1.0DESIGN OF A BUS STOP KIOSK1.1 Introduction to design of Bus-stop KioskWe 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 DesignThe 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 available603719240859Fig.1.1.1 details of the bus stop kiosk design section1.2Structural Analysis1.

2.1Roof Beam StructureThe 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 bhShape Material E (MPa) V P(kg/m3)Hollow section Structural steel 193000 0.

27 7850B = 0.05, h=0.07, d =0.005m, Mass = 8.13kg/m, Length = 1.2mArea of the beam = AA=A1 – A2= (0.07 x 0.05) – (0.

06 x 0.04) = 0.0011m2Y axis centroid = 0.

07/2 = 0.035mX axis Centroid = 0.05/2 = 0.025mb cx cy hArea 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 x10-2m1.

2.2Load AnalysisDesigning 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 codeFor every m2 of snow will weigh about 95.5kgDimension of Roof 4.5 x 1.2m, with each section peg at 1.2m x 2.

5m, these will be the basis for our calculationsRoof 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 areaTherefore, the pressure load acting on the roof (KN/m2) 288 x 9.81/3.0 = 941.

76Nm-21.2.2.

1Load Acting on Beam 1.2m941.76Nm-2 x 1.

2m = 1130.112 Nm-1Self weight of the beam = 8.13 x 9.81 = 79.56NTotal Load = 1130.

112 + 79.56 = 1209.672N19050984251.2.

3Shear forceWhen x =0 to x =1.2Taking 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 = 202.2N337103637201.2.

4Bending Moment1.2.4.1Bending momentLet x = 0 to x = 1.2 for bending momentAssuming a cut for 0 ? x ? 1.2:-394970-398145?Mx = 0+202.

2(x-0) – M1(x) = 0M1(x) = +202.2x for equation 0 ? x ? 1Taking a cut for 1? x ? 1.219050-1822For 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 ? 21.

2.4.2Maximum Bending MomentAssuming 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) = 1007.44KNm19050-349885Maximum bending Moment for beam 1.2m1.2.

5DeflectionMaximum deflection ?C = FL3 48EI?C = (1209.64x 1.23) / (48 x 193000 x 7.0917 x 10-7) = 2.3882 x 10-3mIf allowable Deflection = L/240, then 1200/240 = 5mm, hence the materials selected for the design is can withstand deflection1.3Column DesignIf 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.

5m50 70Assuming a lateral section of 50mmRatio = 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 bucklingEuler’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 column1.

3.1Column Joint constantEnd Fixing K Value (Practical)Pinned (End) 1Fixed Ends 0.65Fixed 0.8Free 2.1The column support from our drawing is pinned which means the value of K is 1Therefore, the effective length Le= L x K = 4.

5 x 1 = 4.5mThe 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 x10-2mSlenderness ratio (S) = Le / RTherefore, S = Le / R = 4.5 / 1.9287 x10-2 = 233.

3281.3.2Critical 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 beamProperties ValueMass 8.12 kg/mYoung Modulus 193paLength 4.

5 x 1.2mMoment of Inertia (2nd) 7.10 x 10-7mArea 0.0011 m2Max.

Bending Moment 1007.44KNmYield force 391KNBuckling Force 384.9 KN