Machine Design
Module Leader: Dr Mostafa El shazly
Scissor table
Group Members:
Name
ID
Contribution
Khaled Ahmed Azzazy
Sakr
173734
Calculation of the lower
position & Safety factor
Ahmed Abdelaziem
176253
Solidworks & Videos
171841
Calculation of Higher
Position & section views
167530
Calculation of higher
position& Drawings
Omar abdelaziz
Aly Alaa Elassal
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-
Content
1- Main Dimensions & Weight: ……………………………………………………………3
2- Selection of Wheels Material & Bearing: ……………………………………………….4
3- Description: ……………………………………………………………………………...5
4- Force Analysis for Highest Position: …………………..………………………………..7
5- Stress Analysis At Highest Position: ………………………………………………...…11
6- Force Analysis for Lowest Position: ………………………………….………………..13
7- Stress Analysis At Lowest Position: ………………...…………………………………16
8- References: ……………………………………………………………………………..18
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1- Main Dimensions & Weights
3
. Top Table
. Length = 0.7m
. Width = 0.6m
. Weight on the Table = 200 kg
. Leg
. Length = 0.65m
. Thickness = 0.022m
. α Is the angle between the leg and the horizontal line
. Weight of the legs “WL” = 4 kg
. Lifter
. Weight of the lifter “Ws” = 1.5 kg
3- Selection of Wheels material & Bearings
. Diameter of the Wheels = 80mm
. Material of the wheels is Polyurethane, since it provides nice floor protection and it is cost
effective.
. Type of the wheel is top plate with capacity more than more than 150 kg (for each wheel).
. The bearing used for the wheel is Tapered Roller Bearing.
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2- Description
This type of scissor table depends on the moment generated from piston, where the
piston stays horizontal and has a slightly movement above and below the x-axis. The lifter
is the part which convert the force of the piston and the reaction of the to a force and
moment to leg such as explained in the below picture.
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By hiding some of the component the force will be more clear
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3- Force Analysis for Highest Position
At the highest Position α= 64 (To accomplish the required Height)
& the moment on the leg due to lifter = -0.147P -0.04Ws + 0.123Rx + 0.09Ry
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Σ fx = 0
= Rx – P – Fx =0
Σ fy = 0
= Ry + Dy – Ws – Fy – W/2 – Wl/2 = 0
ΣME = 0
= –Dy.L/4 cosα – Wl/2.L/4 – Fy. L/4 cosα – W/2.3L/4cosα + Fx.L/4 sinα –
0.147P – 0.04Ws +0.123Rx + 0.081Ry = 0
ΣMo = 0
= –DY. L/2 cosα – Ry.L/4 cosα – P. L/4 sinα + Rx.L/4 sinα + Ws. L/4 cosα –
W/2. L/2 cosα – 0.147P –0.04Ws + 0.123Rx + 0.081Ry = 0
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ΣMB = 0
= –Dy. Lcosα – Ry.3L/4 cosα + Ws 3L/4 cosα – P 3L/4 sinα +Rx 3L/4 sinα +
Fy.L/2 cosα + Wl/2. L/2 cosα – Fx.L/2 sinα –0.147P – 0.04Ws + 0.123Rx + 0.081
Ry = 0
ΣMD = 0
= Ry.L/4 cosα – Ws.L/4 cosα + P.L/4 sinα – Rx.L/4 sinα – Fy L/2 cosα –
Wl/2.L2 cosα + Fx.L/2 sinα – W/2.Lcosα– 0.147P – 0.04Ws + 0.123Rx+ 0.081Ry
=0
By substitution, we can find Fx, Fy, Rx, Ry, Dy, & P Results
Fx = 589.87 N
Fy = 226.524 N
Rx = 1703.95 N
Ry = 1210.41 N
Dy = 30.25 N
P = 1114.06 N
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Diagrams
Where,
FDx` = 27.19 N
FDy` = 13.26 N
FEx` = 5.97 N
FDy` = 1333.37 N
FOx` = 479.805 N FOy` = 421.71 N
FBy` = 429.58 N
FBx` = 880.77 N
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4- Stress Analysis at Highest Positon
The used Material for Legs is AISI 1020
Thickness = 22mm
Yield Strength = 351.571 N/mm^2
σall =
- At point E (stress concentration section)
Existing loads on section
FN= 1360.56 N
M= 141.122 N.mm
Fs= 429.58 N
Area properties:
A=37.62*7=263.34 mm^2
I= (1/12)*b*h^3 = [(7*52.62^3)-(7*15^3)]*2/12=166043.0855
Stress concentration:
Normal: P/7*(52.62-15)
Bending: 6M/(52.62-15)*7^2
Shear: P/7*(52.62-15)
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Stresses
Normal.. δx = [1360.56/7*(37.62)] + [6*141.122/(37.62)*7^2]= 5.63
N/mm^2
Shear= 429.58/7*(37.62)=1.63 N/mm^2
δ1 = 0.5*(δx + δy) + [(0.5*(δx+δy))^2+(ᴌ x,y)^2]^0.5
= 0.5*(5.63+ (0)) + [(0.5*(5.63+0))^2+(1.63)^2]^0.5=6.1 N/mm^2
δ2 = 0.5*(δx + δy) - [(0.5*(δx+δy))^2+(ᴌ x,y)^2]^0.5
= 0.5*(5.63+ (0)) - [(0.5*(5.63+0)) ^2 + (1.63)^2]^0.5= -0.44
N/mm^2
Thus, δ3=0
USING MSST THEORY
δall.=Sy/n= 351.571/3= 117.19<6.1 N/mm^2
ᴌall.= 0.58*δall.= 0.58*117.19=67.97 < 1.63 N/mm^2
Thus, Structure is SAFE.
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5- Force Analysis for Lower Position
At the lowest position, α = 9
& the moment on the leg due to lifter = – 0.046P – 0.0725Ws + 0.145Ry Σfx
=0
= Rx – P – Fx =0
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Σ fy = 0
= Ry + Dy – Ws – Fy – W/2 – Wl/2 = 0
ΣME = 0
= –Dy.L/4 cosα – Fy.L/4cosα – Wl/2.L/4cosα + Fx.L/4sinα – W/2. 3L/4 cosα
– 0.046P – 0.0725Ws + 0.145 Ry=0
ΣMO = 0
= –Dy.L/2 cosα – P.L/4sinα –Ry.L/4 cosα + Ws.L/4 cosα – W/2.L/2 cosα +
Rx. L/4 sinα – 0.046P – 0.0725Ws + 0.145Ry = 0
ΣMB = 0
= –Dy.L cosα – P.3L/4 sinα – Ry.3L/4 cosα + Ws. 3L/4 cosα + Rx.3L/4 sinα
+ Fy.L/2 cosα + Wl/2. L/2 cosα – Fx. L/2 sinα – 0.046P – 0.0725Ws + 0.145Ry = 0
ΣMD = 0
= P.L/4 sinα + Ry.L/4 cosα – Ws.L/4 cosα – Fy.L/2 cosα – Wl/2. L/2 cosα +
Fx.L/2 sinα – W/2.L/2 cosα – Rx.L/4 sinα – 0.046P – 0.0725WS + 0.145Ry = 0
By substitution, we can find Fx, Fy, Rx, Ry, Dy, & P
Fx= -2296.01N
Fy= 2026.3N
Rx= 2067.1N
Ry = 5065.21 N
Dy= -2024.6N
P = 4363.11 N
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Diagrams
Where,
FDx` = 316.72 N
FDy` = 1999.66 N
FEx` = 1477.67 N
FEy` = 5347.5 N
FOx` = 1947.7 N
FOy` = 2380.128 N
FBx` =153.3 N
FBy` =967.93 N
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7- Stress Analysis at Lowest Position
The used Material for Legs is AISI 1020
Thickness = 22mm
Yield Strength = 351.571 N/mm^2 σall =
Sy /n = 351.571 /3 = 117.19 N/mm^2 τall
= 0.58 σall = 67.97 N/mm^2
- At point E (stress concentration section)
Existing loads on section
FN= 1794.4 N
M= 856.28 N.mm
Fs= 3347.84N
Area properties:
A=37.62*7=263.34 mm^2
I= (1/12)*b*h^3 = [(7*52.62^3)-(7*15^3)]*2/12=166043.0855
Stress concentration:
Normal: P/7*(52.62-15)
Bending: 6M/(52.62-15)*7^2
Shear: P/7*(52.62-15)
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Stresses
Normal…δ x = [1794.4/7*(37.62)] + [6*856.28/(37.62)*7^2]= 9.6N/mm^2
Shear= 3347.84/7*(37.62)=12.71 N/mm^2
δ1 = 0.5*( δx +δ y) + [(0.5*(δx+δy))^2+( ᴌx,y)^2]^0.5
= 0.5*(9.6+ (0)) + [(0.5*(9.6+0))^2+(12.71)^2]^0.5=18.39 N/mm^2
δ2 = 0.5*(δx +δ y) - [(0.5*(δx+δy))^2+( ᴌx,y)^2]^0.5
= 0.5*(9.6+ (0)) - [(0.5*(9.6+0))^2+(12.71)^2]^0.5=-8.79 N/mm^2
Thus, δ3=0
USING MSST THEORY
δall.=Sy/n= 351.571/3= 117.19<18.39 N/mm^2
ᴌall.= 0.58* δall.= 0.58*117.19=67.97 < 12.71 N/mm^2
Thus, Structure is SAFE.
17
References
Fundamentals of Machine Design Hamrock, Schmid, and Jacobson 2nd Edition,
McGraw-Hill, 2006
Motion Industries. (2013). Retrieved from https://www.motionindustries.com/
Olenin, G. (2016). Design of hydraulic scissors lifting platform. Saimaa University of Applied
Sciences.
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