AECT250-Lecture 29

AECT250-Lecture 29 - Lecture 29 Shear in Beams (cont.) As a...

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Lecture 29 - Page 1 of 6 Lecture 29 – Shear in Beams (cont.) As a continuation of our discussion of shear in concrete beams, let’s look at an example of a concrete girder with point loads. Example GIVEN : The 20” x 34” concrete girder as shown below. Use the following: Concrete f’ c = 4000 PSI #3 – Grade 60 U-shaped stirrups REQUIRED : Determine the stirrup bar requirements. Assume only one spacing for the beam. 8’-0” 8’-0” 8’-0” 8’-0” 8’-0” 20” d = 32” h = 34” 40’-0” w u = 1.2 KLF #3 U-shaped stirrup bars Beam Cross-Section Conc. column 25 KIPS 25 KIPS 25 KIPS 25 KIPS
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Lecture 29 - Page 2 of 6 Step 1 – Draw shear diagram : Step 2 – Determine V u at distance “d” from face of support : V u = 74 KIPS - ) 2 . 1 ( / " 12 " 32 KLF ft = 70.8 KIPS Step 3 – Determine shear capacity of concrete φ V c : φ V c = 0.75( d b f w c ' 2 ) = 0.75( )) " 32 )( " 20 ( 4000 2 PSI = 60,716 Lbs. φ
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AECT250-Lecture 29 - Lecture 29 Shear in Beams (cont.) As a...

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