Annex C(Clause 6.1.9)

NBC 2016 (SP 7:2016) Volume 1

2016 · PDF page 1210

Hover any section to copy its link or text, ask Copilot about it, or bookmark it.

Procedure for Calculating the Approximate Deflection at Centre of Glass Panel

C-1 Maximum glass deflection as a function of plate geometry and load may be calculated using the following:

Deflection, $w = t$ . exp $(r_0 + r_1X + r_2X^2)$

where

\begin{array}{r l} {X} & {= \ln \{\ln [ q (a. b) ^ {2} / E t ^ {4} ] \}} \end{array}

r _ {0} = 0. 5 5 3 - 3. 8 3 (a / b) + 1. 1 1 (a / b) ^ {2} - 0. 0 6 9 (a / b) ^ {3}

r _ {1} = - 2. 2 9 + 5. 5 3 (a / b) - 2. 1 7 (a / b) ^ {2} +

0. 2 0 6 7 (a / b) ^ {3}

maximum lateral deflection (w) of a vertical 1 200 mm × 1 500 mm × 6 mm rectangular glass plate subjected to a uniform lateral load of 1.80 kPa. The actual thickness of the glass is 5.60 mm as determined through direct measurement.

r _ {0} = - 2. 6 8 9

r _ {2} = 1. 4 8 5 - 1. 9 0 8 (a / b) + 0. 8 1 5 (a / b) ^ {2} -

a = 1 5 0 0; b = 1 2 0 0

r _ {1} = 2. 0 1 1

0. 0 8 2 2 (a / b) ^ {3}

r _ {2} = 0. 2 1 3

q = 1. 8 0 \times 1 0 ^ {- 3} \mathrm{N/mm} ^ {2}

where

E = 71.73 \times 10^{3} \mathrm{N/mm}^{2}

$E =$ Young's modulus of glass, $\mathrm{N / mm^2}$ ;

t = 5. 6 0

$q =$ net pressure on the pane, $\mathrm{N / mm^2}$ ;

$a =$ longer dimension, mm;

X = 1. 4 9 0

$b =$ shorter dimension, mm; and

$t =$ thickness of the glass pane calculated.

Example:

Lateral deflection calculation — Determine the

Therefore maximum deflection at the centre of glass is:

\begin{array}{c} w = 5. 6 \mathrm{x} \exp (- 2. 6 8 9 + 2. 1 1 1 \times 1. 4 9 0 + \\ 0. 2 1 3 \times 1. 4 9 0 ^ {2}) \end{array}

w = 12.2 \mathrm{mm}

PART 6 STRUCTURAL DESIGN — SECTION 8 GLASS AND GLAZING

Model Calculation of Glass Thickness

D-1 Model Calculation

D-1.1 Glass panel dimension width 1 200 mm and height 1 800 mm for location Chennai for annealed glass type.

D-1.1.1 General Checks has to be Done for Safety Performance

a) The maximum area of glass panel is restricted to $15\mathrm{m}^2$ .

b) The maximum span of window is restricted to $4\mathrm{m}$ .

c) Applicable to normal, reflective, laminated, tempered and insulating glass.

d) Applicable to rectangular panels properly fixed.

The procedure for calculating glass thickness is described below:

Step 1 — Calculate aspect ratio ( $AR_{max}$ ) from Table 16.

4

]}

2

−

0.069(a/b)3

Aspect ratio $(AR_{\mathrm{max}}) =$ Longer side/shorter side of glass

Aspect ratio $= 1800 / 1200 = 1.5$

Step 2 — Wind load calculation

P _ {\mathrm{z}} = 0. 6 V _ {\mathrm{z}} ^ {2}

V _ {z} = V _ {\mathrm{b}}. k _ {1}. k _ {2}. k _ {3}

where

$V_{\mathrm{b}} =$ basic wind speed based on location {see Fig. 1 of good practice [6-8(7)]}.

$k_{1} =$ risk co-efficient factor taken as per good practice [6-8(7)].

$k_{2} =$ terrain factor $\{\text{see Table 2 of good practice [6 - 8(7)]}\}$

$k_{3} =$ topography factor $\{\text{see Table 3 of good practice [6 - 8(7)]}\}$

$V_{\mathrm{z}} =$ design wind speed.

$P_{z}$ = design wind pressure, N/m $^{2}$ , at height, h.

Design wind pressure for Chennai location {Basic wind speed = 50 m/s taken from good practice [6-8(7)]} = 1 750 N/m²

3) Step 3 — Strength factor

Based on the types of glass (Laminated/Tempered/Insulating Glass) pressure factor is calculated. To determine the thickness of laminated/tempered/insulating glass, the

design wind pressure $P_{\mathrm{net}}$ is modified as below:

P _ {\mathrm{net}} = P _ {\mathrm{z}} / P _ {\mathrm{f}}

where

$P_{\mathrm{f}} =$ pressure factor dependent on the type of glass.

The values of the pressure factor $P_{\mathrm{f}}$ can be taken from Table 2.

For annealed units, $P_{\mathrm{f}} = 1.0$ , therefore,

P _ {\mathrm{net}} = 1 7 5 0 / 1. 0 = 1 7 5 0 \mathrm{N} / \mathrm{m} ^ {2}

At this stage the glass thickness value shall be calculated from the table and then the deflection and stress check be done as described below:

Deflection check — The deflection of the glass can be calculated using the formula in Annex C. Actual deflection for the pane shall be obtained as below:

\text {Deflection,} w = t. \exp (r _ {0} + r _ {1} X + r _ {2} X ^ {2})

X = \ln \left\{\ln [ q (a ^ {*} b) ^ {2} / E t ^ {\prime} ] \right\}

\begin{array}{c} r _ {0} = 0. 5 5 3 - 3. 8 3 (\mathrm{a} / b) + 1. 1 1 (\mathrm{a} / b) ^ {2} - \\ 0. 0 6 9 (\mathrm{a} / b) ^ {3} \end{array}

\begin{array}{c} r _ {1} = - 2. 2 9 + 5. 5 3 (\mathrm{a/b}) - 2. 1 7 (\mathrm{a/b}) ^ {2} + \\ 0. 2 0 6 7 (\mathrm{a/b}) ^ {3} \end{array}

\begin{array}{c} r _ {2} = 1. 4 8 5 - 1. 9 0 8 (a / b) + 0. 8 1 5 (a / b) ^ {2} - \\ 0. 0 8 2 2 (a / b) ^ {3} \end{array}

where

E = \text {Young's modulus} (7 1. 7 \times 1 0 ^ {3} \mathrm{N/mm} ^ {2}),

q = \text {net pressure on the pane (N / mm} ^ {2}),

a = \text { longer dimension (mm) },

b = \text {shorter dimension (mm), and}

$T =$ thickness of the glass pane calculated.

The actual deflection is, $4\mathrm{mm} < 11\mathrm{mm}$ maximum allowable.

Result: $1800 / 175 = 10.3\mathrm{mm}$ , which is the maximum allowable deflection.

Hence the glass thickness calculated is safe against deflection.

Glass thickness calculation:

P _ {\mathrm{net}} A = 2 0 0. 0 T ^ {k} (\text {let} T \leq 6 \mathrm{mm})

NATIONAL BUILDING CODE OF INDIA 2016

P _ {\text {net}} = 1 7 5 0

A = 2. 1 6 \mathrm{m} ^ {2}

K = 1. 7 6 5

$T = 5.28 \mathrm{~mm}$ , therefore choose the thickness of glass as $6 \mathrm{~mm}$ .

r2

X2

)

}

/

b

)2

−

0.069(a/b)3

2.17

(

a/b

)2

+

0.2067(a/b)3

)2

−

0.0822(a/b)3