Stress and strain of uncracked sections - Concrete Structures: Stresses and Deformations - page 67

Forces necessary to restrain relaxation (Equation 2.44) are:

( ∆ N ) relaxation =− 84 kip; ( ∆ M ) relaxation = 4074 kip-in.

The total restraining forces are:

∆ N = 3177 kip; ∆ M =− 154.1 × 10 3 kip-in.

Properties of the age-adjusted transformed section are:

¯ A = 10 170 in 2 ; B =− 242.1 × 10 3 in 3 ; = 16.29 × 10 6 in 4 .

M on the age-adjusted section and use Equation
(2.19) to calculate the changes in strain between t 0 and t :

Apply −∆ N and −∆

∆ ε O ( t , t 0 ) =− 86.66 × 10 −6 ; ∆ ψ ( t , t 0 ) = 4.784 × 10 −6 in −1 .

Adding the change in strain to the initial strain in Fig. 2.16(b) gives
the total strain at time t , shown in Fig. 2.16(c).
The time-dependent change in stress in concrete is calculated by
Equation (2.46):

[ ∆ σ c ( t , t 0 )] top = 0.729 + 1558[ − 86.7 × 10 −6

+

4.784 × 10 −6 ( − 56)] = 0.177 ksi.

[ ∆ σ c ( t , t 0 )] bot = 0.729 + 1558[ − 86.7 × 10 −6

+

4.784 × 10 −6 ( − 40)] = 0.296 ksi.

Adding these stresses to the initial stress (Fig. 2.16(b) ) gives the
total stress at time t shown in Fig. 2.16(c). It is interesting to note the
change in the resultant force on the concrete (the area of the concrete
cross-section multiplied by the stress at its centroid). The values of the
resultants are 1130 kip at time t 0 and t respectively. The

substantial drop in compressive force is due to the fact that the time-
dependent shortening of the concrete is restrained after its attachment
to a relatively sti

− 2180 and −

ff structural steel section.