Self-weight of the steps measured horizontally as a horizontal slab with thickness
Live load taken as it is, since the load always applies to the plan dimension.
Activity 6.1
Design a longitudinally spanning stair with a span of 2975 mm.
Given:
Nominal axial live load, 3
Span of stair, L 2975 mm
Thickness of finish, 15 mm
Stair dimension rise, R 175 mm
Tread, T 300 mm
Nosing, N 25 mm
Going, G = T-N 275 mm
Waist, W 170 mm
Initial assumption:
Assume unit width of stair is taken for analysis
Concrete unit weight 24
Characteristic concrete cube strength, 30 MPa
Yield strength of steel in tension, 450 MPa
Flexural reinforcement diam. 12 mm
Distribution reinforcement diam. 8 mm
Concrete cover 20 mm
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RCD3700 STUDY GUIDE
UNISA 6-4 ENGINEERING
Design:
Loading:
Design loading: SLS
Self-weight: = 4.836
= 2.1
= 0.393
Total self-weight: = + + = 7.329
Live load: 3
Consider a unit width of stair b = 1 m
Design loading: ULS
Total design axial load, = 13.59
Design moment:
= 12.03 kN.m
Design reinforcement:
Assume effective depth, = 145 mm
Required reinforcement
= 0.016 < 0.156. Therefore, no compression reinforcement is required.
= 142.32 mm
= 0.981 > 0.95. Therefore take = 0.95
137.75 mm
= 223.1
Using bar 10 mm 78.54
No. of bars required = 2.84. Take 3
Spacing required = 333.3. Say S = 300 mm. Since the calculated spacing gives
, spacing is therefore reduced.
Provide Y10 @ 300 mm c/c per metre width of the stair.
= 261.8 =223.1 . Therefore, it is okay.
RCD3700 STUDY GUIDE
UNISA 6-5 ENGINEERING
Span-effective depth ratio:
Assume
= 1- conservative
Service stress
= 250.23 MPa
Therefore,
Calculating the actual = 32.212 < . Therefore, it is okay.
Distribution reinforcement:
= 221
Using bar 8 mm 50.27
No. of bars required = 4.40. Take 5
Spacing required = 200. Say S = 200 mm
Provide Y8 @ 200 mm c/c per metre width of the stair.
= 251.330 =221 . Therefore, it is okay.
RCD3700 STUDY GUIDE
UNISA 6-6 ENGINEERING
