WHAT ARE THE TYPES OF PROBLEMS IN DESIGN OF COMPRESSION MEMBER AND STEP BY STEP PROCEDURE TO SOLVE THE PROBLEMS.
There are basically two types of problems we have two types
of problems for compression member
And they are
Type -1 – find the strength of given compression member (axial compression member)
Type -1 – find the strength of given compression member (axial compression member)
Type -2 – design of compression member or design of axial
compression member
Now let’s take the steps involved in both the types of problems so all the students can have the easy understanding of what we are actually going to do to solve this type of problems.
Note* - basically below steps mentioned are the best practice to solve any kind of problems related to compression member.
And yes it’s a very simple steps and easy to remember if you know all the terminology related to compression members.
Now let’s take the steps involved in both the types of problems so all the students can have the easy understanding of what we are actually going to do to solve this type of problems.
Note* - basically below steps mentioned are the best practice to solve any kind of problems related to compression member.
And yes it’s a very simple steps and easy to remember if you know all the terminology related to compression members.
Type – 1 - find the
strength of given compression member (axial compression member)
Data given –
a. Sections and its arrangements
b. Moment of inertia
c. Length and its end conditions
Solution –
1. find the effective area of the given section (Ae)
The effective area of the section can be found in steel
table.
2. find moment of inertia and radius of gyration (M.I. and r)
The value of moment of inertia (Ixx and Iyy) is given in steel table
2. find moment of inertia and radius of gyration (M.I. and r)
The value of moment of inertia (Ixx and Iyy) is given in steel table
And the radius of gyration with respect to X-axis and Y-axis can be found with the help of formula
3. find the effective length (Le) of the member from the actual length and end condition table.
4. fin the maximum slenderness ratio with the help of below
formula (λmax).
\[{λ_{max} = {L_{eff.} \over r_{min.}}}\]
\[where,\]
\[r_{min.} = ~minimum~ value~ of~ rxx~ and~ ryy\]
5. find the strength of axial compression member (Pac) with
the help of below formula
Type -2 – design of axial compression member
Data given / assume data –
a. Axial compressive load
b. Section and their arrangements
c. Length and its end conditions
d. Yield stress of steel (Fy)
e. Permissible stress in axial compression (σac)
Solution / steps –
1. assume the value of (σac) / approximate value of (σac)
Guide lines for assuming the value of (σac) Permissible
stress in axial compression.
Below is the table in which some range is given for assuming
the value of (σac) Permissible stress in axial compression.
Sr. No.
|
Sections details
|
Range of σac
|
1
|
Single angle sections
|
0.15 Fy to 0.25 Fy
|
37.5 Mpa to 62.5 Mpa
|
||
2
|
For other rolled steel sections
|
0.25 Fy to 0.35 Fy
|
62.5 Mpa to 87.5 Mpa
|
||
3
|
Built-up sections / compound sections
|
0.35 Fy to 0.45 Fy
|
87.5 Mpa to 112.5 Mpa
|
||
4
|
Built-up sections having equal strength
|
0.45 Fy to 0.55 Fy
|
112.5 Mpa to 137.5 Mpa
|
||
Note* since we are assuming the value of (σac) from the
above given table. So at first we have to do all calculation on the basis if
assumptions that’s why at the time of writing we write (σac) i.e. (σac.
Assumed/approx.) And accordingly we do the calculation unless we get the value
of strength (Pac) calculated more the than value of (Pac) required.
2 find the approximate cross section area required (Ae)
3 choose the suitable section from the steel table on the
basis of approximate area required (Ae approx.)
\[A_{e~ approx.} = {P_{ac} \over σ_{ac~approx.}}\]
4 calculate the value of maximum slenderness ratio (λmax) with the help of below formula
\[A_{e~ approx.} = {P_{ac} \over σ_{ac~approx.}}\]
4 calculate the value of maximum slenderness ratio (λmax) with the help of below formula
\[{λ_{max} = {L_{eff.} \over r_{min.}}}\]
5 now find out the value of Permissible stress in axial compression (σac) for the above value of maximum slenderness ratio (λmax)
And the value of Yield stress of steel (Fy) for the give steel or assumed steel
5 now find out the value of Permissible stress in axial compression (σac) for the above value of maximum slenderness ratio (λmax)
And the value of Yield stress of steel (Fy) for the give steel or assumed steel
By doing the 5th steps we basically fining the value of (σac)
which is based on (λmax) so we name it as (σac calculated)
6. find the value of strength (Pac calculated)
\[{P_{ac~ calculated } = ~A_{e~calculated}~X~σ_{ac}}~\]
6. find the value of strength (Pac calculated)
\[{P_{ac~ calculated } = ~A_{e~calculated}~X~σ_{ac}}~\]
If the value of Pac calculated is more than Pac required the
our design is okay otherwise we have to revised our section.
7. after the checking the condition of strength we have to check the section for slenderness ratio.
7. after the checking the condition of strength we have to check the section for slenderness ratio.
I.e. the value of slenderness ratio should be more than the required
slenderness ratio.
8. and at the end we also have to check the section for end conditions on the basis of effective length we have to compare the end conditions.
8. and at the end we also have to check the section for end conditions on the basis of effective length we have to compare the end conditions.

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