HomeWork1 - u w is 29 2 1 7.5 2 2 37.5 2 3 u u w w × × =...

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Home Work #1 An equilateral triangle ABC is to be installed at a corner of a tunnel complex at a municipal wastewater treatment plant to facilitate a passage of a utility vehicle. The equivalent uniformly distributed load intensity of the vehicle may be assumed to be 300 psf. Both fixed edges, AB and AC, are 15 ft long and the free edge BC is computed to be 15 2 (21.21) ft. Assume an identical reinforcement is placed in both x and y directions so that nx M and ny M are the same. Assuming a load factor of 1.7 for both the live load (without an impact allowance) and the dead load of the concrete slab, determine the ultimate bending moment capacity required for the slab. 15’ C A B 15’ 1
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C 15’ 7.5’ 7.5 2 7.5' D A B E Solution, HW#1 Assume the thickness of the slab to be 8 in. Then, the dead load of the slab is 150o8/12=100 psf. ( 29 300 100 1.7 680 psf 0.68 ksf u w = + × = = Since nx M = ny M , nb nx ny M M M = = , 7.5 2 AD BD = = 7.5 DE = The segments ADB and ADC rotate 1 θ about AB and AC and 2 about AD. 1 7.5 = and 2 7.5 2 = The external virtual work done by the ultimate load,
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Unformatted text preview: u w , is ( 29 2 1 7.5 2 2 37.5 2 3 u u w w × × = The internal virtual work done is ( 29 ( 29 ( 29 ( 29 15 2 7.5 2 2 6 7.5 7.5 2 P P P M M M + = Equating these two yields, 37.5 6 u P w M = ⇒ ( 29 37.5 0.68 4.25 ft-kips/ft 6 P M = = Assuming ' 4 ksi, 60 ksi c y f f = = and a minimum concrete cover of 2 in and #6 [email protected] in., d=8-.38-2=5.62”. 60 0.44 0.65 0.85 4 12 a × = = × × in. ( 29 ( 29 .44 60 5.62 .65 / 2 /12 11.65 ft-kips/ft n M =-= 0.9 10.5 ft-kips/ft > 4.25 ft-kips/ft u n M M = = Alternatively, if one assumes a unit width strip at the free edge, the span length may be assumed to be 19 ft (14o √ 2 ≈ 19). Ext work: 0.68o19o ∆ /2, Int. work: P M ( ∆ /9.5 o4). P M =15.34, n M =15.34/0.9=17.05 ft-kips/ft ⇐ 4 times larger than that required for a triangular plate. 2...
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HomeWork1 - u w is 29 2 1 7.5 2 2 37.5 2 3 u u w w × × =...

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