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### BRAE433 - Chapter 2b

Course: BRAE 433, Fall 2011
School: Cal Poly
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Word Count: 432

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2 Chapter RC Design Evaluation and Design RC Rectangular Beam Problems Part 2 ITRC Cal Poly Ultimate Moment, Mn Fig 1-1 As = Area of steel Tables A-1 and A-2 Also called the nominal moment strength Point just prior to failure of the beam BRAE/ITRC Cal Poly 1 Ultimate Moment Theoretical internal resultant forces ,NC and NT internal resultant forces Parallel, equal and opposite Separated by distance Z...

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2 Chapter RC Design Evaluation and Design RC Rectangular Beam Problems Part 2 ITRC Cal Poly Ultimate Moment, Mn Fig 1-1 As = Area of steel Tables A-1 and A-2 Also called the nominal moment strength Point just prior to failure of the beam BRAE/ITRC Cal Poly 1 Ultimate Moment Theoretical internal resultant forces ,NC and NT internal resultant forces Parallel, equal and opposite Separated by distance Z internal resisting couple = ultimate moment (or nominal moment strength) BRAE/ITRC Cal Poly Ultimate Moment To solve, need to calculate the resultant force NC in the concrete determine its location in the beam BRAE/ITRC Cal Poly 2 Equivalent Stress Distribution BRAE/ITRC Cal Poly Beta vs. fc' 0.900 1 = 0.85 for fc 4000 psi 0.850 1 = 0.85 0.05 (fc-4000)/1000 (fc for fc > 4001 psi to 0.65 Beta 0.800 0.750 0.700 0.650 0.600 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 fc' (psi) BRAE/ITRC Cal Poly 3 Isometric View of the Internal Couple Ready to Solve for Mn BRAE/ITRC Cal Poly BRAE/ITRC Cal Poly 4 Ultimate Moment (nominal moment strength) - Mn Example 2-1ss Given: fc = 5,000 psi, A615 grade 75 steel 12 24 4 - #6 BRAE/ITRC Cal Poly Steps to Solve Ultimate Moment, Mn 1. NC = NT Nc = 0.85 fc a b, NT = As fy 2. 0.85 fc a b = As fy fc 3. solve for a = As x fy/ (0.85 x fc x b) 4. Z = d a/2 5. solve Mn for = NC (d a/2) Strain Check (did steel yield?) Check (did steel yield?) 1. solve for c = a / 1 2. solve actual steel strain t = (d c) x 0.003 / c 3. ck against yield strain y = fy/Es = fy / 29,000,000 BRAE/ITRC Cal Poly 5 Did the Steel Yield? Perform a strain check... Is actual strain in steel less than yield strain? If concrete were to fail (exceed 0.003 strain). Not good... BRAE/ITRC Cal Poly BRAE/ITRC Cal Poly 6 Values in Table A-1 0.002 10 BRAE/ITRC Cal Poly Limiting the steel strain and determining the required steel Example 2-2ss Given: fc = 5,000 psi, A615 grade 75 steel 12 0.003 24 4 - #6 0.005 BRAE/ITRC Cal Poly 7 Solving for Required As Example 2-2ss 1. steel strain t = 0.005 st 2. solve for c = 0.003 (d) / 0.008 3. solve for NC = 0.85 fc (c 1) b 4. NC = NT = As fy 5. solve for As required = NC / fy for fy 6. Check As maximum against As supplied 7. Comment on adequacy BRAE/ITRC Cal Poly Reinforcement Ratio Design As must be lowered to ensure failure in steel t limit at 0.005 Balanced As = bd BRAE/ITRC Cal Poly 8 Minimum Steel Balanced As , min = 3 f 'c 200 bwd bwd fy fy BRAE/ITRC Cal Poly Minimum Steel Need to make sure there is enough steel Rectangular beams: bw = b Table A5 Values for As,min and recommended As , min = 3 f 'c 200 bwd bwd fy fy BRAE/ITRC Cal Poly 9 End ITRC Cal Poly 10
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