# Beam 82

Extracts from this document...

Introduction

Assignment by:

Sean Lynch

2nd yr HNC student

Assignment i) Beam 82

Total UDL = 17.6 + 7.2 + 0.7 = 25.5 KN/m

## Reaction at supports

RA = 25.5 x 8 = 102 KN RB = 25.5 x 8 = 102 KN

2 2

## Bending Moment

When shear force is at zero maximum bending moment occurs.

Mmax = ½ x 102 x 4 = 204KNm

## Beam 82 with roof addition

#### Total UDL for whole beam

UDL = 0.7 + 7.2 + 17.6

UDL = 25.5 KN/m

Total load of Trapezium

L = (1/2x1.5x18)2 + (18x1) = 45 KN

## Taking moments @ B

8RA = (45 x 2) + (204 x 4)

8RA = 90 + 816

RA = 906/8 = 113.25 KN

## Taking moments @ A

8RB = (204 x 4) + (45 x 6)

8RB = 816 + 270

RB = 1086/8 = 135.75 KN

Calculations for Shear Force diagram (KN)

A-C = 113.25 – (25.5 x 4) = 11.25

C-D = 11.25 – (25.5 x 1.5) – (1/2 x 1.5 x 18) = -40.5

D-E = -40.5 – (25.5 x 1) – (18 x 1) = -84

E-B = -84 – (25.5 x 1.5) – (1/2 x 1.5 x 18) = -135.75

## Calculations for Bending Moments diagram (KNm)

Maximum Bending Moment occurs when Shear Force = 0. Calculations for the distance where SF = 0 can be found below:

Shear Force at centre = 11.25 KN

Distance of Mmax from centre (z):

##### Total weight acting down over length z = 11.25

Therefore:

(½z x 12z) + (z x 25.5) = 11.25

6z² + 25.5z = 11.25

6z² + 25.5z - 11.25 = 0

Middle

## Assignment i) Beam 84

## Total UDL = 5.42 + 0.7 = 6.12 KN/m

## Total load of Trapezium = (1/2 x 1.5 x 18) + (18 x 1) = 45 KN/m

## Reaction at A

## RA = 45 + (6.12 x 4) = 69.48 = 34.74 KN

## 2

## The beam is simply supported and all loads are uniformly distributed:

## RA = RB

## RB = 34.74 KN

## SHEAR FORCE DIAGRAM CALCS (KN)

## A-C = 34.74 – (1/2 x 1.5 x 18) – (6.12 x 1.5) = 12.06

## C-D = 12.06 – (18 x 1) – (6.12 x 1) = -12.06

## D-B = -12.06 – (1/2 x 1.5 x 18) – (6.12 x 1.5) = -34.74

## Bending Moment diagram calculations (KNm)

## @ C = (34.74 x 2.5) – (6.12 x 2.5 x 2.5/2) – (13.5 x 1.5) – (18 x 0.5) = 38.48

## @ middle = (34.74 x 2) – (18 x 0.25) – (13.5 x 0.75) – (6.12 x 2 x 1) = 42.62

## @ D = (34.74 x 1.5) – (13.5 x 0.5) – (6.12 x 1.5 x 0.75) = 38.48

### Assignment ii) Beam 41

## Total UDL = 0.7 + 7.224 = 7.924 KN/m

## Taking Moments about A

9 RB = (143.6 x 3) + (72 x 4.5) + (86.7 x 6) 9 RB = 430.8 + 324 + 520.2

RB = 1275 / 9 = 141.7 KN

## Taking Moments @ B

9 RA = (86.7 x 3) + (72 x 4.5) + (143.6 x 6) 9 RA = 260.1 + 324 + 861.6

RA = 1445.7 / 9 = 160.6 KN

Calculations for Shear Force diagram (KN)

A-C = 160.6 – (7.924 x 3) = 136.6, @ C = 136.6 – 143.6 = -7

C-D = -7 – (7.924 x 3) = -31, @ D = -31 – 86.7 = -117.7

D-B = -117.7 - (7.924 x 3) = -141.7

Calculations for Bending Moment diagram (KNm)

Bendimg mom.

Conclusion

m = 12.656 – 2.103 – 2.25 – 8.303

m = 0 KNm

## When z = 3m

H = 3.554, y = ((4 x 2.5)/6²) x 3(6 – 3) = 2.5

m = (5.625 x 3) – (1.5 x 0.75 x 4.125) – (2 x 2.25 x 1.25) - (3.554 x 2.5)

m = 16.875 – 2.953 – 5.062 – 8.86

m = 0 KNm

## When z = 3.75

H = 3.554, y = ((4 x 2.5)/6²) x 3.75(6 – 3.75) = 2.343

m = (5.625 x 3.75) – (1.5 x 0.75 x 3.375) – (2 x 3 x 1.5) - (3.554 x 2.343)

m = 21.093 – 3.796 – 9 – 8.303

m = 0 KNm

## When z = 4.5m

H = 3.554, y = ((4 x 2.5)/6²) x 4.5(6 – 4.5) = 1.875

m = (5.625 x 4.5) – (1.5 x 0.75 x 4.125) – (2 x 3.75 x 1.875) - (3.554 x 1.875)

m = 25.313 – 4.64 – 14.062 – 6.645

m = -0.035 KNm,

The bending moment values for z = 1.5 and 4.5m are the same because they are in exactly the same place but on the opposite sides of the arch. This principle also applies to the bending moment values of z = 5.25 and 6m:

When z = 5.25m, m = -0.078 KNm

When z = 6m, m = 0 KNm

Distance along arch (m) | Bending Moment (KNm) |

0 | 0 |

0.75 | -0.078 |

1.5 | -0.0345 |

2.25 | 0 |

3 | 0 |

3.75 | 0 |

4.5 | -0.0345 |

5.25 | -0.078 |

6 | 0 |

Calculation of internal forces for the arc shown below.

Slope = 4h(l – 2x)

L²

= 4 x 2.5(6 – ((2 x 0.75))

6²

Slope = 1.25

Therefore the angle to the horizontal = tan-1(1.25) = 51.34º

Resolving parallel to N:

N + (1.5 x 0.75)(sin51.34º) = (3.554 cos51.34º) + (5.625 sin51.34º)

N = 2.213 + 4.392 – 0.878

N = 5.727 KN

Resolving parallel to Q:

Q + (3.544 sin51.34º) + (1.5 x 0.75)( cos51.34º) = (5.625 cos51.34º)

Q = 3.513 – 0.702 – 2.767

Q = 0.044 KN

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