lecture19 - Lecture19OnewaySlab Design July21,2003 CVEN444...

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    Lecture 19 -  One-way Slab  Lecture 19 -  One-way Slab  Design Design July 21, 2003 CVEN 444
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    Lecture Goals Lecture Goals One-way slab design Joist Slab
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    Example: One-way Slab Design Example: One-way Slab Design Design a one way slab with a clear span of 10 ft. Use 12 in. wide segment b w . f y = 40 ksi, f c = 3 ksi and a live load of 325 psf.
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    Example: One-way Slab Design Example: One-way Slab Design Determine the preliminary slab depth for an internal section correction * factor 28 l h y 60 ksi f
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    Example: One-way Slab Design Example: One-way Slab Design The correction factor is: CF = 0.4 + (f y /100) Self-weight of the element for a 1 ft. segment 12 in. 10 ft 40 ft * 0.4 3.43 in. 3.5 in. 28 100 h + = ( 29 3 DL 1 ft 3.5 in. 1.0 ft *0.15 k/ft 0.04375 k/ft 12 in. w = =
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    Example: One-way Slab Design Example: One-way Slab Design The factored load Factored moments demanded ( 29 ( 29 u DL LL 1.2 1.6 1.2 0.04375 k/ft 1.6 0.325 k/ft 0.5725 k/ft w w w = + = + = 2 u u 2 u u 16 critical 11 w l M w l M + - = =
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    Example: One-way Slab Design Example: One-way Slab Design The factored moment The depth of the reinforcement (assume #6 bar) is ( 29 ( 29 2 2 u u 0.5725 k/ft 10 ft 5.205 k-ft 11 11 62.5 k-in. w l M - = = = = b 0.75 in. cover 3.5 in. 0.75 in. 2 2 2.375 in. d d h = - - = - - =
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    Example: One-way Slab Design Example: One-way Slab Design Obtain a set of the equations to find (c/d) ( 29 2 u u u c u 1 2 c 0.85 1 2 2 / 1 1 0.85 M R bd k R f k M c k f bd d φ β = = - = - - =
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    Example: One-way Slab Design Example: One-way Slab Design Find (c/d) for the given slab Therefore, the slab is in the compression control region. ( 29 ( 29 ( 29 ( 29 2 2 62.5 k-in / 0.9 1 1 0.558 0.85 3 ksi 12 in. 2.375 in. 0.558 0.657 0.85 k c d = - - = = =
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    Example: One-way Slab Design Example: One-way Slab Design Need the steel to be in the tension controlled region. So chose an arbitrary k, k=0.3 and k’=0.255 ( 29 ( 29 ( 29 ( 29 ( 29 u u 2 u u u 0.255 0.85 3 ksi 0.255 1 0.567 ksi 2 62.5 k-in. 0.9 3.19 in. 0.567 ksi 12 in. R M M R bd d R b φ = - = = = = =
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    Example: One-way Slab Design Example: One-way Slab Design Determine the depth of the slab with the new d. with h =5 in., then d= 3.875 in. Go back and check (c/d) ratio b 0.75 in. cover 3.19 in. 0.75 in. 2 2 4.318 in. Use 5 in. d h d = + + = + + =
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  Example: One-way Slab Design Example: One-way Slab Design The new factored load is The maximum moment is ( 29 ( 29 u DL LL 1.2 1.6 1.2 0.0625 k/ft 1.6 0.325 k/ft 0.595 k/ft w w w = + = + = ( 29 ( 29 2 2 u u 0.595 k/ft 10 ft 5.409 k-ft 11 11 64.9 k-in. w l
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lecture19 - Lecture19OnewaySlab Design July21,2003 CVEN444...

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