Check Beam Shear

1 / 42

# Check Beam Shear - PowerPoint PPT Presentation

Check Beam Shear. 245. 166. 245. Bending moment design. 2.5m. Short direction. Central band ratio =. Central band of short direction = 0.83 As = 0.83 (10.1)=8.6cm 2. Footing Design Part II Combined footing. Example 1. Design a combined footing As shown. Dimension calculation

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Check Beam Shear' - minna

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

245

166

245

Bending moment design

2.5m

Short direction

Central band ratio =

Central band of short direction = 0.83 As = 0.83 (10.1)=8.6cm2

Footing Design

Part II

Combined footing

Example 1

Design a combined footing As shown

Dimension calculation

The base dimension to get uniform distributed load

2000kN

800 kN

1200 kN

A

x1=0.2m

x2=6.2m

x

800(0.2)+1200(6.2)=2000(x)

x = 3.8m

800 kN

1200 kN

Try thickness =80cm

2x =7.6 m

Check for punching Shear

d = 730 mm

1.13m

0.765

A

Draw S.F.D & B.M.D

Stress under footing

= 190 *1.8 = 342 kN/m

Check for beam shear

b = 1800mm, d = 730mm

Under Column B

Shrinkage Reinforcement in short direction

Footing Design

Part III

Combined footing, strip footing, & Mat foundation

Example 2

Design a combined footing As shown

1950kN

1200 kN

750 kN

A

x2=4.2m

x

Dimension calculation

The base dimension to get uniform distributed load

750(4.2)+1200(0.2)=1950 (x)

x = 1.75m

x1=0.2m

Check for punching Shear

h= 750mm

d = 732 mm

A

B2=1m

B1=4m

Empirical S.F.D & B.M.D

m

Convert trapezoidal load to rectangle

Mmax

Clear distance between column

B in moment design = ave. width = 2.5m

b

Y=1.5m

1m

x

4m

Check for beam shear

d = 665mm

Under Column B

Shrinkage Reinforcement in short direction

Example 3 (Strip footing)

Design a combined footing As shown

Dimension calculation

The base dimension to get uniform distributed load

3040kN

800 kN

1280 kN

960 kN

A

Assume

L1=0.6

x1=5.2m

L2

x2=10.7m

x

800(0.6)+1280(5.1)+960(10.6)=

3040 (x)

x = 5.65m,

2(x)=11.3m

L2=11.3 -(10.6)=0.7

Check for punching Shear

h = 700 mm

d=630mm

Example

B

You can check other columns

Draw S.F.D & B.M.D

Stress under footing

= 195 *1.8 = 351 kN/m

Check for beam shear

b = 1800mm, d = 630mm

Bending moment Long direction

Design Short direction as example 1 (lecture 11)