PLAIN AND CIVIL: INTERIOR BAYS: STRENGTH ONE DESIGN FOR NEGATIVE LIVE LOAD EFFECT

After determining the reinforcement used for positive moments, move further in determining the effect of negative live load.

The procedure will be similar only that you are going to use the negative moment instead of the positive moment. The other thing you might be changing is the different diameter of the reinforcing bar you might be using.

For the negative moment, using the AASHTO table:

Using the same S, the next thing you would need to consider is the flange width of the beam.

Assuming that you are going to use W250x200x67.

Properties are as follows: depth, h=257mm
width, w=204mm
web thickness = 8.9mm

The negative moment will be taken from the tip of the flange or 1/2 of the top flange width, which is 100mm for the specified W section. By interpolation, we would be able to identify the value.

s= 2650mm
distance from CL = 100mm

at s=2600: 26470 (@75mm) and 22730 (@150mm)
interpolate for 100mm,

$-M_{2600}=25223.33\frac{N-mm}{mm}$

at s=2700: 26920 (@75mm) and 23170 (@150mm)
interpolate again for 100mm,

$-M_{2700}=25670\frac{N-mm}{mm}$

Then the negative moment due to live load:

$-M_{2650}=25223.33+\frac{25670-25223.33}{2}={\color{Red} 25446.665\frac{N-mm}{mm}}$

Assume #19 bars:

Diameter = 20mm
Area = 284 sqmm

The procedure in determining the spacing of rebars will be the same:

1.   $M_{neg}=\gamma _{p,DC}M_{DLslab}+\gamma _{p,DC}M_{DLparapet}+\gamma _{p,DW}M_{DLfws}+\gamma _{LL}M_{LL}$

2. From the assumed bar size for thee negative moment. solve the area of steel and then the depth.

$d_{e}=t_{slab}-C_{top}-\left ( \frac{bardia}{2} \right )-t_{IWS}$

3.    $R_{n}=\frac{M_{neg}}{\varphi _{f}bd_{e}^{2}}$

4.    $\rho =0.85\left ( \frac{f'c}{fy} \right )\left [ 1-\sqrt{1-\frac{2R_{n}}{0.85f'c}} \right ]$

5. $A_{s}=\rho bd_{e}$

6. $Required bar spacing=\frac{bar dia}{A_{s}}$

SERVICE I LIMIT STATE - NEGATIVE-MOMENT FLEXURE CRACKING

1. $M_{NEG}=\gamma _{DC}M_{DLslab}+\gamma _{DC}M_{DLparapet}+\gamma _{DW}M_{DLfsw}+\gamma _{LL}M_{LL}$

2. $d_{c}=C_{top}+\frac{bar dia}{2}$

3. $\beta _{s}=1+\frac{d_{c}}{0.7\left ( h-d_{c} \right )}$

4. $\rho =\frac{A_{s}}{bd_{s}}$

5. Using n=8

$k=\sqrt{\left ( \rho n \right )^{2}+2\rho n}-\rho n$

$j=1-\frac{k}{3}$

$f_{s}=\frac{M_{NEG}}{A_{s}jd_{s}}$

$s\leqslant \frac{123000\gamma _{e}}{\beta _{s}f_{s}}-2d_{c}$

PLAIN AND CIVIL

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Friday, 24 April 2020

INTERIOR BAYS: STRENGTH ONE DESIGN FOR NEGATIVE LIVE LOAD EFFECT

SERVICE I LIMIT STATE - NEGATIVE-MOMENT FLEXURE CRACKING

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