Abstract In this paper describes the design as well as analysis of actuator scissor lift. The scissor lift used to lift the body to appreciable height, and many other applications also such lifts can be used for various purposes like maintenance and many material handling operations. It can be of mechanical, pneumatic or hydraulic type. The Nurnberg scissors” type load lifting mechanisms are produced abroad. But such load lifting platforms are very dangerous and might overturn since equipments’ stiffness is low. Also these load lifting platforms are very heavy so design described in the paper is developed keeping in mind that the lift can be operated by mechanical means by using actuator so that the overall cost of the scissor lift is reduced. …show more content…
A lift table is defined as a scissor lift used to stack, raise or lower, convey and/or transfer material between two or more elevations. The main objective of the devices used for lifting purposes is to make the table adjustable to a desired height. A scissor lift provides most economic dependable & versatile methods of lifting loads; it has few moving parts which may only require lubrication. This lift table raises load smoothly to any desired height. The scissor lift can be used in combination with any of applications such as pneumatic, hydraulic, mechanical, Scissor lift design is used because of its ergonomics as compared to other heavy lifting devices available in the market. Scissor lifts typically operate in two axis of movement and are designed for applications where people and material need only up and down travel (stationary lift), where the lift needs to be moved around to perform work (manually positioned lift), or to access work along a fixed area of travel (rail guided lift). Utilized in applications from wash, prep and paint, to metal blasting, welding, and assembly; scissor lifts offer a safer, more ergonomic and cost effective solution for positioning workers. Systems utilize actuator power units for hazardous areas such as paint …show more content…
= 1040.86 N ≈ 1041 N From free body diagram, R1=R2 =1041 N Bending moment equation, M/I=σ/y=E/R M=F1*(L-d) =736*(610-25) = 430560 N-mm I = B/12* D^3 - b/12* d^3 =30/12* 〖60〗^3 - 27/12* 〖57〗^3 = 540000-416684.25 = 123315.75 〖mm〗^4 According to bending moment equation, M/I=σ/y=E/R σ=M*y/I (y/I=Z=section
Contents TASK 1 1 1.1)TYPICAL AXIS CONVENTION 1 1.2) operations of types of drives and axis control system. 4 1.3) SIX DEGREES OF FREEDOM. 8 1.4] WORK HOLDING DEVICE FOR LATHE 8 TASK 2. 12 2.1) Assess the suitability of machine tools for the production of following components 12 2.2) SEQUENCE OF OPERATIONS TO PRODUCE THE GIVEN COMPONENT 13 2.3) MACHINING AND FORMING PROCESS 13 TASK 3
A line of best fit was drawn from the results which gave the power of the unknown lenses. The equation of the line of best fit was[Y = 100x]
The coordinates of the system is defined by , θ = angle of the chassis from vertical, α = angle of tread assemblies from vertical, Ø = rotation angle of tread sprockets from vertical, mc = mass of chassis, mT = mass of tread, ms = mass of sprocket, Lc = length from centre of sprocket to centre of chassis, LT = length from centre of sprocket to centre of tread assembly. The kinetic energies of the sprocket, chassis and tread assemblies are given respectively , T_S=1/2[m_c x ̇^2+J_S φ ̇^2] (1) T_C=1/2 [〖m_c (x ̇-L_c θ ̇ cosθ)〗^2+m_c (〖L_c θ ̇ sin〖θ)〗〗^2+J_c θ ̇^2 ] (2) T_T=1/2[m_T (〖x ̇-L_T α ̇ cos〖α)〗〗^2+m_T (〖L_T α ̇ sin〖α)〗〗^2+J_T α ̇^2] (3) The gravitational potential energy is given by ,
The intent of this paper is to take a constructive view of Wedge Recovery Center South with applying the System approach. The System method would assist us with formative the structure organization regarding to the Wedge Recovery Center South. To address this unbiased investigation, the focal system, subsystems and supra-systems would provide the surveillance that are require with constructive observation of given structure within the Wedge Recovery Center. System terminology would be applied, along with the definitions. It is imperative for the Wedge Recovery Center South be establishes as the focal system (object of attention) in order for the system methodology to be unitize.
Using FIT with theses adjusted patterns I feel confident that I will complete this assignment
Sturdy and well designed Surprise, Your Sawzall Can Do One More Job You Never Thought About Milwaukee 6509-22 Sawzall 11 Amp Reciprocating Saw Milwaukee 6509-22 Sawzall 11 Amp Reciprocating Saw 10A, Lightweight Sawzall, 3/4″ Blade Stroke, Quik-Lok Blade Clamp, Trigger Speed Control 0-2800 SPM, 8′ Rubber Cord, Adjustable Pivot Shoe, Includes Carrying Case. Milwaukee 6509-22 Sawzall 11 Amp Reciprocating Saw Use them for work and own one for me I can’t really elaborate much more on the Milwaukee Sawzalls beyond what has been said, but for professional use I highly suggest one of these over nearly all of the others (I haven’t seen anyone actually test the Hilti saws over a year or two, but they are a bit too heavy for continual use). I will say this – for a lot less money the Milwaukee beats the saw Makita just recently put out in as far as reliability. Product name This set was called “Ultimate Demolition” I wrongly assumed it was just a hype type name.
FOV Dia.(mm) FOV Area (mm^2) 10 10 10 100 2mm 4mm^2 20 20 10 200 1mm 1mm^2 40 40 10 400 .5 mm
1. Why is he well-known? Mention some of his accomplishments. Kip Cullers is highly intelligent Missouri Farmer, who has set world records of soybean yields. He has several key pointers to improve soybean yields on farms.
0 0 0 0 0 15 33.6 33.6 0.1867 -0.286 0.242207344 30 22.9 56.5 0.3139 -0.54 0.406195492 60 40.4 96.9 0.5383 -1.26 0.592397947 120 20.2 117.1 0.6506 -2.02 0.716882124 240 16.4 133.5 0.7417 -4.5 0.747840076 Explorated Rinf=0.75 Calculated Rinf=0.75 6*10^(-5) 150 0
The measured tensions were normalized relative to the maximum tension and the lengths were normalized relative to the length at which maximum force was generated (Ln = 29 mm). The measured data and expected data were plotted together (Figure 3). Correlation between measured and expected tensions at the same normalized lengths was determined for the three expected segments of the length-tension relationship.2,3,4 This was done using piecewise linear regression and yielded r2isometric = 0.956033. The fitted curve for the isotonic experiment resulted in r2isotonic = 0.960557. The F0 was 19.5 N for the fit and 6.35 N for the guess.
Material weighs 100 lbs = Sand based material = Temporary lifting solution = Requires a series of large drilled holes, so the thick material can be pumped under the slab Polyjacking - Uses a two component pump that injects polyurehane under the sunken slab - Smaller and fewer are drilled through the slab =
Back in 2006, Daktronics faced a strong three-year growth period since 2003. To-tal sales increased by 74% from $177M to $309M. To maintain this growth, Daktronics set the goal of eliminating manufacturing and capacity constraints. Before 2006, Daktronics followed the main strategy of replication (increasing number of facilities, equipment and people). They decided to expand their first facility in Brookings and add two more (Redwood Falls and Sioux Falls). Increas-ing pressure on cost reduction led the company to think of different methods of growth management.
A jack A lug wrench An inflated spare tire Wheel wedges A small cut wood of 2’’*6’’
Physics, period 3 Malak Mokhles Data collection: Jan To measure the period of a swinging stopper for three selected radii in order to calculate the centripetal force Data Table Calculations Calculate the centripetal force acting on the stopper. (Fc=mac) 50 cm radius: (0.025kg)(50m/s2)=1.3N 35 cm radius: (0.025kg)(43m/s2)=1.1N 25 cm radius: (0.025kg)(39m/s2)=1.3N State the weight of the washers 50 cm radius: 15 washers=0.75N 35 cm radius: 15 washers=0.75N 25 cm radius: 10 washers=0.50N Calculate the percent error for each radius (% error =|theoretical - experimental /( theoretical ) | × 100%) 50 cm radius: |0.75 – 1.3 /(0.75) | × 100% = 73% 35 cm radius: |0.75 – 1.1 /(0.75) | × 100% = 47% 25 cm radius: |0.50 – 1.0 /(0.50) | × 100% = 100% Analysis/Discussion
Figure 1 shows the load versus extension and Figure 2 shows the stress strain curve. Figure 1: Load as a Function of Extension of AA 2024T351 Figure 2: Stress as a Function of Strain Graph of AA 2024T351 AA 5052-0 The material properties of alloy AA 5052-0 is shown in Table 1. Table 2: 5052-0 Specimen Experimental