# Deflections Due To Shrinkage | Calculation | Formulas | Example

Shrinkage in concrete produces a compressive stress in the longitudinal reinforcement of beams and slabs. Tensile stress crops in response to balance the compressive stress. Usually, reinforcements are not placed symmetrically to the concrete centroid. And if the same in case of shrinkage, then shrinkage produces curvature in the concrete beams, slabs and this results in corresponding deflection. This deflection will be in the same line as those that are produced because of the loads, unless if the steel reinforcement is on the side then it leads the member in flexural tension.

## Calculation of Shrinkage Deflection

Shrinkage deflection is calculated along with the creep deflection. ACI has defined a procedure in code 6.7d. There are also certain cases where a separate calculation of shrinkage deflection is required. For example, in case of thin and lightweight slabs. In concrete, compression steel reduces the deflection because of the loads are very minute. But, it plays a very significant role in reducing the deflections that occur due to shrinkage and creep. In most of the cases compression steel is only added for this purpose.

## Formulas For Shrinkage Deflection

### 1. Curvature Due To Shrinkage

The analysis gets more complicated when considering the effect of cracking and creep. By using un-cracked gross concrete section, good results can be obtained.

$varnothing _{sh} = frac{2T_{sh}e}{E_{c}I_{g}}$

### 2. Deflection Due To Shrinakage

$bg_green Delta _{sh} = k_{sh}o _{sh}I^{2}$

## Example

### Statement

Calculate the deflection due to shrinkage in a simply supported beam having 20ft span.
$small dpi{100} epsilon _{sh} = 780*10^{-6}$

Width of concrete cross-section = b = 10in.

Effective depth = d = 20in.

$small dpi{100} A_{s} = 3.00 in^{2}$

Compression Steel = A’s= 0

$small dpi{100} E_{c}= 3.6*10^{6} Psi$

$small dpi{100} E_{s}= 29*10^{6} Psi$

## Solution

Tensile force = (3.00)*(780*10^-6)*(29*10^6) = 67900lb.

Gross moment of inertia = 6670

place the value in formula 1 (Curvature due to shrinkage)
$dpi{100} bg_red phi _{sh} = 42.4*10^{-6}$

Now deflection can be easily calculated by using the second formula for deflection.

Here, Ksh = 0.125 for simply supported beam
$dpi{100} bg_green Delta = 0.125*42.4*10^{-6}*240^{2} = 0.31 in.$