quiz_rev_det

# quiz_rev_det - 1.051 Structural Engineering Design Prof...

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1.051 Structural Engineering Design Prof. Oral Buyukozturk Fall 2003 1.051 Structural Engineering Design QUIZ 1 REVIEW Example 1 (Flexural Strength of a Given Member) b = 12” d = 17.5” A s = 4.00 in 2 f y = 60,000 psi f c ’ = 4000 psi d A s b Find M n , M u M u φ M n ; φ = 0.9 for flexure Therefore, with M n , M u can be calculated = ' 2 7 . 1 1 c y y n f f bd f M ρ ρ Æ Equation (1) 019 . 0 5 . 17 12 00 . 4 = × = = bd A s ρ Æ Equation (2) Check 0033 . 0 200 min = = y f ρ Æ Equation (3) y y c b f f f + = 87000 87000 85 . 0 4 3 4 3 ' 1 β ρ = 0.0214 Æ Equation (4) ∴ρ min ρ ¾( ρ b ) Æ Equation (5) M n = 3487 kips.in and M u = 0.9 x M n = 3138 kips.in

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1.051 Structural Engineering Design Prof. Oral Buyukozturk Fall 2003 Example 2 (Section Design with a Given Moment) Unknowns: b, d, h, A s Given: l = 15 feet DL = 1.27 kips/ft LL = 2.44 kips/ft f c ’ = 4000 psi f y = 60,000 psi γ c = 150 psf 1. Assume b and h for self-eight determination: Let b = 10 in and h = 18 in d = 18 – 2.5 = 15.5 d/b = 1.5 Minimum depth for simply supported beam = l/16 = 15/16 . 12 = 11.25; OKAY! 2. Find the applied moment to be resisted W = 150 . (10/12) . (18/12) . (1/1000) = 0.1875 kips/ft (this is to be revised) Therefore, W u = 1.4(1.27 + 0.1875) + 1.7(2.44) = 6.19 kips/ft M u = w u l 2 /8 = 6.19 (15) 2 /8 . 12 (ft/in) = 2089 kips.in 3. Compute ρ min & ρ b ; and choose ρ ρ min = 200/f y = 0.0033 ¾ ρ b = 0.0214; use ρ = 0.0214 (Not economical, but adequate for demonstration purpose) Find the required bd 2 = ' 2 7 . 1 1 ) ( c y y u f f bd f M ρ φρ bd 2 = 2229 in 2 Actual bd 2 = 10. (15.5) 2 = 2403 in 2 > Required bd 2 = 2229 in 2 ; OKAY! 4. Assign rebar arrangement ρ = A s /bd = 0.0214 A s = 0.0214.b.d = 0.0214.10.(15.5) = 3.32 in 2 (required) Provide 2#10 + 1#8 A s provide = 3.32 in 2 Note: ρ can be smaller and a larger section may be needed to improve cost and deflection performance. However, if there is architectural restrictions on sizes, a ρ with a value closer to the upper bound is normally used (to reduce section size as much as possible)
1.051 Structural Engineering Design Prof. Oral Buyukozturk Fall 2003 Example 3 (Crack Width Determination) Given: b = 12” h = 20” A s = 4.00 in 2 (4 #9) f y = 60,000 psi Exposure = external 3 000091 . 0 A d f w c s = Æ

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