1.0DESIGN OF A BUS STOP KIOSK

1.1 Introduction to design of Bus-stop Kiosk

We present a simple design of a bus – stop kiosk using the simplest design methods. A steel plate will be used for the roof panel because still have the capacity to withstand load from rain and wind, and also because it is easy to bend and drill for further connections. The plate will consist of four beam at the edges to act as a support to the load on top of the plate and also to act as an attachment for the roof mainframe. Paints will be applied to the entire structure to protect the component from weather effects and corrosion, we believe it will also help in the modification of the entire structures.

Details of the Design

The structure is rectangular and will be designed with a gentle slope. It has a vertical beam (4 pieces) acting as a structural support to the horizontal column holding the beam together. The bus stop kiosk has a dimension of 4.5 x 1.2 x 2.5m with an inner space for user’s. It consists of seat for inner fixtures for the purpose of those who would want to rest till the next bus is available

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Fig.1.1.1 details of the bus stop kiosk design section

1.2Structural Analysis

1.2.1Roof Beam Structure

The structural analysis was focused on beam structure of length 1.2m for the trusses (Modular) to resist any lateral load acting on it. The structural steel beam will beam will be designed to S355 eurocode with anti-corrosive properties

b

h

Shape Material E (MPa) V P(kg/m3)

Hollow section Structural steel 193000 0.27 7850

B = 0.05, h=0.07, d =0.005m, Mass = 8.13kg/m, Length = 1.2m

Area of the beam = AA=A1 – A2= (0.07 x 0.05) – (0.06 x 0.04) = 0.0011m2

Y axis centroid = 0.07/2 = 0.035m

X axis Centroid = 0.05/2 = 0.025m

b

cx

cy h

Area moment of inertia about Y- axis (Iy) = (bh3 – (b-2d) (h-2d) 3)

12

= (0.05 x 0.073) – (0.04 x 0.063)/ 12 = 7.0917 x 10-7 m4

Area moment of inertia (Z-axis) (IZ) = (b3h–(b-2d) 3 (h-2d)/12

= (0.053 x 0.07) – (0.043 x 0.06)/ 12 = 4.0917 x 10-7 m4Sectional modulus, Z = Iy /y = 7.0917 x 10-7 / 0.035 = 2.0262 x 10-5 m3Radius of gyration, R = (Iy/AA) 1/2= (7.0917 x 10-7 / 0.0011)1/2= 2.5391 x10-2m

1.2.2Load Analysis

Designing based on the Eurocode, consideration will be given to snow, since Bus-stop Kiosk belongs to category C, and the following assumption will be made based on the code

For every m2 of snow will weigh about 95.5kg

Dimension of Roof 4.5 x 1.2m, with each section peg at 1.2m x 2.5m, these will be the basis for our calculations

Roof area = 1.2 x 2.5 = 3.0 m2If 1m2= 96kg then, 3.0m= 3.0 x 96 = 288kg of snow mass will act on the roof area

Therefore, the pressure load acting on the roof (KN/m2) 288 x 9.81/3.0 = 941.76Nm-2

1.2.2.1Load Acting on Beam 1.2m

941.76Nm-2 x 1.2m = 1130.112 Nm-1

Self weight of the beam = 8.13 x 9.81 = 79.56N

Total Load = 1130.112 + 79.56 = 1209.672N

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1.2.3Shear force

When x =0 to x =1.2

Taking Reaction from the left support RBSum of forces acting along the right support: +? ? f (y) = 0RA+ RB – 1209.64 = 0RA + RB= 1209.64N ———————————————————–(1)Sum of moment about the left support equals zero for static equilibrium:?M = 0RB (1.2-0) + (-1209.64) = 0——————————————————— (1) RB = 1007.44N (2)From equation (1)

RA + RB= 1209.64N, RA = 1209.64 – RBRA = 1209.64 – 1007.44 = 202.2N

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1.2.4Bending Moment

1.2.4.1Bending moment

Let x = 0 to x = 1.2 for bending momentAssuming a cut for 0 ? x ? 1.2:

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?Mx = 0+202.2(x-0) – M1(x) = 0M1(x) = +202.2x for equation 0 ? x ? 1Taking a cut for 1? x ? 1.2

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For x = 0 to x = 2:?Mx = 0+202.2(x-0) + (-1209.64) (x-1) – M2(x) = 0M2(x) = +1209.64-202.2x equation for 1 ? x ? 2

1.2.4.2Maximum Bending Moment

Assuming a maximum Bending at the middle of the steel beam where x =1Using previously generated equation for the steel beam; Mmax(x) = +1209.64-202.2xTherefore, Mmax(1) = +1209.64-202.2(1) = 1007.44KNm

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Maximum bending Moment for beam 1.2m

1.2.5Deflection

Maximum deflection ?C = FL3 48EI

?C = (1209.64x 1.23) / (48 x 193000 x 7.0917 x 10-7) = 2.3882 x 10-3m

If allowable Deflection = L/240, then 1200/240 = 5mm, hence the materials selected for the design is can withstand deflection

1.3Column Design

If the ratio of the radius of gyration to the effective length of column is greater than 50, then we regard it has a long column but if less than 50, it is then called a short column based on Eurocode. For the purpose of this calculation, the column (rectangular hollow) is having a length of height of 4.5m

50

70

Assuming a lateral section of 50mm

Ratio = 4.5/0.05 = 90 ; 15.

Therefore this column is a slender column (long column). Since this column is slender (long) we can apply Euler’s Formula to check for buckling

Euler’s Formula for Critical force Pcr = ?2EI / Le2Pcr = Critical loadE = Modulus of elasticity of the materialI = Moment of inertia of the materialLe= Effective length of the column

1.3.1Column Joint constant

End Fixing K Value (Practical)

Pinned (End) 1

Fixed Ends 0.65

Fixed 0.8

Free 2.1

The column support from our drawing is pinned which means the value of K is 1

Therefore, the effective length Le= L x K = 4.5 x 1 = 4.5m

The moment of inertia with least axis will most likely have bucklingmoment of inertia about this axis aremoment of inertia Y-axis (Iy) = 7.0917 x 10-7 m4moment of inertia Z-axis (IZ) = 4.0917 x 10-7 m4buckling will occur about the Z- axis.Radius of gyration about the Z-axis, R = (IZ/AA) 1/2= (4.0917 x 10-7 / 0.0011)1/2= 1.9287 x10-2m

Slenderness ratio (S) = Le / RTherefore, S = Le / R = 4.5 / 1.9287 x10-2 = 233.328

1.3.2Critical forceCritical force Pcr = ?2EI / Le2For column 4.5m, Pcr= (?2 x 193000x 4.0917 x 10-7x 1012) / 4.52= 384989.094N = 384.9 KN (Buckling force for 4.5m column)

1.3.3Critical stressCritical Stress (?critical) = ?2E / (Le/ R) 2?critical = ?2 x 193000/ (4.5 / 1.9287 x10-2)2= 33.98MPaResult show that the critical stress is less than the material yield stress.

1.3.4Yield forceYield Stress ?yield = Yield force (Pyield) / Area (A)Pyield = ?yield x A. The yield stress for S355 Pyield = 355 x 0.0011 x 106 = 390500N = 391KN (355 yield force)Since the material yield force (391KN) is greater than the buckling force(384.9KN), the material is good for bus stop koisk structure.

1.5Materials properties for the S355 beam

Properties Value

Mass 8.12 kg/m

Young Modulus 193pa

Length 4.5 x 1.2m

Moment of Inertia (2nd) 7.10 x 10-7m

Area 0.0011 m2

Max. Bending Moment 1007.44KNm

Yield force 391KN

Buckling Force 384.9 KN