Examples DRCB Tutorial.pptx
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This document contains examples of the analysis and design of doubly reinforced concrete beams. In the first example, the reinforcement ratios and stresses are calculated for a given beam and found to satisfy requirements. The moment capacity is determined as 507 kN-m. The second example involves designing reinforcement for a beam to resist a factored moment of 740 kN-m. The reinforcement amounts and spacing are determined and checked against criteria. The third example also involves sizing reinforcement for a beam, checking that strain and spacing requirements are met.
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- 1. Example1 : (Analysis of Doubly reinforced concrete Rectangular beams)
- 2. Solution 𝜌 = 3217 300∗450 =0.0238 𝜌′ = 𝐴𝑠 ′ 𝑏𝑑 𝜌 = 𝐴𝑠 𝑏𝑑 𝐴𝑠(4 32)=3217 𝑚𝑚2 𝐴𝑠 ′ (2 29)=982 𝑚𝑚2 𝜌′ = 982 300∗450 = 0.0072 𝜌𝑚𝑎𝑥=0.85 27.6 413 ×0.85 × 0.003 0.003+0.005 = 0.0181 𝜌𝑚𝑎𝑥=0.85 𝑓𝑐 ′ 𝑓𝑦 ×𝐵1 × 0.003 0.003+0.005 𝜌 > 𝜌𝑚𝑎𝑥
- 3. 𝜌′cy=0.85 𝑓𝑐 ′ 𝑓𝑦 × 𝑑′ 𝑑 0.85 × 𝜖𝑢 𝜖𝑢−𝜖𝑦 +𝜌′ 𝜖𝑦= 𝑓𝑦 𝐸𝑠 = 413 200000 = 0.0021 𝜌′cy = 0.85 × 27.6 413 × 63 450 ×0.85 × 0.003 0.003−0.0021 + 0.0072 𝜌′ cy = 0.0289 > 0.0238 (𝜌) 3217 ∗ 413 = 0.85 ∗ 27.6 ∗ 0.85 𝐶 ∗ 300 + 982 ∗ (600 ∗ ( 𝑐−63 𝑐 )- 0.85*27.6)
- 4. 𝐶2-127.45 C-6204.9= 0 𝐶 = 127.45± 127.452+4∗1∗6204.9 2∗1 C = 165 mm 𝑓𝑠 ′ =600 ( 𝐶−𝑑′ 𝐶 ) 𝑓𝑠 ′ =600 ( 165−63 165 )= 370.9 MPa < 𝑓𝑦 =413 MPa a =𝛽1*C a =0.85*165 =140.25 mm 𝑀𝑛=0.85 *27.6*140.25 * 300*(450 - 140.25 2 )+982 (370.9-0.85*27.6)(450-63) 𝑀𝑛=507 KN.m
- 5. 𝜀𝑠 =0.003 ( 450−165 165 ) =0.00518 > 0.005 So = 0.9 𝑀𝑢= . 𝑀𝑛 𝑀𝑢=0.9 . 507 = 456.31 𝐾𝑁. 𝑚
- 6. Example 2 : (Design of Doubly reinforced concrete Rectangular beams)
- 7. Solution 𝑎𝑠𝑠𝑢𝑚𝑒 𝜖𝑡 0.005 =0.9 𝑀𝑛= 𝑀𝑢 0.9 = 740 0.9 = 822.2 KN.m 𝜌𝑚𝑎𝑥=0.85 𝑓𝑐 ′ 𝑓𝑦 ×𝐵1 × 0.003 0.003+0.005 𝜌𝑚𝑎𝑥=0.85 24 400 ×0.85 × 0.003 0.003+0.005 = 0.0162
- 8. 𝑅𝑛= 𝑀𝑢 ∅ 𝑏 𝑑2 = 740 × 106 0.9× 360 × 5352 = 7.98 MPa 𝑚 = 𝑓𝑦 0.85𝑓𝑐 ′ = 400 0.85×24 = 19.6 𝜌 = 1 19.6 (1 − 1 − 2×19.6×7.98 400 ) = 0.0272 > 0.0162 𝐴𝑠1= 𝜌𝑚𝑎𝑥 × b × d = 0.0162 × 360 × 535 = 3120.12 𝑚𝑚2 𝑎= 𝐴𝑠1×𝑓𝑦 0.85×𝑓𝑐 ′ ×𝑏 = 3120.12×400 0.85×24×360 = 169.94 mm
- 9. 𝑀𝑛= 𝐴𝑆 ×𝑓𝑦 (d - 𝑎 2 ) 𝑀𝑛1= 3120.12 ×400 (535 - 169.9 2 ) 𝑀𝑛1= 561.7 KN.m C = a 𝛽1 = 169.9 0.85 = 199.88 mm 𝜀𝑠 ′= ( 𝑐−𝑑′ 𝑐 ) 0.003 = ( 199.88−60 169.88 ) 0.003 = 0.0021> 𝜀𝑦 = 𝑓𝑦 𝐸𝑠 = 400 200000 = 0.002
- 10. 𝑀𝑛2= 𝑀𝑛- 𝑀𝑛1 𝐴𝑠 ′ =𝐴𝑠2 = ( 𝑀𝑛2 𝑓𝑦 (𝑑−𝑑′) ) = 260.5 × 106 400 (535−60) = 1371 𝑚𝑚2 𝑀𝑛2= 822.22- 561.7 = 260.5 𝐾𝑁. 𝑚 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝐴𝑠 ′ (2 30)=1414 𝑚𝑚2 𝐴𝑠=𝐴𝑠1 + 𝐴𝑠2 𝐴𝑠=3120.12 +1371 = 4491.12 𝑚𝑚2 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑𝐴𝑠(10 24)= 4524 𝑚𝑚2 o𝑟 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝐴𝑠(6 25) +(2 32)= 4554 𝑚𝑚2
- 11. 𝑆 = 360 −2×40 −2×10 −5×24 4 = 35 > 25 𝑚𝑚 𝑜𝑘 𝜀𝑠 =0.003 ( 535−199.8 199.8 ) =0.005033 > 0.005 𝑜𝑘
- 12. Example 3 : (Design of Doubly reinforced concrete Rectangular beams)
- 13. Solution