Non-linear analysis of plane frames - Concrete Structures: Stresses and Deformations - page 438

Table 13.1 Strain parameters at any section of the cantilever of Example 13.2 due to
unit force F 1 *, F 2 * or F 3 * applied at the free end

Force Strain Uncracked Fully cracked Mean Multiplier

applied parameters

F 1 * = 1 − 5405 − 25780 − 21700 10 − 3 ( h 2 E c ) − 1

Ou1

7886 30130 25680 10 − 3 ( h 3 E c ) − 1

u 1

F 2 * = 1 7886 x 30130 x 25680 x 10 − 3 ( h 3 E c ) − 1

Ou2

− 15770 x − 41530 x − 36380 x 10 − 3 ( h 4 E c ) − 1

u2

F 3 * = 1 − 7886 − 30130 − 25680 10 − 3 ( h 3 E c ) − 1

Ou3

15770 41530 36380 10 − 3 ( h 4 E c ) − 1

u3

in Equations (8.43) and (8.44). The results of these calculations are
presented in Table 13.1. We give below, as example, the calculations for
F 1

1.
For, F 1

=

= 1, N u1 = − 1 and M u1 = 0 at any section. Apply Equation

(13.12) for uncracked section:

0.2344
E c h 2 [0.6840(0.2344)

ε O = −

(0.3420) 2 ]

= − 5405 × 10 −3 ( E c h 2 ) −1

−

0.3420
E c h 3 [0.6840(0.2344)

ψ =

(0.3420) 2 ]

= 7886 × 10 −3 ( E c h 3 ) −1

−

Apply the same equation for a fully cracked section:

0.1582
E c h 2 [0.2549(0.1582)

ε O = −

(0.1849) 2 ]

= − 25780 × 10 −3 ( E c h 2 ) −1

−

0.1849
E c h 3 [0.2549(0.1582)

ψ =

(0.1849) 2 ]

= 30130 × 10 −3 ( E c h 3 ) −1

−

Mean parameters (with ζ = 0.8):

ε O ε O ε O − 21700 h

25680

ψ mean = (1 − ζ ) ψ uncracked + ζ ψ fully-cracked =

( E c h 3 ) −1

The cients are determined by Equation (13.8), with the

integrals evaluated explicitly giving:

fl exibility coe