Week 11 Challenge

Aim : To calculate the temeprature loads on the given structures at the given place and to analyse the structure in Staad

Procedure :

Given

width of carriage way = 7m

Span of bridge = 9m

Thickness of deckslab = 0.55m

Assuming the deck is covered with 50mm wearing course on the top

Calculations

(i) Maximum air shade temperature is 47.5

minimum air shade temperature is -2.5

the differrence is 50 i.e., > 20 c

bridge temperature is 22.5+10 = 32.5 C

Positive temperature Differences

h1 = 0.3*0.55 = 0.15m

h2 = 0.3*0.55 = 0.165m

h3 = 0.3*0.55 = 0.15 m

Reverse temperature Differences

h1 = 0.2*0.55 = 0.110m

h2 = 0.2*0.55 = 0.110m

h3= 0.25*0.55 = 0.1375m

h4= 0.25*0.55 = 0.1375m

Let us consider the positive rise case

Total axial force = 5.71+71.27 = 76.98 t = 769.8 KN

Total Moment = 10.824-1.285 = 9.539 t-m = 95.39 KN-m

stress due to axial force = 769.8/(0.55*0.5) = 2799.27KN/m^2

stress due to Moment force = 95.39/((0.5*0.55^2)/6) = 3784.07 KN/m^2

Since Rise rate the stress compression

Let us consider the Reverse temperature

Total Axial force due to fall = 16.76 + 24.29 = 41.05 t =410.5 KN

Total moment = 5.001 -3.697 = 1.304 t-m = 13.04 KN.m

stress due to axial force = 410.5/(0.55*0.5) = 1492KN/m^2

stress due to Moment force = 13.04/((0.5*0.55^2)/6) = 517.289 KN/m^2

Since fall rate the stress tensile

Staad file generation

since the temperature loads are prodominant on the deck slab

The following model is generated

the plate thickness is assumed as 0.55 m

The following temperature loads are generated

in the temperature rise case due to variation of temperature vertically downwards the compressive stress is developed in the members as shown in the figure

In temperature inverse case due to variation of temperature vertically downwards the tensile stresses are generated

The figure shows that there is no error in the model

The figure showing the maimum absolute pressure on plates