RCD3700 Assignment 01 Answers Year 2025

Reinforced Concrete Design IV

Assessment 01 Answers

Year 2025




0027 63 985 5033

,
,
,QUESTION 1

BEAM DESIGN



SANS10100 1.1 Consider the 3 –span beam given:

Loading

Total dead load, Gn equals self-weight, Gself, plus superimposed dead loads,Gsdl

Self-weight of beam, Gself = 24 x 0.5 x 0.4 = 4.8 kN/m

Gsdl = 5 kN/m

Total dead load = 5 + 4.8 = 9.8 kN/m

Live load = 12 kN/m

Design Load, wu = (1.2 x 9.8) + (1.6 x 12) = 30.96 kN/m



Load case 1




Load case 2




Load case 3

, 1.2 Load case 1




Using Moment Distribution Method

Fixed End Moments

30.96 × 72
𝑀𝐹𝐴𝐵 = − = −126.42 𝑘𝑁𝑚
12

30.96 × 72
𝑀𝐹𝐵𝐴 = = 126.42 𝑘𝑁𝑚
12

30.96 × 52
𝑀𝐹𝐵𝐶 = − = −64.50 𝑘𝑁𝑚
12

30.96 × 52
𝑀𝐹𝐶𝐵 = = 64.50 𝑘𝑁𝑚
12

30.96 × 42
𝑀𝐹𝐶𝐷 = − = −41.28 𝑘𝑁𝑚
12

30.96 × 42
𝑀𝐹𝐷𝐶 = = 41.28 𝑘𝑁𝑚
12

Joint Member 𝒌 ∑𝒌 Distribution Factor
𝒌
𝑫𝑭 = ∑
𝒌

4𝐸𝐼 0.571𝐸𝐼
B BA = 0.571𝐸𝐼 = 0.416
7 1.371𝐸𝐼
0.571EI+0.8EI
4𝐸𝐼 = 1.371EI 0.8𝐸𝐼
BC = 0.8𝐸𝐼 = 0.584
5 1.371𝐸𝐼

4𝐸𝐼 0.8𝐸𝐼
C CB = 0.8𝐸𝐼 0.8EI+0.75EI = 0.516
5 1.551𝐸𝐼
= 1.55EI
3𝐸𝐼 0.75𝐸𝐼
CD = 0.75𝐸𝐼 = 0.484
4 1.551𝐸𝐼



Note: Stiffness coefficients
4𝐸𝐼
Far end is fixed =
𝐿

3𝐸𝐼
Far end is hinged or with roller support =
𝐿

4𝐸𝐼
Far end is continuous =
𝐿