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# Engineering Mechanics: Statics Design Problem 3.2

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Engineering Mechanics: Statics
Design Problem 3.2
A plate Storage System

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### Engineering Mechanics: Statics Design Problem 3.2

1. 1. Abstract The project to design is the chute of concrete truck for delivering wet concrete to a construction site as shown in the diagram bellow. This article shows that a concrete chute could be able to design in order to accessible a wet concrete through using an average industry truck on the ground. While the segments BC and CD either added or removed would result a change in the length or size of the chute. The shape of the chute, the hydraulic cylinder that being used to raise and lower the chute and also the weight of each segments of the chute needs to be consider in to design this project. The scope of this design project includes the design of the construction site in its entirety including the chute concrete truck, the concrete, the concrete mixer and also the concrete chute itselfin order to specify the force capacity of the hydraulic cylinder GH. Technical Report Subject Date Name Project Design # 5.4; Design of spring of A Plate Storage System March 5, 2012 KehaliBekele Haileselassie TR Location UW Milwaukee
2. 2. Figure 1 Introduction “Concrete is a composite construction material made primarily with aggregate, cement, comprising sand, comprising gravel, crushed stone and water”.(Wikipedia)Each of these materials is produced separately and mixed at the site to make concrete. Concrete can be found in many forms (such as ready-mix concrete, wide mix concrete) with varied properties.Concrete is the most popular product in the world today. “Concrete is widely used for making architectural structures, foundations, brick/blocks wall, pavements, bridges/overpasses, motorways, runways, parking structures, dams, pools/reservoirs, pipes, footings for gates, poles and even boats.”(Wikipedia) Concrete is manufactured in a factory with a set recipe and then delivered to a work area, by using the truck mounted concrete mixer and the chute of concert truck. Concrete chutes are manufactured in different forms. In most case, we find a concrete chutes either with a flare or without a flare. See the picture bellow for more elastration about the two different forms of concrete chute.
3. 3. Concrete chute without flare Figure 2. (“The concrete supply house”) Concrete chute with flare Figure 3 (“The concrete supply house”)
4. 4. In general, Concrete chutes are specifically produced for the purpose of delivering and transporting a concrete (specially wet concrete) from the concert chute truck to the construction site. In general, Concrete chutes are the most useful product in engineering and construction projects. Free Body Diagram Figure 4 (“Plesha,P319”) Road Map: The following needs must be satisfied by the force capacity of the hydraulic Cylinder GH: The force capacity should be between the capacity we determine using angle zero and angle fifty to transport the concrete safely and successfully through the concrete chute. The concrete chute should not be touching the ground while it’s delivering the wet concrete. Concrete surface should be visually examined for a safety purpose due to weathering, stress, chemical attack, erosion, destructive forces and so on. As we see in the given diagram (figure 1) and the problem stated, All what we need toanalyze is to specify the force capacity of the hydraulic cylinder GH while it raised or lowered
5. 5. the chute in the range of angle zero and fifty. We should consider the maximum length of the segments, the weight of each segments, the angle that the hydraulic cylinder GH makes while it raised or lowered the chute and also the weight of the chute segments (segment AB, segment BC and Segment CD). Furthermore, we should assume the concrete has a constant density about 2400kg/m3 . This will add another load component to the problem beside the weight of the chute segments. To be more accurate, we need also to consider the shape of the chute in order to specify the weight of the concrete. The given values of the problem are specified in details in table one (1) below for further clarification: The given values of the problem SAB 48 in. SBC 48 in. SCD 48 in. WAB 50 lb. WBC 50 lb. WCD 50 lb. Dcon(S) 2,400kg/m 3 ( 0.086706 lb./in 3 ) Rchute 8 in. Smax 144 in. Where, W stands for the weight S stands for the length of the segment Dcon is the density of the concrete Rchute is the radius of the chute Smaxstands for the maximum length of the chute
6. 6. Governing Equations in details : WAB = 50 lb. , WBC = 50 lb. , WCD = 50 lb. SMax= 48 in. + 48 in. + 48 in. = 144 in. Since 1kg = 2.2046lb., 1m = 39.3701in. Dcon= 2,400kg/m3 * (2.2046lb. / 1kg) *(1m/ 39.3701in.)3 = 0.086706lb./in3 Dcon= 0.0867lb./in3 Since Density is mass divided by volume, Mass is density multiplied by volume and Volume of semicircle is half of the area of a circle. Weight = D * (0.5 * π r3 ) = 0.086706lb./in3 * (0.5 * π * (8 in.)2 ) = 8.717lb. Steps 1 At θ=00 Taking a moment at position A ∑MA = 0 Angle AGH = Tan-1 (24/30) = 38.650 (WABCos(00 ))(24)+(WBCCos(00 ))(48+24)+(WCDCos(00 ))(48+48+24)+(WconCos(00 ))(144/2)- (FGHCos(38.650 )(7)-(FGHSin(38.650 )(30)=0 50(24)+50(72)+50(120)+8.717(72)-5.467FGH-18.737FGH = 0 50(24)+50(72)+50(120)+8.717(72) = 5.467FGH+18.737FGH 11427.6 = 24.204FGH ,FGH = 472.138lb. FGH = 472.138lb.
7. 7. Steps 2 At θ=50 0 Based on the figure, we can see that angle HAG = 90 0 -50 0 = 40 0 Determine angle (AGH) From Triangle HAG, we can develop an equation using cosine law: (GH)2 = (AH)2 + (AG)2 - 2*(AH)*(AG)*Cos(400 ) (GH)2 = (30)2 + (24)2 – 2*(30)*(24)*Cos(400 ) (GH)2 = 372.896 (GH) = 19.31in. By using sin law, we can determine angle (AGH) (30/Sin(AGH)0 ) = (19.31/Sin(400 )) → (AGH)0 = Sin-1 (30*Sin(400 )/19.31) → (AGH)0 = 87.0050 (AGH)0 = 87.0050 Taking a moment at position A using 87.0050 (WABCos(500 ))(24)+(WBCCos(500 ))(48+24)+(WCDCos(500 ))(48+48+24)+(WconCos(500 ))(144/2) -(FGHSin(87.0050 )(7)-(FGHCos(87.0050 )(30) = 0 50(0.642)(24)+50(0.642)(72)+5090.642)(120)+8.717(0.642)(72)-6.99FGH-1.567FGH = 0 50(0.642)(24)+50(0.642)(72)+50(0.642)(120)+8.717(0.642)(72) = 6.99FGH+1.567FGH 7569.73 = 8.5579FGH →FGH = 884.531lb. FGH = 884.531lb. So, the loading Capacity of the Chute is in between 472.138lb. and 884.531lb.
8. 8. Conclussion Aim: The aim of this deign project is to analysis the chute of concrete truck for delivering wet concrete to a construction site. In this report, a design for a concrete chute has been analyzed. The report also presented enough information about mounted concrete mixer, concrete truck and also about concrete chute. It also presented the materials, uses and structures of concrete in general. In order to designs and presents the chute of concrete truck for delivering wet concrete to a construction site in this project, Two alternative steps have been made:Step 1, the hydraulic cylinder GH that attaches to the concrete chute is considered as it was at zero degree, and steps 2, the hydraulic cylinder GH that attaches to the concrete as it was at fifty degree. In each steps, we specify the force capacity of the hydraulic cylinder GH as it was determined in our calculation. The weight of the concrete has been determined using the given radius of the chute and the constant density of the concrete. However, the force capacity of the hydraulic cylinder is recommended to be in the range between the capacities of the force that we found using zero degree and fifty degree (472.138lb. and 884.531lb.) in order to delivering wet concrete using the chute.
9. 9. Reference 1. Plesha, Michael. Engineering Mechanics. 2nd . McGraw-Hill, 2013. Print. 2. Wikipedia ( The Free Encyclopedia ), 2 March 2013. Article 3. “The concrete supply house”, Kraft Tool Co. , http://www.concretesupplyhouse.com/concrete-tools/concrete-chutes/ 4. Stephan Guedon, Concrete Building Design, VIA University, 2011, Interdisciplinary Project 5. Bruce A. , Concert Placement for Small Pour , University of Colorado, http://www.concreteconstruction.net/images/Concrete%20Placement%20for%20Small% 20Pours_tcm45-342184.pdf