BRAE433 - Chapter 1b

BRAE433 - Chapter 1b - Chapter 1 – Bending Stresses 1 2 3...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 1 – Bending Stresses 1. 2. 3. 4. 5. 6. 7. 8. Review on Stress/Strain Bending formula for concrete only or steel only formula for concrete only or steel only Rebar Internal Couple Problem with bending and internal couple Ck against fr Assumption for Chapter 1 Problem using 3 x 3 x 12 in beam Concrete in Compression • f’c – Compressive strength of concrete – psi (can be ksi also) • ASTM – American Society for Testing and Materials – Standardized test to determine - f’c 1 Max strength Concrete Strength vs. Strain Note: slope = E Hooke’s Law – E = σ/Є Steel Strength vs. Strain 2 Chapter 1 - Bending • From the flexure formula…we use the Cracking Moment formula to evaluate beams for concrete only. – Mcr = fbI/c – Problem using this with concrete plus steel • beam is not homogeneous • concrete does not behave elastically over full range – Can be used for concrete beam only Hoover Dam 700 ft tall bottom thickness = ? Hint: Rewrite terms and solve for ht 3 Actual Hoover Dam 726.4 ft tall 45 ft top thickness 600 ft bottom thickness Steel Bending • Resisting Moment Formula – MR = faI/c = fa x S x – Commonly used in steel design • fa = allowable stress • fa = fy x 0.66 ( or 0.6 depends on compact or not) • steel does behave elastically over range to fa 4 RC - Steel • Rebar – #4 and #5 common for ag – 1/2” diameter = #4 – 5/8” diameter = #5 – 3-8 based on diameter – 9-11: 1 in sq, 1.125 in sq and 1.25 in sq in sq, 1.125 in sq and 1.25 in sq – 14 and 18: 1.5 in sq and 2.0 in sq – Table A-1 and A-2 RC - Steel • Rebar – fy = yield stress (psi) – See Table A-1 for values • 40,000 psi • 60,000 psi – E = Modulus of Elasticity (psi) – E = 29,000,000 psi 5 RC - Steel • Welded Wire Reinforcing (WWR) – fy = yield stress (psi) = 70,000 psi yield stress (psi) 70 psi – WWR6 x 12-W16 x W8 • plain wwr with 6 x 12 traverse spacing • cross sectional area = 0.16 sq in for longitudal wires • = 0.08 sq in for transverse wires Internal Couple Couple Internal Couple Approach Method 6 7 Given: Concrete beam (no steel) 6 in width x 12 in depth 4 ft beam length, wt = 75 lb/ft P = 4,500 lb at midspan Find: a) max tensile strength using internal couple method b) same using flexure formulas c) compare against ACI code Solution: Mmax = PL/4 + wL2/8 PL/4 = 4500 lbx4 ft/4 + 75 lb/ft x (4 ft)2/8 = 4650 ft-lb or 55,800 in-lb ftin1. Stress and strains vary linearly about neutral axis 2. Plot C – compressive force and T - tensile force about neutral axis force located at their respective centroids (1/3b for triangle) 3. Hint: Z = 2/3 x (depth of beam) Example 1-1 1Solution: Mmax = PL/4 + wL2/8 = 4500 lbx4 ft/4 + 75 lb/ft x (4 ft)2/8 4500 lbx4 ft/4 75 lb/ft (4 ft) = 4650 ft-lb ftM = CZ = TZ (External Moment = Internal Couple) 4650 ft-lb x 12 in/ft = C (8 in) ftsolve C = 6975 lb C = average stress x (area of beam on which stress acts) = ½ ftop (6in) x (6in) = 6,975 Answer: ftop = 388 psi solve ftop = 388 psi 8 Example 1-1 • b) Flexure formula I = bh3/12 = 6in x (12in3)/12 = 864 in4 bh 6in (12in 864 in ftop = fbottom = Mc/I = 4650 ft lbs (12in/ft)(6in)/864 iin4 n = 388 psi (same as part a) • c) Modulus of Rupture 0.5 for 7.5 • ACI code fr = 7.5 (f’c) 0.5 for 3,000 psi f’c = 411 psi (rupture will occur) • d) Therefore: Okay Assumptions made for Ex. 1-1 elastic theory -theory 1. A plane section before bending remains a plane after bending (variation in strain remains linear) 2. The Ec (modulus of elasticity) remains constant (stress distribution is linear) 9 Lab Problem for Non-Reinforced Beam • Find: the load before cracking of a 3 in by 3 in square beam 12 in length P = ??? • Assume f’c = 3000 psi • Use the flexure formula only • Assume concrete = 144 pcf P = ??? Soln: MR = Mcr = Cracking Moment = frI/c Cracking Moment I = bh3/12 = 3 in x (3in3)/12 = 6.75 in4 0.5 fr = 7.5 (f’c) 0.5 = 7.5 (3,000 psi)^0.5 = 410.8 psi 7.5 7.5 Mcr = fr I/c = 410.8 (6.75 in4)/(1.5 in x 12 in/ft) = 154.1 ft lbs 154 ft lbs M = PL/4 + wL^2/8 w = .25 x .25 x 144 pcf = 9 Lb/ft P = (M – wL^2/8)/(L/4) = = (154.1 – 9x12/8))/(1/4) = 612 # 10 P = ??? C T Soln 2: Using Internal Couple Method C = average stress x area of beam on which stress acts C = ½ x ftop x 1.5 in x 3 in 0.5 Use ftop = fr = 7.5 (f’c) 0.5 = 7.5 (3,000 psi)^0.5 = 410.8 psi 7.5 7.5 C = ½ x 410.8 #/in2 x 1.5 in x 3 in = 924.3# Z = 2/3 x depth of the beam = 2/3 x 3 in = 2 in M = CZ = TZ M = 924.3# x 2 in x ft/12 in = 154.1 ft-lbs M = PL/4 + wL^2/8 w = .25 x .25 x 144 pcf = 9 Lb/ft P = (M – wL^2/8)/(L/4) = = (154.1 – 9x12/8))/(1/4) = 612 # Non-Reinforced Concrete Beam Problems • Assume as homogeneous • Flexure formula will apply – fb = Mc/I (this will NOT work for reinforced concrete problems since the Modulus of Elasticity is different between steel and concrete) • Internal Couple method will yield same result as flexural formula 11 Possible Exam Questions • Problem format: format: – a) Known beam size, load P at failure, Check ftop against ACI code for fr – b) Known beam size, find Cracking moment or load P that will cause cracking moment by substituting ACI code fr for ftop – c) Known load P at failure, find a beam size that will work given a depth constraint 12 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online