z – section (assume b = 30cm) (checkered plate 6/8)
WDL = 20 + 150 * 0.3 = 65 kg/m
WLL = 400 * 0.3 = 120 kg/m
WT = 120 + 65 = 185 kg/m
M = 0.185 *
8
25
.
1 2
= 0.036 mt = 3.6 cmt
Q = 0.185 * (1.25/2) = 0.11 t
Ix =
12
6
.
0
*
30 3
(neglected) + 2








+ 2
3
5
.
2
*
4
.
4
*
6
.
0
12
4
.
4
*
6
.
0
= 41.52 cm4
Checks:
1. f =
52
.
41
7
.
4
*
6
.
3
= 0.41 t/cm2
< 1.4 t/cm2
(Lip is small, so we can neglect L.T.B)
2. q =
6
.
0
*
5
*
2
11
.
0
= 0.02 t/cm2
<<< 0.84 t/cm2
3. .
.L
L
δ = 3
4
10
*
52
.
41
*
2100
25
.
1
*
120
*
384
5
= 0.04 cm < =
300
125
0.42 cm
0.5
1.25
(6.00)
1.25
(4.50)
(3.00)
3.0
1.25
1.25
c
d
b
a
Beam B
1.0
1.0
Lip = 5cm
4.4
t = 0.6
30

5. Stairs
5/7-Stairs
Design of stair beam:
Assume o.w. = 25 kg/m
Weight of other steel structure is 70 kg/m
WDL = 25 + (70 + 150) * (1.25 / 2)
= 162.5 kg/m
WLL = 400 * (1.25 / 2) = 250 kg/m
Wtotal = 250 + 162.5 = 412.5 kg/m
M =
8
5
.
5
*
41
.
0 2
= 1.56 mt
Q = 0.41 * (5.5 / 2) = 1.13 t
Assume Fbcx = 1.4 t/cm2
Sx = 156 / 1.4 = 111 cm3
Choose channel 160
Checks:
Lu act = zero So No L.T.B.
f
t
c
=
05
.
1
05
.
1
75
.
0
5
.
6 −
−
= 4.48 < 10.9
w
w
t
d
= 11.6 / 0.75 = 15.5 < 82
Therefore, the section is not slender (simply symmetric channel)
1. f = 156 / 116 = 1.34 t/cm2
< 1.4 t/cm2
2. q =
75
.
0
*
16
13
.
1
= 0.09 t/cm2
< 0.84 t/cm2
3. 6
4
.
. 10
*
925
*
2100
5
.
5
*
25
.
0
*
384
5
=
L
L
δ = 1.53 cm < 550/300 = 1.83cm
1.13
w = 0.41 t/m'
1.56
1.41

6. Stairs
6/7-Stairs
Min channel:
94
.
9
2
2
=
− t
h
(From drawing)
cm
t 1
≈ h = 22 cm
Choose channel minimum 220
Design of connection butt weld:
M connection = 1.13 * 1.25 = 1.41 mt ο
25
.
13
2
=
θ
hweld =
25
.
13
cos
22
= 22.6 cm
Ix = 2
3
)
2
25
.
1
2
6
.
22
(
*
)
9
.
0
8
(
*
25
.
1
*
2
12
6
.
22
*
9
.
0
−
−
+ = 2888.4 cm4
f = 55
.
0
2
6
.
22
*
4
.
2888
141
= t/cm2
< 0.7*1.4 = 0.98 t/cm2
Design of connection between channel & beam at 1/2 floor beam:
R = 1.13 t (use only one angle)
Rss = 4
*
25
.
0
*
6
.
1
*
4
2
π
= 2.01t
tmin is smaller of 0.7 or 0.9 (tw of channel)
Rb = (0.8 * 3.6)*1.6*0.7 = 3.22 t
n = 1.13 / 2.01 = 0.56 use 2 bolts min
plate
Chequerd L70*7
=9.94
2
h-2t
30cm
5cm
L70*7
26.5
0
t=0.9
22.6
t=1.25
8

7. Stairs
7/7-Stairs
Design of beam B (at mid floor level)
Rtotal stair beam = 1.13 t, O.W. = 50 kg/m'
RLL stair beam = 0.25 * 5.5 / 2 = 0.69 t
(To be used in check of deflection)
Mconc = 2.385 x 2.25 – 1.13 x 1.25 – 0.05 x
1.252
/2= 3.83 mt
Q = 0.03 x 5 / 2 + 1.13 x 4 / 2 = 2.385 mt
Assume Fbcx = 0.64 Fy = 1.54 t/cm2
Sx = 383/ 1.54 = 259 cm3
choose IPE 220
Checks:
f
t
c
=
92
.
0
2
.
1
*
2
59
.
0
11 −
−
= 8.7 < 10.9 ,
w
w
t
d
= 21.2 / 0.59 =35.9 < 82
Assume using checkered plate welded to upper flange of beam
So Lu act = zero Fbcx = 0.64 Fy
1. f = 383 / 252 = 1.52 t/cm2
< 1.54 t/cm2
2. q =
2.38
22*0.59
= 0.18 t/cm2
< 0.84 t/cm2
3.51
0.69 1.725
0.54 0.56
0.69
0.69 0.69
1.25 1.25
3. deflection due to live load only " RLL stair beam "
ε
M = 3.51 * 2.5 – 0.56 * (0.25/2) – 0.54 * (0.5 + 1.25/3) – 1.725 * (0.25 +
1.25/2) – 0.69 * (0.25 + 1.25 + 1/3) = 5.43 m3
t
2770
*
2100
10
*
43
.
5 6
.
. =
L
L
δ = 0.93 cm < 550/300 = 1.67cm
1.25 1.25
1.13 1.13 1.13
1.00 1.00
0.5
0.05