RCD3700
Assignment 1
DUE: 22 May 2026
, UNIVERSITY OF SOUTH AFRICA
RCD3700 – Reinforced Concrete Design IV
Assessment 1 – May 2026
Examiner: Mr. Ngonyama | Total Marks: 70
QUESTION 1 [20 Marks] – Continuous Beam Analysis
Problem Description
A two-span continuous beam is fixed at end A, supported on rollers at B (interior) and C
(free end). The beam cross-section is 400 mm × 600 mm.
Spans: L₁ = 7 m (Span AB), L₂ = 5 m (Span BC)
Unfactored Dead Load (excluding self-weight): wDL = 10 kN/m
Unfactored Live Load: wLL = 8 kN/m
1.1 Design Load Calculation [5 marks]
Self-Weight of Beam
Self-weight, wsw = γc × b × h = 24 × 0.4 × 0.6 = 5.76 kN/m
Total Unfactored Dead Load
wDL,total = wDL + wsw = 10 + 5.76 = 15.76 kN/m
Factored Design Loads (BS8110 / SABS 0100 load combinations)
Load factors: γDL = 1.4, γLL = 1.6
, Load Case Description Span AB (w₁, kN/m) Span BC (w₂, kN/m)
LC1 Maximum load – full 1.4(15.76) + 1.6(8) = 1.4(15.76) + 1.6(8) =
LL on both spans 34.86 34.86
LC2 Max sagging in span 1.4(15.76) + 1.6(8) = 1.4(15.76) = 22.06
1 – full LL on AB, DL 34.86
only on BC
LC3 Max sagging in span 1.4(15.76) = 22.06 1.4(15.76) + 1.6(8) =
2 – DL only on AB, 34.86
full LL on BC
1.2 Moments and Shear at Critical Sections [12 marks]
Method: Three-Moment Equation (Clapeyron's Theorem)
Boundary conditions: Fixed end at A (slope = 0), rollers at B and C (M_C = 0).
To model the fixed end, an imaginary span of zero length is introduced to the left of A.
Equation (i) – from imaginary span and Span AB:
2·M_A·L₁ + M_B·L₁ = −w₁·L₁³/4
=> 14·M_A + 7·M_B = −w₁·(343)/4
Equation (ii) – from Span AB and Span BC:
M_A·L₁ + 2·M_B·(L₁+L₂) + M_C·L₂ = −w₁·L₁³/4 − w₂·L₂³/4
=> 7·M_A + 24·M_B = −w₁·(343)/4 − w₂·(125)/4 [since M_C = 0]
LOAD CASE 1 – Full Design Load on Both Spans
w₁ = w₂ = 34.86 kN/m
Eq(i): 14·M_A + 7·M_B = −1.4 × 34.86 × 343/4 = −2,989.6
Eq(ii): 7·M_A + 24·M_B = −34.86×343/4 − 34.86×125/4 = −4,079.1
Solving simultaneously:
Assignment 1
DUE: 22 May 2026
, UNIVERSITY OF SOUTH AFRICA
RCD3700 – Reinforced Concrete Design IV
Assessment 1 – May 2026
Examiner: Mr. Ngonyama | Total Marks: 70
QUESTION 1 [20 Marks] – Continuous Beam Analysis
Problem Description
A two-span continuous beam is fixed at end A, supported on rollers at B (interior) and C
(free end). The beam cross-section is 400 mm × 600 mm.
Spans: L₁ = 7 m (Span AB), L₂ = 5 m (Span BC)
Unfactored Dead Load (excluding self-weight): wDL = 10 kN/m
Unfactored Live Load: wLL = 8 kN/m
1.1 Design Load Calculation [5 marks]
Self-Weight of Beam
Self-weight, wsw = γc × b × h = 24 × 0.4 × 0.6 = 5.76 kN/m
Total Unfactored Dead Load
wDL,total = wDL + wsw = 10 + 5.76 = 15.76 kN/m
Factored Design Loads (BS8110 / SABS 0100 load combinations)
Load factors: γDL = 1.4, γLL = 1.6
, Load Case Description Span AB (w₁, kN/m) Span BC (w₂, kN/m)
LC1 Maximum load – full 1.4(15.76) + 1.6(8) = 1.4(15.76) + 1.6(8) =
LL on both spans 34.86 34.86
LC2 Max sagging in span 1.4(15.76) + 1.6(8) = 1.4(15.76) = 22.06
1 – full LL on AB, DL 34.86
only on BC
LC3 Max sagging in span 1.4(15.76) = 22.06 1.4(15.76) + 1.6(8) =
2 – DL only on AB, 34.86
full LL on BC
1.2 Moments and Shear at Critical Sections [12 marks]
Method: Three-Moment Equation (Clapeyron's Theorem)
Boundary conditions: Fixed end at A (slope = 0), rollers at B and C (M_C = 0).
To model the fixed end, an imaginary span of zero length is introduced to the left of A.
Equation (i) – from imaginary span and Span AB:
2·M_A·L₁ + M_B·L₁ = −w₁·L₁³/4
=> 14·M_A + 7·M_B = −w₁·(343)/4
Equation (ii) – from Span AB and Span BC:
M_A·L₁ + 2·M_B·(L₁+L₂) + M_C·L₂ = −w₁·L₁³/4 − w₂·L₂³/4
=> 7·M_A + 24·M_B = −w₁·(343)/4 − w₂·(125)/4 [since M_C = 0]
LOAD CASE 1 – Full Design Load on Both Spans
w₁ = w₂ = 34.86 kN/m
Eq(i): 14·M_A + 7·M_B = −1.4 × 34.86 × 343/4 = −2,989.6
Eq(ii): 7·M_A + 24·M_B = −34.86×343/4 − 34.86×125/4 = −4,079.1
Solving simultaneously:
