A Tale Of The Hilly Chair

The boys loved the hilly chair so much, oh did they love that chair. Theboys wanted everyone to have a hilly chair so they decided to figure out how the hilly chair was made so they can built seventy seven hundred and three for the children of the kingdom.

AB

C

D

What is the total weight of the hilly chair?Wtotal= _ KN?

Individual volumes based on the Rhino model

A 1 509 006 mm³

B 1 534 318 mm³

C 1 609 529 mm³

D 463 098 mm³

6.07 KN / m³

Density of Birch plywood

The total weight of the hilly chair is thesum of multiplying each volume by its density.

Wtotal = Vtotal x density

= 0.006042 m³ X 6.07 KN / m³ = 0.036675 KN = 36.7 N

Vtotal = 3X D + A + B + C = 3 ( 463 098 mm³) + 1 509 006 mm³ + 1 534 318 mm³ + 1 609 529 mm³ = 1 389 294 mm³ + 4 652 853 mm³ = 6 042 147 mm³ = 0.006042 m³

The total weight of the hilly chair is 36.7 N, or 3.74 Kg.

Ftotal = MA36.7N = M x 9.8 m/s ² = 3.74 Kg

Vtotal = 3X D + A + B + C = 3 ( 463 098 mm³) + 1 509 006 mm³ + 1 534 318 mm³ + 1 609 529 mm³ = 1 389 294 mm³ + 4 652 853 mm³ = 6 042 147 mm³ = 0.006042 m³

What are the chair’sreactions ?

Reaction of the front right leg when Fuzzy cub stands on the hilly chair.

When Fuzzy cub stands above the right leg of the hilly chair, he sets the chair up for the potential of reaching its worst case loading scenario. The worst case loading scenario for the hilly chair is one chair leg carrying 1/3 of the weight of the chair and all of Fuzzy’s weight.

Therefore Ftotal = W/3 + FFuzzy

F = ( 36.7 N/3 ) + ( 9.8 m/s ² x 68 kg ) = 678.6 N

Thus the reaction force at the right leg in this loading scenario is 678.6 N.

Area of Fuzzy’s distributed load

Fuzzy Cub

Reaction of each leg when no one is sitting on the Chair.

The seat of the hilly chair can be considered as an equilateral shape. This makes �nding the reaction forces at each leg fairly straightforward. Each leg must support 1/3 of the weight of the seat.

Therefore F = W/3

F = 36.7 N/3 = 12.23 N

Thus the reaction force at each leg in this loading scenario is 12.23 N.

Reaction of each leg with load at the centre of the Chair.

If a load is placed at the centre of the chair. The reaction forces are still F = W/3 . For example, a load of 68 kg will be distributed equally to the 3 legs.

Therefore F = W/3

F = [ 36.7 N + (9.8 m/s ² x 68 kg ) ] /3 = 234.4 N

Thus the reaction force at each leg in this loading scenario is 234.4 N.

How much load does each leg of the hilly chaircarry?

VIEW 4VIEW 3

VIEW 6

VIEW 5

VIEW 2

VIEW 1

LEG A LEG B

LEG C

DL= 36.7 N

LL= 667 N

LL = 667 N

66.7 mm 313.3 mm

FRONT ELEVATIONVIEW 1

LEG A LEG B

PLAN VIEW

What is the dead load?

Since the load is located at the centre of the seat, the reactions should be symetrical at legs A, B and C.

Therefore F = W/3

F = 36.7 N/3 = 12.23 NThus each leg in this loading scenario carries 12.23 N of dead load.

What is the live load?

Leg A = 667 N ( 313.3 mm / 380 mm ) = 549.9 N

The live load carried by Leg A can be calculated through the principle of superposition.

Based on VIEW 1

LL = 549.9 N

316 mm 64 mm


Leg A = 667 N ( 316 mm / 380 mm ) = 457.3 N

Based on VIEW 2

457.3 NLEG C LEG A

LL = 667 N

313.3 mm 66.7 mm

LL = 549.9 N

249.3 mm 130.7 mm

LEG B LEG C

LEG C



69.7 NLEG A

Based on VIEW 4

Leg C = 667 N ( 313.3 mm / 380 mm ) = 549.9 N


Leg C = 549.9 N ( 130.7 mm / 380 mm ) = 69.7 N

Based on VIEW 3

LL = 667 N

130.7 mm 249.3 mm


LL = 229.4 N

64 mm 316 mm


LEG A LEG B

LEG B LEG C140 N

Based on VIEW 5

Leg B = 667 N ( 130.7 mm / 380 mm ) = 229.4 N


Leg C = 229.4 N ( 316 mm / 380 mm ) = 140 N

Based on VIEW 6

To find the overturning force of the hilly chair, we must find its centerof gravity.

A1

A2

A3

A4

319 mm

100 mm

30 mm

A1 = 30 mm (380 mm) = 11 400 mm ²

A1, A2, A3, A4 must be calculated to �nd y.

Lets find y

A2 , based on the Rhino model = 19 000 mm ²

A3 = 38 mm (460 mm) = 17 480 mm ²

A4 = 38 mm (460 mm) = 17 480 mm ²

y1, y2, y3, y4 must be calculated to �nd y.

y1 = 1/2 (30 mm) = 15 mm ( From the top of the seat )

y2 = 1/3 ( 100 mm) = 33.3 mm ( From the top of the terrain ) y2 = 449 mm - 30 mm - 33.3 mm = 385.7 mm

y3 = 2/3 (449 mm) = 299.3 mm ( From the ground ) y4 = 2/3 (449 mm) = 299.3 mm ( From the ground )

y can now be found since we now have A1, A2, A3, A4 and y1, y2, y3, y4.

y = y1A1 + y2A2 + y3A3 + y4A4 A1 + A2 + A3 + A4

= (385.7 mm)(19 000 mm ²) + 2(299.3 mm)(17 480 mm ²) + (434 mm)(11 400 mm ²)

19 000 mm ² + 2( 17 480 mm ² ) + 11 400 mm ²

= 347.92 mm

Y3 Y4

Y2

Y1

y

LEG A LEG B

FRONT ELEVATION

LEG A LEG B

LEG C

x

y

319 mm

x = 1/2 (380mm) = 190 mm

190 mm

347.92 mm

LEG A

LEG B

Foverturn = ? N

LL = 667 N

468.8 N

F

190 mm

FRONT ELEVATION

PLAN VIEW

Now we can find the overturning force.

Foverturn = 667N (190 mm) = 282.2 N

Now we know how much force is required to tip the chair over. Lets �nd out how much force one needs to make the chair slide.

what’s the friction force ?

Ffriction = μ ( normal force ) = 0.25 ( 468.8 N) = 117.2 N

μ = 0,.25 For wood sliding on wood

117.2 N is the force required to slide the chair without tipping it over.

If a person is trying to tip over the seat while sitting on it, the force required to tip the chair would be:

εm = 0 A = A

667N ( 190mm ) = 380mm ( Force from feet pushing down ) Force pushing down = 333.5 N

Therefore 333.5 N is the vertical force required to �ip the chair. The force can be reducedif one’s feet apply force on an angle. The vertical force required to �ip the chair is decreased when is increased.

Φ

333.5 N

x

333.5 N

Φ

Construct a FBDfor each member

LEG A LEG B

DL = 36.7 N2.6 N/ mm

LL = 2.6 N/mm x (252 mm) = 667 N

252 mm64 mm 64 mm

Area of distributed load

LEG A

LEG B

FRONT ELEVATIONOF SEAT

PERSPECTIVEVIEW

LEG A LEG B

DL = 36.7 N and LL = 667 N

252 mm64 mm 64 mm

Find reaction at leg A , to be used in FBD εm = 0 A = A

667N ( 126mm ) + 36.7N ( 126mm ) = B ( 252mm ) B = 351.9 N

ε Y = 0 667N + 36.7N - 351.9N - A = 0 A = 351.9 N

εm = 0 B = B

100 mm

449 mm

351.9 N

DL = 2.8 N

LEG B

DL = 9.8 m/s ² (0.29 kg) = 2.8 N

B = 351.9N ( 100mm ) + 2.8N ( 50mm ) B = 35 330 N.mm

Since the weight is equally distributed, the bending moment for legs A,B and C are 35 330 N.mm.


LEG A LEG B

DL = 36.7 N6.67 N/ mm

LL = 6.67 N/ mm (100 mm) = 667 N

100 mm64 mm 64 mm


LEG A

LEG B

126 mm 26 mm


PERSPECTIVEVIEW

LEG A LEG B

DL = 36.7 N

Find reaction at leg A , to be used in FBD

εm = 0 A = A

667N ( 202mm ) + 36.7N ( 126mm ) = B ( 252mm ) B = 553 N

ε Y = 0 667N + 36.7N - 553N - A = 0 A = 150.7 N

εm = 0 =

100 mm

449 mm

553 N

DL = 2.8 N

LEG B

351.9N ( 100mm ) + 2.8N ( 50mm ) = A 15 210 N.mm = A

The load is equally distributed between leg A and leg C because it lies on the mid line between leg A and leg C.

100 mm64 mm 64 mm126 mm 26 mm

LL = 667 N

100 mm

449 mm

150.7 N

DL = 2.8 N

LEG A

A A


351.9N ( 100mm ) + 2.8N ( 50mm ) = A 15 210 N.mm = A

100 mm

449 mm

LEG A

460

mm

LEG A

457.3 N

Φ Φ

= tan ( 100mm/460mm)=12.3 degrees

Analyze compression for this leg

Fx = 457.3N cos (12.3 degrees) = 446.8 N

Therefore 220 kpa is the compression force for leg A.

Find strain for this leg

σ = 446.8 N area

446.8 N π ( 25.4mm)

= ²

= 220 kpa

ε = σ E

220 kpa11 030 mpa

= = 0.000015321

= 0.0015%

E for birch plywood = 11 030 mpa

δ = ε . L = 0.000015321 ( 0.460 m) = 0.007047 mm

Therefore leg A will compress by 0.007047 mmdue to the axial load.

Δ max

LEG A

FRONT ELEVATION

Analyze bending for this leg

Fy = 457.3N sin (12.3 degrees) = 97.4 N

De�ection will occure at the free end, in this case,the ground touching end of the leg.

Δ max = fy . L³ 3 EI

97.4N (0.460m)³3[141030 x 10 pa][π/4(0.025m ]

=

= 0.93mm

46

Analyze shear for this leg

Fy will cause shear stress in the beam. Shear, τ , for a circular section is:

τmax= 4 3 A

V

= 4 3 π(0.025m)²

97.4 N

= 66.2 kpa

Yield Strength of birch plywood: 143.3 kpa

66.2 kpa < 143.3 kpa

τmax is less than the yield strength. Therefore, leg A is ok.

Max Compression

Max Compression

THREADEDCONNECTIONDETAIL

The bending moment M will cause compression zones where the leg is screwed into the seat.

σ max =

= 97.4N (0.460 m)(0.0254m)

max stress caused by bending moment σ max = My

I

3.06 x 10 m4-7

= 3.72 mpa

Therefore the maximum stress is 3.72 mpa. Since the ultimate tensile strength of pine is 40 mpa.The leg is ok.

3.72 mpa < 40 mpa


LEG A

LEG B

LEG C

457.3 N

69.7 N

140 N

PERSPECTIVEVIEW

We’re calculating for the worst case scenario. In this case, the worst case scenario is Prince Charming stepping onto the stool. All of the weight of Prince Charming is distributed ontoa small area.

fy

Max Compression

Max Compression

fx

High-load leg Leg A

From the beam and column analysis

fx = 446.8 N fy = 97.4 N

Low-load legs Legs B & C

From the beam and column analysis

fx = 159 N fy = 35 N

High-load leg Leg A

τ = fy πd²/4

= 97.4N π(50)²/4

= 49.6 kpa

Low-load legs Legs B & C

τ = fy πd²/4

= 35N π(50)²/4

= 17.8 kpa

Yield Strength of birch plywood: 143.3 kpa

49.6 kpa < 143.3 kpa

The stresses on both the high and low load legs are below the yield strength. Therefore, legs A, B and C are ok.

fx

Section of connectionplan view

Let’s analyze the bearing stress

fy causes bearing stress on the legs of the chair, bearing stress can be calculated by :

σB = fytd

t= thickness of pin connection d= diameter of the pin (leg)

From Drawings: t= 38.1mm d= 50mm

High-load leg Leg A

σB = fytd

= 97.4N(38.1mm)(50mm)

= 51.1kpa

Yield Strength in crushing of birch plywood: 2448 kpa

51.1 kpa < 2448 kpaSince the bearng stress of of the high-load leg is under that of the yield strength in crushing of the wood, therefore the bearing stress is acceptable.

High-load leg Leg A

σB = fytd

= 35N(38.1mm)(50mm)

= 18.4 kpa

18.4 kpa < 2448 kpa Therefore the bearing stress is acceptable.

Axial load is transferred to the legs through the threads. The axial load in the seatmust be transferred to the chair. The only path for this is through the surface of the threaded connections. The threaded connections are 50mm in diameter, there are 12 threads per inch. Each thread has a depth of 3mm. Therefore the axial load is transferredthrough an area of :

=Area [ (38.1mm)(12 threads per inch / 25.4 mm per inch)] . [(2π x 50mm)(3mm)]= 16 905 mm²= 0.017 m² Perimeter Thread depth

Let’s analyze the axial stress

High-load legLeg A :

σ = 446.8N0.017 m²

= 26.3 kpa

Low-load legLeg B and C :

σ = 97.4N0.017 m²

= 5.7 kpa

Therefore the legs are ok !

fx = 446.2 N

LEG A

Pin

The diameter of the connecting pin is 1/2” and the force in each leg is transfered through the pin(not directly between the leg and the chair). The area of the pin is :

πr²= π (6.35mm)² = 127 mm²

The shear stress in the pin of the lightest compression member is the compression in that leg divided over twice the pin area since the pin goes through both sides of the chair.

τ = 446.2 N 2 (127mm²)

= 1.76 mpa

Let’s analyze the pin connection

So the boys built seventy seven hundred and three hilly chairsfor the children of the kingdom. Indeed, everyone lived happily ever after.

A Tale Of The Hilly Chair

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