A simply-supported rectangular beam with an 8 m span has a self-weight of 5.5 kN/m and an imposed load of 12 kN/m. In addition to this, the beam supports a brickwall with a thickness of 115 mm and a height of 3.2 m. Given: density of brick wall, ρbrickwall = 19 kN/m3 ; ? = 0.8; breadth of beam, b = 175 mm; concrete strength = 25 MPa; and fsy = 500 MPa. Assume the concrete cover is 30 mm and R10 shear links are used. a. Recommend a suitable depth for the beam and determine the area of reinforcement, Ast, according to minimum ductility requirements from AS3600 (ku = 0.36) by using N32 rebar. Detail the cross-section of the beam. b. Using the same beam dimensions from (a), design the steel requirement if N25 rebar is to be used instead. c. Based on your designs using N25 and N32 rebar, which rebar size would you recommend to the client? Justify your answer.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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A simply-supported rectangular beam with an 8 m span has a self-weight of 5.5 kN/m and an imposed load of 12 kN/m. In addition to this, the beam supports a brickwall with a thickness of 115 mm and a height of 3.2 m. Given: density of brick wall, ρbrickwall = 19 kN/m3 ; ? = 0.8; breadth of beam, b = 175 mm; concrete strength = 25 MPa; and fsy = 500 MPa. Assume the concrete cover is 30 mm and R10 shear links are used. a. Recommend a suitable depth for the beam and determine the area of reinforcement, Ast, according to minimum ductility requirements from AS3600 (ku = 0.36) by using N32 rebar. Detail the cross-section of the beam. b. Using the same beam dimensions from (a), design the steel requirement if N25 rebar is to be used instead. c. Based on your designs using N25 and N32 rebar, which rebar size would you recommend to the client? Justify your answer.
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Step 1: Introduce the problem statement

L equals 8 m
S e l f space w e i g h t equals 5.5 space bevelled fraction numerator k N over denominator m end fraction
B r i c k w a l l space h e i g h t equals 3200 space m m
B r i c k w a l l space t h i c k n e s s equals 115 space m m
U n i t space w e i g h t space o f space w a l l equals 19 space bevelled fraction numerator k N over denominator m cubed end fraction
Q equals 12 space k N divided by m
ϕ equals 0.8
b equals 175 space m m
f apostrophe subscript c equals 25 space M P a
f subscript s y end subscript equals 500 space M P a
R 10 space a s space s h e a r space l i n k s
c o n c r e t e space c o v e r equals 30 space m m

bottom enclose bold To bold space bold determine bold colon end enclose
1 right parenthesis space D e p t h space a n d space r e i n f o r c e m e n t space f o r space m i n i m u m space d u c t i l i t y space i. e. space k subscript u equals 0.36 space w i t h space N 36 space b a r s
2 right parenthesis space R e d e s i g n space t h e space r e i n f o r c e m e n t space a s space N 25 space b a r s space k e e p i n g space t h e space d i m e n s i o n s space s a m e
3 right parenthesis space S u i t a b l e space b a r space s i z e space r e c o m m e n d a t i o n space t o space t h e space c l i e n t

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