The boys loved the hilly chair so much, oh did they love that chair. The boys 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.
What is the total weight of the hilly chair? Wtotal=
A B C D
_ 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
Density of Birch plywood
6.07 KN / m³
The total weight of the hilly chair is the sum of multiplying each volume by its density.
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 Wtotal = Vtotal x density = 0.006042 m³ X 6.07 KN / m³ = 0.036675 KN = 36.7 N
Ftotal = MA
36.7N = M x 9.8 m/s ² = 3.74 Kg
The total weight of the hilly chair is 36.7 N, or 3.74 Kg.
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.
Fuzzy Cub
Area of Fuzzy’s distributed load
What are the chair’s reactions ?
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 finding 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.
VIEW 3
VIEW 4 PLAN VIEW
LEG C
DL= 36.7 N LEG A
LEG B LL= 667 N
VIEW 2
VIEW 6
VIEW 1
VIEW 5
66.7 mm
313.3 mm
LL = 667 N
FRONT ELEVATION VIEW 1
How much load does each leg of the hilly chair carry?
LEG A
LEG B
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 N
Thus each leg in this loading scenario carries 12.23 N of dead load. What is the live load?
The live load carried by Leg A can be calculated through the principle of superposition. Based on VIEW 1 Leg A = 667 N ( 313.3 mm / 380 mm )
= 549.9 N
Based on VIEW 2 Leg A = 667 N ( 316 mm / 380 mm )
= 457.3 N
316 mm
64 mm LL = 549.9 N FRONT ELEVATION VIEW 2
LEG C
457.3 N LEG A
313.3 mm
66.7 mm LL = 667 N
FRONT ELEVATION VIEW 4
LEG B
LEG C
249.3 mm
130.7 mm LL = 549.9 N
FRONT ELEVATION VIEW 3
69.7 N LEG C
LEG A
Based on VIEW 4
The live load carried by Leg A can be calculated through the principle of superposition.
Leg C = 667 N ( 313.3 mm / 380 mm )
= 549.9 N
Based on VIEW 3
Leg C = 549.9 N ( 130.7 mm / 380 mm )
= 69.7 N
130.7 mm
249.3 mm
LL = 667 N
FRONT ELEVATION VIEW 5
LEG A
64 mm
LEG B
316 mm
LL = 229.4 N
FRONT ELEVATION VIEW 6
140 N LEG B
LEG C
Based on VIEW 5
The live load carried by Leg A can be calculated through the principle of superposition.
Leg B = 667 N ( 130.7 mm / 380 mm )
= 229.4 N
Based on VIEW 6
Leg C = 229.4 N ( 316 mm / 380 mm )
= 140 N
To find the overturning force of the hilly chair, we must find its center of gravity.
Lets find y
A1, A2, A3, A4 must be calculated to find y. A1 = 30 mm (380 mm) = 11 400 mm ²
A3 = 38 mm (460 mm) = 17 480 mm ²
A2 , based on the Rhino model = 19 000 mm ²
A4 = 38 mm (460 mm) = 17 480 mm ²
y1, y2, y3, y4 must be calculated to find y. y3 = 2/3 (449 mm) = 299.3 mm ( From the ground ) y4 = 2/3 (449 mm) = 299.3 mm ( From the ground )
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
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
A1 y
A2
Y1
30 mm
Y2
100 mm
A3 A4 FRONT ELEVATION
Y3
LEG A
319 mm
Y4
LEG B
PLAN VIEW
LEG C
x = 1/2 (380mm)
= 190 mm
x
319 mm
LEG A
LEG B
190 mm
LL = 667 N
Foverturn = ? N
y
FRONT ELEVATION
347.92 mm
F
LEG B
468.8 N LEG A
190 mm
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 find 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
Φ
Therefore 333.5 N is the vertical force required to flip the chair. The force can be reduced if one’s feet apply force on an angle. The vertical force required to flip the chair is decreased when is increased.
Φ 333.5 N
x
DL = 36.7 N 2.6 N/ mm FRONT ELEVATION OF SEAT
LEG A
64 mm
LEG B
252 mm
64 mm
LL = 2.6 N/mm x (252 mm)
= 667 N
Area of distributed load
PERSPECTIVE VIEW
Construct a FBD for each member
LEG A
LEG B
DL = 36.7 N and LL = 667 N
FRONT ELEVATION OF SEAT
LEG A
LEG B
64 mm
252 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
εY=0
351.9 N
667N + 36.7N - 351.9N - A = 0 A = 351.9 N
DL = 9.8 m/s ² (0.29 kg)
= 2.8 N
εm = 0
B = B B = 351.9N ( 100mm ) + 2.8N ( 50mm ) B = 35 330 N.mm
DL = 2.8 N 449 mm
Since the weight is equally distributed, the bending moment for legs A,B and C are 35 330 N.mm.
100 mm
LEG B
DL = 36.7 N 6.67 N/ mm FRONT ELEVATION OF SEAT
LEG A
64 mm
LEG B
126 mm
26 mm
100 mm
LL = 6.67 N/ mm (100 mm)
= 667 N
PERSPECTIVE VIEW
Area of distributed load
LEG A
LEG B
64 mm
DL = 36.7 N
LL = 667 N
FRONT ELEVATION OF SEAT
LEG A
64 mm
LEG B
126 mm
26 mm
100 mm
64 mm
Find reaction at leg A , to be used in FBD
εm = 0
A = A
667N ( 202mm ) + 36.7N ( 126mm ) = B ( 252mm ) B = 553 N
εY=0
667N + 36.7N - 553N - A = 0 A = 150.7 N
εm = 0
150.7 N
553 N
A = A
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. DL = 2.8 N
DL = 2.8 N
449 mm
449 mm
100 mm
LEG B
100 mm
LEG A
Φ
Φ
= tan ( 100mm/460mm) =12.3 degrees
LEG A
460
mm
457.3 N
100 mm
449 mm
LEG A
Analyze compression for this leg
Fx = 457.3N cos (12.3 degrees)
= 446.8 N
σ = 446.8 N = area
446.8 N = 220 kpa π ( 25.4mm)²
Therefore 220 kpa is the compression force for leg A.
Find strain for this leg
ε=
σ
E = 0.0015%
=
220 kpa = 0.000015321 11 030 mpa
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 mm due to the axial load.
FRONT ELEVATION
Δ max LEG A
Analyze bending for this leg
Fy = 457.3N sin (12.3 degrees)
= 97.4 N
Deflection will occure at the free end, in this case, the ground touching end of the leg.
fy . L³ =
Δ max =
3 EI
97.4N (0.460m)³ 6 4 3[141030 x 10 pa][π/4(0.025m ]
= 0.93mm
Analyze shear for this leg
Fy will cause shear stress in the beam. Shear, τ , for a circular section is:
τmax= 4 3
V A
= 4 97.4 N 3 π(0.025m)² = 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
THREADED CONNECTION DETAIL
The bending moment M will cause compression zones where the leg is screwed into the seat.
Ď&#x192; max = max stress caused by bending moment Ď&#x192; max = My I = 97.4N (0.460 m)(0.0254m) 3.06 x 10 -7 m4 = 3.72 mpa
3.72 mpa < 40 mpa Therefore the maximum stress is 3.72 mpa. Since the ultimate tensile strength of pine is 40 mpa. The leg is ok.
Area of distributed load
PERSPECTIVE VIEW
LEG C
69.7 N LEG A
457.3 N LEG B
140 N
Weâ&#x20AC;&#x2122;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 onto a small area.
Max Compression
Max Compression
fx fy
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 = 97.4N = 49.6 kpa πd²/4 π(50)²/4
Low-load legs Legs B & C
τ=
fy = 35N = 17.8 kpa πd²/4 π(50)²/4
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 connection plan view
Let’s analyze the bearing stress
fy causes bearing stress on the legs of the chair, bearing stress can be calculated by :
fy td
σB =
t= thickness of pin connection d= diameter of the pin (leg)
From Drawings:
t= 38.1mm
d= 50mm
High-load leg Leg A
fy td
σB = =
97.4N (38.1mm)(50mm)
= 51.1kpa Yield Strength in crushing of birch plywood: 2448 kpa
51.1 kpa < 2448 kpa Since 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 = =
fy td 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 seat must 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 transferred through an area of : Area = [ (38.1mm)(12 threads per inch / 25.4 mm per inch)] . [(2π x 50mm)(3mm)] = 16 905 mm² Perimeter Thread depth = 0.017 m² Let’s analyze the axial stress
High-load leg Leg A : Low-load leg Leg B and C :
σ = 446.8N
= 26.3 kpa
σ = 97.4N
= 5.7 kpa
0.017 m² 0.017 m²
Therefore the legs are ok !
Pin
LEG A
fx = 446.2 N
Let’s analyze the pin connection
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
So the boys built seventy seven hundred and three hilly chairs for the children of the kingdom. Indeed, everyone lived happily ever after.