STRUCTUAL GLASS DESIGN OF HAND RAIL – BALCONY
I.
General 1. Design Philosophy - PREAMBLE The purpose of this calculation is to verify the thickness of glass and its supporting structure for its integrity, strength and stability verification. The result of this conclusion is that the structure considered adequate in meeting the required of design criteria. The glass used is 13.52mm thick laminated glass fixed using a bracket. 2. Unit of Measurement Unit of measurement in design shall be in Metric system.
II.
Design Calculations 1. Design Code and reference: BS 5950 1190: Structural use of steel work in building BS 6399 Part 1: 1996 – Staircase loading Wind loads as per UBC 1997
2. Materials The modulus of elasticity of Steel E = 210000 MPa Ultimate Bending stress steel – Po - σ = 275 MPa = 275 N/mm2 Ultimate Tensile Stress steel – Pa = 435 MPa = 435 N/mm2 Ultimate Shear Stress steel – Pv = 0.6X Po = 0.6X275 = 96 MPa = 96 N/mm2 All bolts used shall be grade 8.8 or high yield. GLASS PROPERTIES 13.52mm thick Laminated Glass The modulus of elasticity E = 72.7 MPa The modulus of Shear G = 29.6 MPa The modulus of Rupture = 41.4 MPa Poisson Ratio = 0.23 Density = 24.525 N/mm2 per m thick sheet Flexure Strength = 166 MPa = 166 N/mm 2
3. Loading Dead Load – Self weight is considered The Forces acting on the Handrails apart from self weight are as follows which is complying with the required standards:
As per BS 6399 Part1 1996, Handrail shall be designed to resist a load of 0.74 KN/m applied in any direction at the top and to transfer this load through the supports to the structure.
As per BS 6399 Part1 1996, Handrail shall be designed to resist a load of 1.0 KN/m2 applied to the infill.
Horizontal Load and moment acting at the support is calculated as mentioned below: Case 1:
Uniform loading of 0.74 KN/m @ top
Width of the loading to be considered as 120 mm Horizontal load = 0.74 X 0.12 = 0.0889 KN Maximum Moment developed by the above forces from the bottom surface i.e. at a distance of 1.10m. Mx1 = (0.0889) X 1.10 Mx1 = 0.0977 KNm Case 2:
Loading of 1.0 KN/m2 on the surface / infill
Area of the loading to be considered as 0.120m X 1.10 = 0.132 m2 Horizontal load = 1.0 X 0.132 = 0.132 KN Distance from the centre of loading to the bottom surface = 1.10/2 = 0.55m Maximum Moment developed by the above forces from the bottom surface i.e. at a distance of 0.55m. Mx1 = (0.132) X 0.55 Mx1 = 0.0726KNm
Wind load calculations as per UBC 1997: Design Wind loads (WL) calculated as per UBC 1997 P = Ce Cq qs Iw Where, Ce : The combined height, exposure, and gust factor coefficient, which depends on the height above the ground, Cq : Pressure co-efficient, which depends on the type of building, qs : Wind stagnation pressure at 10m, which depends on the maximum gust wind speed for the region, and Iw : Wind importance factor, which depends on the function of the structure. The structure is located at a height of less than 5.0m. Basic wind speed assumed as 100 mph.
The co-efficients are as follows: Ce : The height considers at 5m with exposure type B – hence the coefficient is 0.63, - Table 16 G Cq : Pressure co-efficient for height of structure are 0.8 windward, and 0.5 as leeward . – Table 16 H qs : Wind stagnation pressure = 0.00256 V2 , qs = 0.00256 (100)2, qs = 25.6 psf, Iw : Wind importance factor = 1.0. Table 16 K Design wind load P = Ce Cq qs Iw P = 0.63 X 0.80 X 25.6 X 1.0 (Windward direction) (1psf = 47.88 N/m2)
P = 12.902psf (Windward direction) P = 0.618 KN/m2 (Windward direction) P = 0.63 X 0.5 X 25.6 X 1.0 (Leeward direction) P = 8.064psf (Leeward direction)
(1psf = 47.88 N/m2)
P = 0.386KN/m2 (Leeward direction) Total Ultimate Load on Aluminium = 1.2 X 0.618 UL = 0.742KN/m2 IV.
Design Method and Calculations Basic data and Calculations The analysis is carried out manually to verify the sections used are safe. The width of the glass railing structure is considered as 120mm, Hence, the wind load acting is calculated as mentioned below WL = 0.120 X 0.742 = 0.089 KN/m. Factored Load = 1.4 X 0.089 = 0.125 KN/m The uniform loads are used to check the moment and deflections. Check for Moment: Maximum Bending Moment due to the above loading criteria for a span of 1.10 m
B.M (max) = w l2 /2 B.M (max) = 0.125 X (1.10)2 /2 B.M (max) = 0.090KNm The Moment of resistance of the section must be larger than above value. Calculation of Moment of Resistance (MR) MR = Z x Po Z = Sectional Modulus Z=I
xx
/ŷ
The glass section used is 120 X 13.52mm, Moment of Inertia & Sectional Modulus are calculated as below: The moment of inertia Ixx = bd3/12 = 120 X 13.523 / 12, I
xx
= 24713 mm4
Z = bd2/6 = 120 X 13.522 / 6, Z = = 3655 mm3 MR = 3655 X 166 MR = 0.606KNm which is greater than the 0.090KNm The Moment of Resistance is greater than the Maximum Bending moment. Hence the section used to resist the ultimate loads supported is safe. Check for Deflection: δ = WL4 / (8 EI) δ < L / 180 E = 72000 MPa I xx = 24713 mm4 as shown above,
Loading shall be considered as 80% of the Ultimate load in calculating the deflection, W = 0.80 X 0.089 = 0.0712 KN/m δ = (0.0712 X 1.10 X 1000) X (1100)3 /(8 X 72000 X 24713) Note: for long term deflection un-factored load is considered. δ = 7.32 mm L/180 = 1100 / 180 L/180 = 6.11mm δ < 6.11mm Hence the deflection is more than the permissible values. Therefore, you are requested to revise the thickness of the Glass with 15.52mm thick, The moment of inertia Ixx = bd3/12 = 120 X 15.523 / 12, I
xx
= 37383 mm4
δ = (0.0712 X 1.10 X 1000) X (1100)3 /(8 X 72000 X 37383) δ = 4.84 mm Hence the deflection is less than the permissible values. Therefore, please use the glass thickness 15.52 laminated.
Conclusion: All sections used are structurally sufficient and meet its intended purpose.
