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# Stiffness Distribution. Metallic Posts

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To see Figures in more detail, please see pdf below.

Stiffness Distribution. Metallic Posts Figure 0.1

Stiffness Distribution. Metallic Posts Figure 0.2

On some occasions, when designing a metal pole for a railing, the contractor had sections with dimensions smaller than those required by the specifications. In order to use the existing sections, it was decided to add a section with dimensions that will allow it to move inside the existing one, but as tight as possible. This was intended to increase the capacity of the assembly in order to satisfy the design conditions.

The load under question was a horizontal point load in the upper end of the posts.

Neither elements would be connected to each other, there would only be partial contact through points welded to the inner element so that there would be no clearance and the deformations would be transmitted directly.

The problem consisted of determining what part of the horizontal load was transmitted to each element and determining if they were capable of supporting said actual loads.

Since the horizontal deformation in each section had to be the same, we proceeded to define the part of the load that each section takes by means of the stiffness distribution of the component elements.

The work attached “Stiffness Distribution. Metallic Posts” aims to show with an example the stiffness distribution of the horizontal load in a metal post of a railing and to verify that the design complies with established regulations. Additionally, we proceed to deduce the formulas that result from the stiffness distribution for this case.

## Stiffness Distribution. Metallic Posts

On some occasions, when designing a metal pole for a railing, the contractor had sections with dimensions smaller than those required by the specifications. In order to use the existing sections, it was decided to add a section with dimensions that will allow it to move inside the existing one, but as tight as possible. This was intended to increase the capacity of the assembly in order to satisfy the design conditions.

Fig. 1 shows the dimensions of an example that will be expanded on to explain the analysis process and the design of the component elements.

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### Fig. 1 Metallic Post. General Dimensions

To see Fig. 1 in more detail, please see pdf below.

Stiffness Distribution. Metallic Posts Figure 1.1

Stiffness Distribution. Metallic Posts Figure 1.2

The section of the exterior post has an interior spacing of 2 in. minus 2 times the width of 1/8 in., meaning, 2 – 2 × 1/8 = 1 3⁄4 in. Then the interior section can be 1 1⁄2 in. × 1 1⁄2 in. × 1/8 in., leaving a space between the walls equal to (1 3⁄4 in. – 1 1⁄2 in.)/2 = 1/8 in. which allows for it to easily slide inside the other.

Solving the geometric problem, we will proceed to prove that both sections working together can support the maximum loads and it will also be necessary to find out how far it is necessary to insert the inner tube such that the outer tube is capable of supporting the loads by itself.

The first problem to solve is that both sections work separately, meaning, they are not integrated into a single section, but rather a distribution of the loads due to the relative stiffness of each section.

At the end, an appendix will be added to demonstrate the formulas that will be used for the stiffness distribution.

Exterior section A     2 ×2 × 1/8                                              b1 ≔ 2 ∙ in                    t1 ∶= 1/8 ∙ in
.                                                                                                   I1 ∶= 0.486 ∙ in^4         S1 ∶= 0.486 ∙ in³
Interior section B     1 1⁄2 × 1 1⁄2 × 1/8                                     b2 ≔ 1.5 ∙ in                 t2 ∶= 1/8 ∙ in
.                                                                                                   I2 ∶= 0.188 ∙ in^4         S2 ∶= 0.251 ∙ in³

The stiffness coefficient for each section is:

K1 ∶= (I1/I1+I2)               K1 ∶= 0.721                              K2 ∶= (I2/I1+I2)               K2 ∶= 0.279

meaning, section A will receive 72.1% of the total moment and section B will receive 27.9% of the total moment.

Where     L ≔ 3.5 ∙ ft and the load     P ≔ 345 ∙ lb

we have that the total moment in the critical section of the base is

.                         Mtotal ≔ P ∙ L                Mtotal = 1207.5 ft ∙ lb

Section A will have a moment of

.                         M1 ≔ Mtotal ∙ K1 = 870.69 ft ∙ lb

with maximum tensions of

.                         fb1 ≔ (M1/S1) = 21498.5 lb/in² < Fb = 0.6 × 36000 = 21600 lb/in²

Section B will have a moment of

.                         M2 ≔ Mtotal ∙ K2 = 336.81 ft ∙ lb

with maximum tensions of

.                         fb2 ≔ (M2/S2) = 16102.5 lb/in² < Fb = 0.6 × 36000 = 21600 lb/in²

which shows that both sections working together are capable of withstanding the maximum loads.

Now, we need to obtain the distance that section B must penetrate the interior of section A. Fig. 2

The maximum moment that section A resists is:

.            Fy ≔ 36000 ∙ lb/in²          Fb ≔ 0.6 ∙ Fy = 21600 lb/in²
.                                                       Ma ≔ S1 ∙ Fb = 874.8 ft ∙ lb

### Fig. 2 Moment Diagram

To see Fig. 2 in more detail, please see pdf below.

Stiffness Distribution. Metallic Posts Figure 2

The distance from the top to the point where section A is capable of resisting by itself is:

.                                              x ≔ Ma/Mtotal ∙ L = 2.536 ft

then, from the base it will be:

.                                            l ≔ L − x = 11.572 in = 12 in

meaning, it is only necessary to place section B 12 in. away from the base

Fig 3. shows the final design for the reinforced metallic post.

### Fig.3 Final Design

To see Fig. 3 in more detail, please see pdf below.

Stiffness Distribution. Metallic Posts Figure 3.1

Stiffness Distribution. Metallic Posts Figure 3.2

## Stiffness Distribution. Metallic Posts. Formula Deduction

To see Figures in more detail, please see pdf below.

Stiffness Distribution. Metallic Posts Figure 4.1

Stiffness Distribution. Metallic Posts Figure 4.2

.            Element 1 E, 11

.            Element 2 E, 12

The deformation at the top must be the same in both elements, then

.            ∆1 ≔ ∆2

.            (P1 ∙ L³)/(E ∙ I1) ≔ (P2 ∙ L³)/(E ∙ I2)

.            (P1 /I1) ≔ (P2/I2)

The horizontal load P is distributed to both, it is fulfilled that

.            P1 + P2 ≔ P     where     P2 ≔ P – P1 P1 ≔ P – P2

Substituting we have:

.            (P1/I1) ≔ (P − P1)/(I2)

.            (P2/I2) ≔ (P − P2)/(I1)

Then, each element will support a load equal to:

.            P1 ≔ P ∙ (I1/(I1 + I2))               P2 ≔ P ∙ (I2/(I1 + I2))

.            K1 ≔ I1/(I1 + I2)                       K2 ≔ I2/(I1 + I2)

Being that K1 and K2 are the stiffness of each element, we finally have

.            P1 ≔ P ∙ K1                              P2 ≔ P ∙ K2

### Stiffness Distribution. Metallic Posts

Section 1     2 ×2 × 1/8                                              b1 ≔ 2 ∙ in                    t1 ∶= 1/8 ∙ in

.                                                                                     I1 ∶= 0.486 ∙ in^4         S1 ∶= 0.486 ∙ in³

Section 2     1 1⁄2 × 1 1⁄2 × 1/8                                     b2 ≔ 1.5 ∙ in                t2 ∶= 1/8 ∙ in

.                                                                                     I2 ∶= 0.188 ∙ in^4         S2 ∶= 0.251 ∙ in³

L ≔ 3.5 ∙ ft               P ≔ 345 ∙ lb               Fy ≔ 36000 ∙ lb/in²               Fb ≔ 0.6 ∙ Fy = 21600 lb/in²

K1 ∶= (I1/I1+I2)               K1 ∶= 0.721

K2 ∶= (I2/I1+I2)               K2 ∶= 0.279

Section 1                                 P1 ≔ P ∙ K1                             P1 = 248.769 lb

.                                                 M1 := P1 · L                             M1 = 870.69 ftlb

.                                                    fb ≔ (M1/S1)              fb ≔ 21498.5 lb/in²    < Fb     OK

Section 2                                 P2 ≔ P ∙ K2                             P2 = 96.231 lb

.                                                 M2 := P2 · L                             M2  = 336.81 ftlb

.                                                    fb ≔ (M2/S2)              fb ≔ 16102.5 lb/in²    < Fb     OK

Section 1                Maximum Moment

.            Mm ∶= S1 ∙ Fb             Mm = 874.8 ∙ lb ∙ ft             Maximum Capacity of Section 1

.            Mtotal ∶= P ∙ L             Mtotal = 1207.5 ft ∙ lb

.            x ∶= (Mm ∙ L)/Mtotal = 2.536 ft             3.5 ∙ ft − x = 11.572 ∙ in

.                       distance where section 2 is needed

Translation By: Ivet Llerena (Eastern Engineering Group).

Structural Drawings By: Sebastian Paz (Eastern Engineering Group).

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