788f74d3-2fd5-468a-a566-96ec1c10b017 17 4 5 false false 0 decimal

&[DATE] &[TIME] - &[FILENAME]

EGB 375 - DESIGN ASSINGMENT

Design of Concrete Structures - According to Australian Concrete Standards (AS3600-2018)

Name : Darcy Marnane

ID Number : n9216570

Beam Cross-Section :

Dimensions and Properties

L.clear_span 5.75 m * :

b.beam 250 mm * :

L.span 6.0 m * :

D.beam 400 mm * :

c.1 30 mm * :

y.max D.beam 2 / : mm 200

γ 25 kN m 3^/ * :

A.g b.beam D.beam * : mm 2^1105^*

Unit Weight of Concrete

I.x b.beam D.beam 3^* 12 / : mm 4^1.3333 109^*

f'.c 32 MPa * :

Z.t I.x y.max / : mm 3^6.6667 106^*

Z.b Z.t : mm 3^6.6667 106^*

Beam Loading:

G.slab 6 kN m / * :

Permanent dead load of terrace slab onto the beam

G.fall 3 kN m / * :

Permanent dead load of terrace fall, and flooring onto the beam

G.rail 1 kN m / * :

Permanent dead load of safety railing onto the beam

Q.balc 2.5 kN m / * :

Imposed live load onto the beam

1. Draw the beam elevation schematically supported on columns with span length and the loading on top.

Assuming - Column Sizes of 250x250mm.

2. Determine the self-weight of the beam in kN/m

G.self_weight γ b.beam * D.beam * : kN m / 2.5

3. Determine the magnitude of the ultimate design load w*

Assuming - Permanant and Imposed Actions Govern the ULS Design Load

AS 1170.0 : 2002 Clause 4.2.2

G.total G.self_weight G.slab + G.fall + G.rail + : kN m / 12.5

Q.total Q.balc : kN m / 2.5

w.uls.design 1.2 G.total * 1.5 Q.total * + : kN m / 18.75

4. Determine the design bending moment M*

Table 1

Note : For simply supported beams with an applied UDL, the maximum bending moment, M*, and Maximum Deflection, δ, can be calculated using the formula provided in Table 1

M.max.uls w.uls.design L.span (2^* 8 / : kN m * 84.375

5. Determine the short-term bending moment, Ms*

From AS1170.0 - Clause 4.3 - Combinations of Actions for Sevicibility Limits and table 4.1

w.short.term G.total 0.7 Q.total * + : kN m / 14.25

Short term load

M.s.max w.short.term L.span (2^* 8 / : kN m * 64.125

Maximum Short term Bending Moment

6. Determine the long-term bending moment, M*sus

From AS1170.0 - Clause 4.3 - Combinations of Actions for Sevicibility Limits and table 4.1

w.long.term G.total 0.4 Q.total * + : kN m / 13.5

M.l.max w.long.term L.span (2^* 8 / : kN m * 60.75

7. Determine the area of steel required for the calculated M*

ϕ.1 0.85 :

Capacity reduction factor (AS3600 Table 2.2.2)

Assuming Tension Steel has Yeilded.

Assuming 16mm Diamter Steel Bars

Steel and Section Properties:

f.sy 500 MPa * :

Characteristic Yeild Strength of Steel Reinforcment

Dia.bar 16 mm * :

Diameter of Reinforcement

d.eff D.beam c.1 - Dia.bar 2 / - : mm 362

Effective Depth of Cross Section

α.2 0.85 0.0015 f'.c (MPa / * - : 0.802

Equation 8.1.3(1) - AS3600

γ.2 0.97 0.0025 f'.c (MPa / * - : MPa 8.9 107 -^* m s 2^* kg / *

Equation 8.1.3(2) - AS3600

M.u M.max.uls ϕ.1 / : kN m * 99.2647

M.u A.st f.sy * d.eff γ.2 d.n * 2 / - (* :

Design Moment Capacity

ϕM.u M.max.uls :

Calculating the Depth of Neutral Axis, in terms of Ast

α.2 f'.c * (b.beam * γ.2 d.n * (*

A.st f.sy * (

C = T

C =

T =

Rearranging to solve for dn

alpha2 and gamma are unitless in this equation

d.n f.sy A.st * α.2 f'.c * b.beam * γ.2 * / : 3125000 A.st * 35689 m * /

Solving for A.st by considering,

ϕM.u M.max.uls :

M.u M.max.uls ϕ.1 / : kN m * 99.2647

Solution with solve:

Sol.1 M.u A.st mm 2^* f.sy * d.eff γ.2 d.n * mm 2^* 2 / - (* ≡ A.st 01000 solve mm 2^* : mm 2^585.2978

Solution with roots:

Sol.2 M.u A.st mm 2^* f.sy * d.eff γ.2 d.n * mm 2^* 2 / - (* ≡ A.st roots mm 2^* : mm 2^585.2978

Solution with maple:

Sol.3 M.u A.st f.sy * d.eff γ.2 d.n * 2 / - (* ≡ A.st solve maple : mm 2^8705.0702 585.2978 21 mat

Solution with al_nleqsolve:

A.st equ A.st f.sy * d.eff γ.2 d.n * 2 / - (* M.u - :

v f v 1 el mm 2^* equ :

Sol.4 0.1 11 mat f al_nleqsolve mm 2^* : mm 2^585.2978 11 mat

Solution with FindRoot:

M.u A.st f.sy * d.eff γ.2 d.n * 2 / - (* ≡ A.st 10 mm 2^* ≡ FindRoot mm 2^585.2978

A.st mm 2^585.2978

Rearranging to solve for Ast