Calculation of Stiffness in Structural elements
Calculation of Stiffness in Structural elements
Question 1:
Compute lateral stiffness of the one-story frame with an intermediate realistic stiffness of the beam. The system has 3 DOFs as shown. Assume L = 2h and Elb = Elc
Answer:
This structure can be analyzed by any of the standard methods, including moment distribution. Here we use the definition of stiffness influence coefficients to solve the problem.
Thus, the structure's 3 × 3 stiffness matrix is known and the equilibrium equations can be written.
For a frame with Ib = Ic subjected to lateral force fS, they are,
The degree of freedom is 3
For the Case 1:
The Elements Ki1 in the first column, we determine the stiffness matrix by imposing
U1 = 1, with U2 and U3 = 0
For the Case 2:
The Elements Ki2 in the second column, we determine the stiffness matrix by imposing
U2 = 1, with U1 and U3 = 0
For the Case 3:
The Elements Ki3 in the third column, we determine the stiffness matrix by imposing
U3 = 1, with U1 and U2 = 0
Since we have three equations, we will be forming this equation as a matrix
Here, we apply the static condensation method by deducting the 3 unknown expressions into 2 unknow equation
From the Case 2 and 3 equations, the joint rotation can be expressed in terms of lateral displacement as follows
Stiffness is the amount of force required for unit displacement.
So interchanging the translation U1 to the left side, we arrive at the stiffness coefficient as follows
Question 2:
For the following structures:
- Determine the number of degrees of freedom for dynamic analysis
- Establish the equation of motion
- Calculate their natural frequencies
Answer:
2(a):
The position of the element can be described by 1 coordinate only i.e., u(t)
The System has a single degree of freedom structure
The column has a fixed end at the bottom, so the stiffness coefficient will be as follow
2(b):
The position of the element can be described by 1 coordinate only i.e., u(t) (because a beam with infinite stiffness will induce uniform translation in all columns)
The System has a single degree of freedom structure
The two columns with both ends fixed and one column have pinned fixed ends. So the stiffness coefficient will be as follow
Question 3:
Consider the propped cantilever shown in the figure below. The beams are made from the same steel section and have lengths as shown on the diagram. Determine the natural period of this system if a large mass, M, is placed at the intersection of the beams at point A. The weight of the beams in comparison with the mass M is very small.
Answer:
Both Elements have the same material and section.
It means they have the same modulus of elasticity and second moment of inertia, but the lengths are different
Both elements will respond simultaneously when a mass 'M' is applied at the point.
It means they are connected in parallel
Question 4
Determine an expression for the effective stiffness of the following systems:
Answer:
4(a):
The Stiffness coefficient of the beam element is 3EI/L^3
The Stiffness coefficient of the spring is K
The Stiffness elements are in series
Hence,
4(b):
The Stiffness coefficient of both column elements = 3EI/h^3
The Stiffness coefficient of the spring is K, but it is inclined
So the horizontal force (P) and deformation (u) has to be resolved into inclined components to arrive actual coefficient of the spring.
Therefore, the effective stiffness of the system,
Ke =
4(c):
The Stiffness coefficient of the pinned beam is K1 = 48EI/L^3
The effective stiffness coefficient for the system has to be formed by adding the relative axial stiffness provided by the inclined strut connected at the right end
Force acting at the right end is 'mg/2' and deformation caused by the strut at the point of 'm' is half of its right end nodal deformation
Hence,
We assumed vertical axial stiffness in our calculation. But we have an inclined axial Stiffness in the given system.
So we have to resolve the (Kea, vert) into (Kea)
Now we have to submit the values of K1 and K2 in the above equation.
So,
