Effects of temperature - Concrete Structures: Stresses and Deformations - page 365

1.02 1.20

∆ N = − (2.652 × 10 6 ) 0.35 y 5 d y + 2.5 y 5 d y

0 1.02

= − 2.229 × 10 6 N.

1.02

∆ M = − (2.652 × 10 6 ) 0.35 (0.769 − y ) y 5 d y

0

1.2

+ 2.5 (0.769 − y ) y 5 d y = 0.7438 × 10 6 N-m.

1.02

M ) on
the cross-section; the resulting axial strain and curvature are (Equation
(2.29) ):

Release the arti fi cial restraint by application of ( −∆ N ) and ( −∆

2.229 × 10 6
0.877 1

30×10 9 ×

∆ ε o 84.73 × 10 −6

∆ ψ

= =

− 153.5 × 10 −6 m −1

0.7438 × 10 6
0.1615

−

The stress in a statically determinate beam (the self-equilibrating
stresses, Equation (2.30) ) is:

y ) 5 ] MPa σ self-equilibrating = [2.542 − 4.606 y − 2.652(0.769 −

for y = − 0.431 to 0.769 m

or

σ self-equilibrating = (2.542 − 4.606 y ) MPa

for y = 0.769 to 0.969 m

where y is the distance in metres measured downwards from the
centroid O. The distribution of the self-equilibrating stress is shown in
Fig. 10.8(c).
We use the force method for the analysis of the statically indeter-
minate forces. The structure is released by the introduction of hinges
at B and C as shown in Fig. 10.8(d). The displacements of the
released structure at the two coordinates indicated are (Equation
(10.26) ):