Mathcad - Midterm Solution 188-D1

# Mathcad - Midterm Solution 188-D1 - Midterm Solution 1...

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Midterm Solution 1. Section Properties See Excel, remove flange area from 10DT34 7DT34 A DT 711in 2 := y t 34in 23.67in := y t 10.33 in = h DT 34in := b DT 7ft := y b 23.67in := I DT 72272in 4 := S t 6994in 3 := S b 3054in 3 := 2. Factored Moment, Mu γ rc 150pcf := w DT A DT γ rc := w DT 741 plf = w SDL 15psf b DT := w SDL 105 plf = w LL 85psf b DT := w LL 595 plf = w u 1.2 w DT w SDL + () 1.6 w LL + := w u 1967plf = L span 70ft := M u w u L span 2 8 := M u 1205 kip ft = M u 14456 kip in = Midterm Solution 188-D1.xmcd Page 1

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3. Estimate Number of Strand w SDL w LL + 7ft 100 psf = Check PCI for approximate number of strand. 208-D1 pattern carries 107 psf superimposed service load with full 10' wide flange. z approx 34in 4in 2 4.25in := z approx 27.75 in = ϕ approx 0.9 := f pu 270ksi := f ps_approx 95% f pu := f ps_approx 256.5 ksi = fps between 95%-99% for ductile DT. Use 95%. A ps_approx M u ϕ approx f ps_approx z approx := A ps_approx 2.257 in 2 = No_Strand A ps_approx 0.153in 2 := No_Strand 14.749 = Round up to 16 required, 8 per web. 4. Total Strand to Provide Good Service Behavior. No_Strand 18 := USE 188-D1 Strand Pattern A ps 18 0.153 in 2 := A ps 2.754 in 2 = y e 15.00in := y c 4.00in := h s y e y c := h s 11in = y s x () y e h s x L span 2 := y s 0ft ( ) 15 in = y s 35ft () 4 i n = CHECK y s 2ft ( ) 14.37 in = y s 0.4L span 6.20 in = Midterm Solution 188-D1.xmcd Page 2
5. Nominal Strength at 0.4L and 0.5L using Approximate Method f' c 6000psi := β 1 0.85 2 0.05 := β 1 0.75 = γ p 0.28 := Low Relaxation Strand b DT 84in = A ps 2.754 in 2 = f pu 270 ksi = Evaluate Midspan d p h DT y c :=

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Mathcad - Midterm Solution 188-D1 - Midterm Solution 1...

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