Seismic forces with STCMB card



Question

1. I have run two separate analysis on my 3-legged structure to compare the STCMB results - Seismic Only and Seismic + Static combination. The base shear output from Seismic + Static analysis is lower than Seismic Only analysis.

2. What is the reason for load cases 1 & 3 (load in tension) and 2 & 4 (load in compression), there aren't difference between these load cases. I'm expecting the vertical forces must be different since one is compression and the other is tension.


Answer

Q1. STCMB combines modal results across multiple directions (Fx, Fy, etc.), including directional load combinations and includes directionality and mode participation factors. It gives more realistic and conservative combination method in SACS. 

Seismic only


Total Force load = sqrt(19905.488^2 + 16590.479^2 + 105784.156^2) = 108,911.7 kN

Total Moment load = sqrt(859409.812^2 + 5731122.0^2 + 697679.938^2) = 5,837,045.64 kN


Seismic + Static

The below table represents the total of the seismic forces in the spring supports. Each spring bears approximately 1/3rd of the total forces.



The table below shows the spring support forces from static analysis. Note that all Z-forces are one-directional (-Z) while X and Y lateral forces are a mixed of positive and negative forces.



The below table is the Seismic+Static total force:



Total Force load = sqrt(6748.773^2 + 5260.627^2 + 301307.594^2) = 301,429.1 kN (>108,911.7 kN)

Total Moment load = sqrt(1003890.812^2 + 14830048.0^2 + 58035.93^2) = 14,864,100.67 kN (>5,837,045.64 kN)

As you can see, the (Seismic + Static) total load is greater than Seismic alone.

Seismic load in the spring support is added to the static load such that the magnitude of the static load increases. Since all Z static forces in the springs are unidirectional, this results in the Z-direction Seismic+Static load being quite larger than Seismic alone (Note that the spring Z static loads are also large).

For static load along either X or Y direction, there are 2 springs with forces along one direction and 1 spring with the opposite force. Hence, the seismic+static total force becomes smaller than that for seismic alone for the X or Y directions (Note that the spring X and Y static loads are quite small).

The seismic forces are added such that the magnitude of the static force increases (i.e. sign remains the same). What does this means? Let's say you have a static axial load on a column of -100kN (compression), and a seismic load calculated from the response spectrum is +30kN. A normal vector addition would yield -70kN. But with STCMB, SACS changes the seismic force to -30kN (same sign as static), and adds it to get -130kN. This makes the total force more conservative by ensuring that the combined load increases the demand on the structure in the same direction as the original load, regardless of the actual direction of seismic response.

Eqk + Static Forces ---> 0.9 and 1.1 defined in dyrinp. as load case factor for seismic and static forces.

Eqk force in X = -22117.21 kN / 3 legs = -7372.40 kN

First leg
Total force = (0.9*-7372.40 kN) + (1.1*-6.68 kN) = -6642.508 kN (Since -6.68 is -ve, hence using -7372.40)
---> Magnitude of static force increases, but sign remains the same i.e. negative in this case

Second leg
Total force = (0.9*7372.40 kN) + (1.1*4.068 kN) = 6639.63 kN (Since 4.068 is +ve, hence using +7372.40)

Third leg
Total force = (0.9*7372.40 kN) + (1.1*2.611 kN) = 6638.03 kN (Since 2.611 is +ve, hence using +7372.40)

Therefore,
Total X force considering all legs = -6642.51+6639.63+6638.03 = 6635.15 kN which is close to -6748.77 kN (negative sign comes from turning force into reaction).

Similar calculations can be performed for Y direction. Note that 1/3rd is an approximation. SACS is using the exact seismic force for its calculations.


Eqk force in Y = -18433.865 kN / 3 legs = -6144.62 kN

First leg
Total force = (0.9*-6144.62 kN) + (1.1* -0.758 kN) = -5530.99 kN (Since -0.758 is -ve, hence using -6144.62)
---> Magnitude of static force increases, but sign remains the same i.e. negative in this case

Second leg
Total force = (0.9*6144.62 kN) + (1.1* 15.863 kN) = 5547.61 kN (Since 15.863 is +ve, hence using +6144.62)

Third leg
Total force = (0.9*-6144.62 kN) + (1.1*-15.105 kN) = -5546.77 kN (Since -15.105 is -ve, hence using -6144.62)
---> Magnitude of static force increases, but sign remains the same i.e. negative in this case.

Therefore,
Total Y force considering all legs = -5530.99+5547.61-5546.77 = -5530.15 kN which is close to 5260.627 kN (negative sign comes from turning force into reaction).

If we check the listing file for the static analysis alone, the summary (total) force in X and Y is zero. Once static is combined with modal forces (seismic), equilibrium is no longer applicable because modal forces are statistically combined estimates of peak responses, not actual time-domain forces.These do not obey the Newton's laws in the same way as static loads do.

 
Q2. The seismic-only forces happen to be the same for all 4 cases. Hence, no matter how they are combined with the static forces, the eqk + static forces will be the same.
 
 
#STCMB causes seismic forces from a response spectrum analysis to be added in the same direction (sign) as the static forces, increasing their magnitude. This leads to more conservative load combinations, ensuring structural safety under worst-case assumptions.