Week 6 Challenge
1) AIM : To design chequered plate.
Introduction :
Uniformly factored distributed live load LL = 15 KN/m2.
Assume factored UDL from Dead Load DL = 0.75 KN/m2.
Dimensions of plate a = 50000 mm = 5m, b = 1200 mm = 1.2m
Assume yield strength of plate fy = 250N/mm2.
Procedure :
Step 1) Check for Bending Stress
Permissible bending stress = 0.66*fy = 0.66*250 = 165 N/mm2
Using Table 11.4 from Roarks chart
a = 5m, b = 1.2m
a/b = 5/1.2 = 4.166
So,
β1 = 0.5, β2 = 0.25, α = 0.0284
Tota load = DL+LL q = 15+0.75 = 15.75KN/m2 = 15.75*1000/10^6 = 0.016 N/mm2
Maximum bending stress acts at centre of long edge and it should be less than permissible bending stress
σmax = σper = β1*q*b^2/t^2, Here 't' is thickness of plate
⇒ 165 = 0.5*0.016*1200^2/t^2
⇒ t = 8.35 mm
Consider thickness of chequered plate = 10mm
Step 1) Check for Deflection :
Maximum deflection = (α*q*b^4)/(E*t^3) = (0.0284*0.016*1200^4)/(2*10^5*10^3) = 4.72mm
Allowable deflection = thickness of plate/2 = 10/2 = 5mm. Hence safe.
RESULT :
Thickness of chequered plate = 10mm.
2) AIM : To design deck slab.
INTRODUCTION :
Dead Load D.L = 5KN/m2
Live Load L.L = 8 KN/m2
PROCEDURE :
Assume thickness of slab = 100mm = 0.1m.
Self Weight of slab = 25*0.1 = 2.5 KN/m2
Total Load = (2.5+5+8)*1.5 = 23.25 KN/m2
Let us try PIL 75/230 profile deck sheeting.
Properties of profile
Thickness of deck sheet = 1.25mm
span = 1.5 m
Self Weight = 0.14 KN/m2
Allowable load = 29.36KN/m2
This profile spans for 2 to 3 supports
Hence allowable load = 0.8*29.36 KN/m2 = 23.48 KN/m2
Total Load = 23.25+0.14 = 23.39 KN/m2 < 23.48 KN/m2
As the maximum spacing of beams allowed is 1.5m due to the selected profile.
The maximum span of slab 1.5m*8m
ly/lx = 8/1.5 = 5.33 > 2. Hence slab is one way slab.
From IS 456, span/d = 25 ⇒ d =1500/25 = 60mm
Depth of deck 75 mm.
Effective depth = 60+(75/2) = 97.5 < 100 mm.
Design of RCc slab as per IS 456.
RESULT :
Thickness of RCC slab = 100 mm
3) AIM : To draw the load path and type of axial forces acting on the different types of bracing system due to the lateral and gravity loads.
PROCEDURE :
