Exam (elaborations)
RCD3700 TEST 3 MEMO SEMESTER 2 2023
- Institution
- University Of South Africa (Unisa)
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RCD3700REINFORCED CONCRETE DESIGN IV
MAJOR TEST 3 MEMO
UNISA
UNIVERSITY OF SOUTH AFRICA
SEMESTER 2 2023
,BEAM DESIGN
SANS10100 Consider the 4 –span beam given:
1.1 Loading
Gself =24 x 0.3 x 0.6 = 4.32 kN/m
Gsdl = 10 kN/m
Gn = G self + Gsdl = 14.32 kN/m
Design Load = (1.2 x 14.32) + (1.6 x 10.00) = 33.184 kN/m
, 1.2 Check whether simplified method can be used
𝑄𝑛 10.00
= 14.32 = 0.698 < 1.25 O. K
𝐺𝑛
The load is uniformly distributed O.K
F = 33.184 x 3 = 99.552 kN
Table 4 Position Moment (kNm) Shear (Kn)
Outer support 0 0.45F = 44.80
Near centre of end 𝐹𝐿𝑒𝑓𝑓 -
span = 27.15
11
First internal support 𝐹𝐿𝑒𝑓𝑓 0.6F = 59.73
− = −33.18
9
Centre of interior 𝐹𝐿𝑒𝑓𝑓 -
span = 21.33
14
At interior supports 𝐹𝐿𝑒𝑓𝑓 0.55F = 54.75
− = −24.66
12
1.3 Design for flexure
d = 600 – 25 – 10 -20/2 = 555 mm
Lz = 0.7 x 3 = 2.1 m
4.3.1.5 beff = 300+ 3000/5 = 900 mm < 3000 mm use b =900 mm
The ultimate bending moment = 263.88 kNm
4.3.3.4.1 𝑀
𝐾=
𝑏𝑑2 𝑓𝑐𝑢
24.66 × 106
𝐾=
900 × 5552 × 30
K = 0.003 < 0.156 (No compression reinforcement is required)
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