5 A reinforced concrete T 20 member 300 mm deep and spannin

5. A reinforced concrete T = 20° member, 300 mm deep and spanning 4.0 m, is fully fixed at the supports. The N member is subjected to a linear thermal gradient such that the top surface temperature is increased by 20°C while the bottom surface temperature is held unchanged. Assume that the member is completely unstressed at the initial condition (ie. At AT = 0°). AT=0 4000 20°C 50 300 200 50 0°C i) Determine the induced reaction forces, M and 4#25 f\'c-30 MPa -e, =-2.0 x 10-3 fy 400 MP:a 10 x 10&PC; 12 x 10 oc ii) Does the member crack? a Hint: Because of the fully fixed conditions, the total strains in the concrete and reinforcement are zero at all locations

Solution

average thermal coefficient = (1+2) /2 = 11 x10^-6 /^oC

A= 300*200= 60000 mm²

compressive stress = N/A= ET

E=4700 f’c Mpa = 4700*30^1/2 = 25742.9 N/mm²

f= 25742.9*11*10^-6*20 =5.66 N/mm²

N=f*A= 5.66*60000 = 339.807 KN

M=EI*

I= 1/12*200*300³ =4.5*10⁸ mm⁴

=* thermal gradient = 11*10^-6 *20/300 = 7.33*10^-7 /mm

M= 4.5*10⁸ *25742.9 *7.33*10^-7 = 8491.3 N-m

stress is only 5.66 N/mm² mamber will not crack