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Max Load & Deflection of RC Beam

The maximum uniform distributed live load the beam can carry is 140.9 kN/m. With a live load of 20 kN/m and total load of 26.6 kN/m, the immediate deflection produced is 2.42 mm. The beam has a reinforced concrete section 330 mm deep with 8 rebars and can support a maximum uniform live load of 140.9 kN/m while satisfying design limits.

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Wissam Nadir
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46 views6 pages

Max Load & Deflection of RC Beam

The maximum uniform distributed live load the beam can carry is 140.9 kN/m. With a live load of 20 kN/m and total load of 26.6 kN/m, the immediate deflection produced is 2.42 mm. The beam has a reinforced concrete section 330 mm deep with 8 rebars and can support a maximum uniform live load of 140.9 kN/m while satisfying design limits.

Uploaded by

Wissam Nadir
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Q4 of final 2016-2017: By using the working stress

design method, find:


a- The maximum uniform distributed live load
can be carry by the simply supported
reinforced concrete beam of section and
details shown in Figure. Use fs=165 MPa,
fc=12.5 MPa, effective depth= 330 mm, dead
load = 30kN/m and n=8.
b- If the live load = 20 kN/m, and 𝑓 =
28𝑀𝑃𝑎,comput the immediate deflection preduce by total load.

note:∆ .=

Solution:
a-
1- Find N.A

𝑦
750 ∗ 100 ∗ (𝑦 − 50) + 250 ∗ 𝑦 ∗ = 𝑛 ∗ 𝐴𝑠 ∗ (𝑑 − 𝑦) → 𝑦 = 117.58 𝑚𝑚
2
2- Find moment of inertia about N.A
𝑏. ℎ
𝐼 = + 𝑛 𝐴𝑠(𝑑 − 𝑦)
3
1000 ∗ 117.6 750 ∗ (117.6 − 100)
𝐼 = − + 8 ∗ 4000 ∗ (330 − 117.6) = 1.98 ∗ 10 𝑚𝑚
3 3
3- Find the moment resisted by section
𝑀. 𝑦 𝑀 ∗ 10 ∗ 117.6
𝑓 = → 12.5 = → 𝑀 = 210.9 𝐾𝑁. 𝑚
𝐼 1.98 ∗ 10
𝑀. (𝑑 − 𝑦) 𝑀 ∗ 10 ∗ (330 − 117.6)
𝑓 = 𝑛. → 165 = 8 ∗ → 𝑀 = 192.27𝐾𝑁. 𝑚 𝑐𝑜𝑛𝑡𝑟𝑜𝑙
𝐼 1.98 ∗ 10

. ∗
4- Find W 𝑀 = → 192.27 = → 𝑊 = 170.9 = 30 + 𝐿. 𝐿 → 𝐿. 𝐿 = 140.9𝑘𝑁/𝑚
b- ∆=
W= self weight +live load
Self weight =((1*0.5)-(0.75*0.3))*24=6.6KN/m

𝑘𝑁 𝑤∗𝑙
𝑊 = (6.6 + 20) = 26.6 , 𝑙 = 6 𝑚 , 𝐸 = 4700√28 = 24870𝑀𝑃𝑎 𝑎𝑛𝑑 𝑀 =
𝑚 8
= 29.9𝑘𝑁. 𝑚
𝑀 𝑀
𝐼 = 𝐼𝑔 + 1 − 𝐼 ≤ 𝐼𝑔
𝑀 𝑀
𝑓 ∗𝑦
𝑀 =
𝐼𝑔
𝑓 = 0.62 ∗ √28 = 3.28 MPa
500
𝑦 = = 250 𝑚𝑚
2
1000 ∗ 500 750 ∗ 300
𝐼𝑔 = − = 8.729 ∗ 10 𝑚𝑚
12 12
3.28 ∗ 8.729 ∗ 10
𝑀 = = 114.52 𝑘𝑁. 𝑚
250
114.52 114.52
𝐼 = 8.729 ∗ 10 + 1 − 1.98 ∗ 10 = 3.81 ∗ 10 𝑚𝑚 ≤ 𝐼𝑔
29.9 29.9
5 𝑤𝑙 5 57 ∗ 6000
∆= = ∗ = 2.42 𝑚𝑚
384 𝐸𝐼 384 24870 ∗ 1.595 ∗ 10

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