| Product(s): | SACS |
| Version(s): | 23 |
| Area: | SACS IV |
Problem
How does SACS calculate member load for temperature difference?
Solution
Referring to Section 2.9.1.3 of SACS IV manual, the typical temperature load application pattern is briefed.
Now let us walk through a sample model to understand how SACS generate the member end loads from given temperature input.
A simple model of beam (5m length) with pinned connections is used here. The cross section of the beam is defined as prismatic member with 45x60 cm^2 dimensional properties. E and α are considered as 20,000 kN/cm^2 and 11.7*10^-6 respectively.
The SACS manual, there is a snap provided to understand the temperature changes.
For Case A, B and D, the temperature gradient change is constant whereas for the other cases it is linearly variable.
Temperature loads in the SACS model are applied as below, considering 4 probable scenarios.
The analysis results show the member end forces as below.
Now let us try to validate the same by hand calculations. You may refer SACV IV manual Section 7.1.2 for a sample problem.
In the SACS model, the case 1 where upper and lower surface temperature change is 100 respectively, the temperature change gradient is linear. Anywhere in between the change is always 100. While applying the temperature load, SACS gives an option of Member Depth when load is applied along member local Y or Z axis. However, for member local X axis, the default depth is set as 1 unit.
When the user defines the depth, based on actual depth of the member SACS internally sub divide the section into number of fiber components. Then it calculates the deflection, strain and consequently the force of each virtually defined segment. Finally, SACS will add up all the forces of the segments to generate the member end forces.
In case of constant gradient, the following calculation is generated.
In case of variable linear gradient, as the number of segments is introduced by defining the Member Depth in Temperature Load, the intermediate temperature change is determined by linear interpolation. In our case, when we divide the member into 2 segments for say 100 degrees vs 20 degrees case, then the intermediate temperature change is average of 100 and 20 which is 60 degrees. But the section is divided into 2 fibers. Top one is having top surface temperature change as 100 degree and 60 degrees. So, average of temperature change (ΔT) for this top fiber is average of 100 and 60, which comes as 80 degrees. Similarly, the ΔT for bottom fiber is 40 degrees.
For the variable gradient cases, the following calculation is generated.