explanation of the load case combination

I have searched through the CAESAR forum, although the topic is discussed , my question is never addressed.

I would like to have clear explanation of the load case combination(algebraic or scalar):

I know that CAESAR recommended load case setup is as follows(except case 3):

case 1 W1+T1+P1(OPE)
case 2 W1+P1(SUS)
case 3 T1(EXP)
case 4 L1-L2(EXP)

It is suggested that Case 4 is the right way to calculate displacement stress. The load from case4 is an algebraic combination of load and displacement from case 1 and case 2.

Q1. As far as non-linear as concerned, I do not think it correct to use two non-linear load results "do algebra" to calculate the load that is the difference of two. On the contrary, I think load case 4 is correct.

Q2. What is the difference between "algebraic or scalar"?

I did not see any difference in restrains reports while outputing case 4 with either setup, but a big difference in stress value.

Q3. To obtain the stress in case 4, CAESAR first solves case 1 and 2 for displacement and restrains, then combine them to calculate the stress in case 4. Is it defined in B31.3 that requires superposition of two non-linear loads?

A1: I do not understand the question.

A2: Press F1 while focused on the Combination Method cell on the Load Case Options tab in the Static Analysis window. You will find the description of algebraic and scalar. Here's an excerpt (parenthetic comments are mine):

"The Displacements and Forces of an Algebraic case and a Scalar case are equivalent. There may be variation at the stress level, since in an Algebraic combination the stresses are calclulated (from the signed loads) and in a Scalar combination they (the unsigned stresses) are combined."

So, structural results (loads and displacements) are the same whether Algebraic or Scalar. Only stresses are affected.

A3: B31.3 paragraph 319.2.3(b) bases the Displacement Stress Range on "the algebraic difference between strains in the extreme displacement condition and the original (as-installed) condition (or any anticipated condition with a greater differential effect)". That's what we do in CAESAR II.

The 2006 Edition of B31.3 will have an updated Appendix S that includes a nonlinear (+Y) support in model to illustrate how this works.

"The Displacements and Forces of an Algebraic case and a Scalar case are equivalent. There may be variation at the stress level, since in an Algebraic combination the stresses are calclulated (from the signed loads) and in a Scalar combination they (the unsigned stresses) are combined."

i think it is reverse Algebraic combination the stresses are (the unsigned stresses)& Scalar combination (from the signed loads).

Remember that we are talking about vectors (X,Y,Z), not just + and -. You can either sum the vectors or sum the vector magnitudes. In CAESAR II summing the magnitudes is Scalar and combining the vectors is Algebraic (I would have prefered the latter to be called Vector, as in Vector or Scalar summation.)

There are other combination methods listed there too.

Dave , I confused.

In my mind,Summation of the magnitude of each freedom x,y,Z,Mx,My and Mz with sign included, which is regarded as "scalar", is the same as Vector Summation.

From http://dictionary.laborlawtalk.com/Scalar :

<font color="#0000ff">
Scalar: (?), n. (Math.) In the quaternion analysis, a quantity that has magnitude, but not direction; -- distinguished from a vector, which has both magnitude and direction.


In physics a scalar is a quantity that can be described by a single number (either dimensionless, or in terms of some physical quantity). Scalar quantities have magnitude, but not a direction and should thus be distinguished from vectors. More formally, a scalar is a quantity that is invariant under coordinate rotations (or Lorentz transformations, for relativity). A scalar is formally a tensor of rank zero.
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I present a test to clarify this issue:

Assume in Global Coordinate there are:

A = 3 (in X direction)
B = -4 (in Y direction)

Algebraic(A+B)= 5 (or -5)
Scalar(A+B)= -1
SRSS(A+B)= 5
ABS(A+B)= 7

Are all the above right?

Take a look at scalar-vector illustration

Dave Diehl Wrote: "The Displacements and Forces of an Algebraic case and a Scalar case are equivalent. There may be variation at the stress level, since in an Algebraic combination the stresses are calclulated (from the signed loads) and in a Scalar combination they (the unsigned stresses) are combined."

This should answer your questions in their entirety. In your example A+B = -1 for both Scalar and Algebraic. But when we talk of Stress these are calculated from the Algebraic case from the resulting forces and moments and in Scalar the stresses are not re-computed, but rather added (or subtracted) from the stress results of the other two cases.

As to why you get a big difference between stresses from load cases T1(EXP) and L1-L2(EXP) it is simple. In T1(EXP) you have not taken into account the possibility of piping moving under the influence of the other loads in the system. For example if your pipe moves to one side of a gap due to weight and then when the system is in operation it moves to the other side of the gap, you now have a movement of twice the gap size. This will cause a much different stress result than T1(EXP) which can at most move only half this distance.

All:

Per you explanantion , I am aware now. Thanks.