The document provides calculations to design the reinforcement for a beam cross-section under maximum positive and negative moments. For the maximum positive moment case, the required reinforcement is calculated as 8 #35 bars and 2 #32 bars. For the maximum negative moment case, the required reinforcement is calculated as 3 #35 bars. Sketches of the reinforced cross-sections are requested to show bar sizes, arrangements, and spacing.

1. 52
T- Beams at negative Moment regions

2. 53
Bundle bars:
When the reinforcement is increased we can use bundle bars to prevent congestion of
reinforcement
Problem 4.4.2 design the beam of cross section shown for maximum positive and negative
moments. Beam weights are not included in the loads given. Show sketches of cross sections
including bar sizes, arrangements, and spacing. Assume concrete
weighs 23.5 kN/m3. fy= 420 MPa and f’c= 28MPa.Place live loads to cause maximum positive and
negative moments. = 30 , = 20
= (30 + (1.5 ∗ 0.1 + 0.5 ∗ 0.4) ∗ 23.5) ∗ 1.2 = 45.87
= 1.6 ∗ 20 = 32
Maximum positive moment case:
= = (45.87 ∗ 18 + 32 ∗ 12)/2 = 604.83
= 1195.25 . ∅ = 0.9 , = 1328 .
ℎ −
Assume a= hf =100 mm

3. 54
= 0.85 −
2
= 0.85 ∗ 28 ∗ 100 ∗ 1500 ∗ 400 −
100
2
= 1249.5 ∗ 10 . = 1249.5 .
< −
=
0.85 ℎ ( − )
=
0.85 ∗ 28 ∗ 100 ∗ (1500 − 400)
420
= 6233
, = =
6233
400 ∗ 400
= 0.039
= −
ℎ
2
= 6233 ∗ 420 400 −
100
2
= 916.25 .
= − = 1328 − 916.25 = 411.75 .
= =
411.75 ∗ 10
400 ∗ 400
= 6.434
=
0.85
1 − 1 −
2
0.85
= 0.85 ∗ 28/420 ∗ (1 − 1 −
2 ∗ 6.434
0.85 ∗ 28
= 0.0183
= = 0.0183 ∗ 400 ∗ 400 = 2928
= + = 2928 + 6233 = 9161
8#35 + 2#32, = 9304
= =
9304
400 ∗ 400
= 0.058
=
0.85
∗
0.003
0.003 + 0.004
+ = 0.059
. =
0.85
∗
0.003
0.003 + 0.005
+ = 0.057 ≈ , ∅ = 0.9
Provide the reinforcement in three layers
H=400+35+35+35/2+50=537.5 mm try h= 550

4. 55
Maximum negative moment case:
= = (45.87 ∗ 18 + 32 ∗ 12)/2 = 700.83
= 350.42 . ∅ = 0.9 , = 390 .
ℎ
− ℎ ℎ
ℎ ℎ ℎ =
= =
390 ∗ 10
400 ∗ 400
= 6.094
=
0.85
1 − 1 −
2
0.85
= 0.85 ∗ 28/420 ∗ (1 − 1 −
2 ∗ 6.094
0.85 ∗ 28
= 0.0171
= = 0.0171 ∗ 400 ∗ 400 = 2736
3#35 = 2886
= =
2886
400 ∗ 400
= 0.018 < . , ∅ = 0.9