In july 1990 mechanical engg news this is the statement : The total force on the element producing stress throughout the element is kx or mx’’. The force at the end of the element should be kx not kx + mx’’.

In november 1994 mechanical engg news, this is the statement: The DLF response spectrum for a given load is generated by solving the dynamic equation of motion for a single degree of freedom system:
Ma(t) + Cv(t) + Kx(t) = F(t)

The force acting on a node in a DYNAMIC problem is whether kx or mx’’ or kx+mx’’ ???

can any one help me, on this topic.

durga,

I think in the concept described in the 90’s article is not so easy to be explained.
I’ll try to reformulate the problem and in order to simplify let’s first say no damping is present.

Let’s consider a spring with a mass "m" attached and a force F produces an elongation "x". Always the "tension" in spring is T=kx where x is the elongation. If you "isolate" the spring this is the mechanical equation.
In the same time, x may be considered as the solution of differential equation which is the motion equation. If you isolate the mass "m" the motion equation is mx"=F-T, but T=kx, so
mx"+kx=F
Just note that the solution of that equation, i.e. "x"- the elongation -depends on the manner in which the force F is applied, so the "tension" in sopring deppends on the "dynamics" of the problem.

To detail:
- In "static" case, the elongation is xs=F/k- this is a very familiar formula for a mechanical engineer. But even here you may consider "static" as "cvasistatic" i.e. a case when you gradually apply the force from 0 to F in an "infinite" time. So here xs may be considered also as the result of limit x(t)-for infinite time, where x(t) is the solution of mx"+kx=F.

- In "dynamic" mode, F is applied "suddenly" at t=0; the motion equation remains mx"+kx=F and for initial conditions x(0)=0, x’(0)=0 this equation has the solution x(t)=xs*[1- cos(sqrt(k/m)*t)].

As you can see, x(t) depends on how we apply the force F, the maximum value of "dynamic" elongation x(t)=xs*[1- cos(sqrt(k/m)*t)] is 2*xs (because there is a property of cosine to have values between -1 and 1), and we say the DLF is 2. But in spring we can say always Tension=kx .
Based on this result I would say:
" the force producing stress in spring is kx, not mx"+kx ".
even in all cases the equation is
mx"(t) + kx(t) = F(t)

mariog,

In dynamic analysis(Seismic analysis)

I studied in dynamic analysis procedure, this force(MASS*ACCELERATION)acting on the pipes and finding out responses.
I think, for that only ZPA concept came after 33 hz (zpa =acceleration)

Then stiffness k= force/deflection

force = k*x

this force is acting in static analysis

In dynamic F= ma and In static F= kx forces are acting like that, if it is like that then in dynamic analysis where the stiffness of the pipe is considered???

F= MA And F=KX Both have same magnitude ???

can you please eloberate and give me some reference also...

durga,

There is a particular explanation in the article you are referring to.
The author says "the mx" and the kx terms represent equal and opposite forces that act on the node at the end of the element".
In this case why is confusing the next statement which is "the total force on the element producing stress throughout the element is kx or mx" ? Are there alternatives to this explanation, i.e. would you claim that force producing stress in element is zero or (kx and mx")=kx+mx" or |kx|+|mx"| or may you offer other explanation?

What I tried to explain previously with mass& spring model is that the force producing stress in spring is kx in both static and dynamic scenarios; the difference between them stays in "x" value and there is a differential equation that gives that value.
The spring may be "isolated" and instead the connection with the mass m we can introduce a "spring tension" T, which correspond to an elastic force kx. So T=kx in all cases, not only in static case and we may say the force producing stress in spring is the spring tension (or compression) T=kx.

I don't know this explanation is satisfactory or not for you but I'm quite unable to further detail it...

I understand what you explained. It’s very clear.
Thank you for your kind explanation

For spring: Force =K*xs in static analysis
For spring: Force=k*x(t) in dynamic analysis

In static analysis F= K*xs (clear from your explanation)

But,

If at the end of the node only k*x(t) acting then what is the use of m*a and vice versa.

Why we are writing in dynamic analysis total force = force acting in spring + due to mass.

If only one force is acting then why we are adding two forces due to spring and due to mass (i.e potential energy= k*x and kinetic energy= m*a)

F = k*x(t) + ma

May be silly doubt, but iam new to this field. I want to know in depth thats why iam asking help from experts like you..

Replace the connection between the mass m and spring k with a pair of internal forces T- this is a very common concept in mechanics.

Now just try to express an equation describing what is happening with the spring. An ideal spring hasn't associated mass, so that equation couldn't include an inertial term; instead it is a simple equation as
T=kx
or more precisely- showing that both T and x depends on t-
T(t)=k*x(t)
In fact this is Hooke's law of elasticity that states that the extension of a spring is in direct proportion with the Load applied to it.

By the other hand, we may say T produces the stress in spring- simply there is not any other force to give stress in spring.

But we don't know yet x(t)- the displacement which is the cause giving T(t) as effect and stress- as effect, again.

Try now to write an equation describing the mass m movement.

Here the equation is the second law of dynamics
F(t)-T(t)=m*a(t)
or
F(t)=T(t)+m*a(t)
but we know that
T(t)=k*x(t), so
F(t)=k*x(t)+m*a(t)
however a(t)=x"(t), so the above equation is a differential equation that would be integrated giving x(t).
In addition, x(t)- as solution of differential equation- depends on F(t) which appears in the differential equation, it follows that T(t)=k*x(t) depends implicitly on F(t); however as expression it remains T(t)=k*x(t).

Note that the above written F(t) is an external force applied to mass m.

Just to finish this long discussion, I think you are confused by the fact the 90s article tries to highlight a simple case when F(t)=0.

In this particular case
k*x(t)+m*x"(t)=0
so "the mx" and the kx terms represent equal and opposite forces that act on the node at the end of the element" as the article says.

Now it makes sense for you the statement valid for that case:
"The total force on the element producing stress throughout the element is kx or mx" "?

As a general statement you may say: The force producing stress throughout the element is Kx(t), where x is the solution of Ma(t) + Cv(t) + Kx(t) = F(t)

So,

If there is no external force applied then ma+kx=0 => ma=-kx, If both applied, it becomes ZERO. thats why only kx or ma is applied.

if there is external force applied then total force F= ma+cv+kx then total F acts on the node (if damping also considered)

My understanding is correct, mariog ?

I have the temptation to say yes; however...

If there is no external force applied to the mass, then ma=-kx.
This is a form of second law of dynamics, where
mass*acceleration= elastic force provided by spring.
If you are interested to evaluate stress spring i.e. the effects of kx in spring, this is mathematically equivalent to work with -ma.
This respond to the question "why does Caesar II output not reflect the mx"=ma term?"

If there is a force applied to the mass, then ma=F-cv-kx.
This is a form of second law of dynamics, where
mass*acceleration= force applied+damping force+elastic force provided by spring.
If you are interested to evaluate stress spring i.e. the effects of kx, this is equivalent to work with F-cv-ma.
Extrapolating to more complex systems than mass m+spring k, this respond to the question "why does Caesar II output not reflect the mx"=ma term?"

thanks Mariq,
its clear now.