Heavy Equipment & Rigging Specialist Training

Transcription

1 Heavy Equipment & Rigging Specialist Training Module 2 Unit 2: Calculating Weight & Center of Gravity Mar08 1

2 Unit Objective Be able to Calculate the Weight of Steel, Concrete & Other Objects Be able to Estimate the Center of Gravity of many different shaped objects 2

3 Enabling Objectives Discuss the Unit Weight (lbs per cubic foot) of materials found in collapsed structures Discuss Volume x Unit Weight Method for calculating weight of simple objects Discuss the Weight per Foot x Length Method for calculating weight of complicated objects Compare to Area Method for Steel Demonstrate quick methods for calculating the weight of non-uniform shaped objects Describe simple Graphical Method for estimating the C/G. of complicated Shapes 3

4 Calculating Weights 4

5 Weight of Common Materials Steel 490 lbs. per cubic foot (pcf) Concrete 150 pcf (145 conc. + 5 rebar) Earth Wood 100 to 125 pcf (clay to rock) 40 pcf (depends on species)

6 Calculating Weight of Common Shapes RECTANGLE Height x Width x Length x Weight 2 x 4 x 20 = 160 cu-ft x 150pcf = 24,000 lbs. 4 feet 2 feet 20 feet 6

7 Calculating Weight of Common Shapes RECTANGLE w/ 30 sq Cutout Solid Slab - Hole 8 /12 ( 6 x x 2.5 ) x 150pcf 60 cu-ft x 150pcf = 8975lbs 6 feet sq 16 feet 7

8 Calculating Weight of Common Shapes ROUND 0.8 x Diameter x Dia x Length x Weight 0.8 X 3 x 3 x 20 x 150pcf 144 cu-ft x 150pcf = 21,600 lbs. 3 feet 20 feet 8

9 Calculating Weight of Common Shapes HOLLOW ROUND (PIPE) Method 1 Weight of Solid Round Weight of Hole 0.8 (4 x 4-3 x 3 ) x 20 x 150pcf 112 cu-ft x 150pcf = 16,800 lbs. 4 ft dia. 6 thick 20 feet 9

10 Calculating Weight of Common Shapes HOLLOW ROUND (PIPE) Method 2 Unfolded Slab Method Sim to Area Method Circumference x thickness x length x 150 pcf 3 x 4 x 0.5 x 20 x 150pcf = 18,000 lbs. (7% higher the thicker the greater error) 4 ft Dia. Circumference = 3 x Dia. 6 thick 20 feet 10

11 Area Method is OK for Thin Pipes What is weight of this 16ft long steel pipe? 12 Diameter x ½ thick ½ Steel = 20 psf Area per ft = 3 x 1 = 3 sq ft Weight per ft = 3 sf x 20 = 60 plf, Total = 960 lb Exact weight = 61.4 plf (only 2% off) For Sq Tube use 4 x Size x Wt per sq ft 11

12 Weight per Foot x Length Method Based on weight of 1 sq-in x 1 foot of length Steel = 490/144 = 3.4 Conc = 150/144 = 1.04 (Use 1.1 since more rebar in columns & beams) Wood = 40/144 = 0.28 Wire Rope = 0.7 x 3.4 = x1 144 sq in 12 12

13 Wt per Foot Method - Example Precast Concrete Example : Double Tee x 60 ft Long (2 x avg. x 30 x 2) x 1.0 (PC Conc) (192 sq-in sq-in) x 1.0 = 492 plf For 60ft long = 492 plf x 60ft = 29,520 lbs Double Tee Single T is similar 13

14 Wt per Foot Method Steel Example Steel weighs 3.4 lbs per sq inch, per foot of length 1 square steel bar weighs 3.4 lbs per ft Example 1: 1 x 12 steel plate x one foot long weighs? 12 sq in x 3.4 = 40.8 lbs per linear foot (plf) (same as 12-1 square bars x one foot long) Example 2: 1 ½ dia. round steel bar x 20 ft weighs 1.5 x 1.5 x.8 (for round shape) x 3.4 x 20ft = 1.77 sq in x 3.4 x 20ft = 120 lbs (Wire Rope weighs abt 2/3 of same size Round Bar) 14

15 16 National Urban Search & Rescue Response System Wt per Foot Method Steel Example Example 3: What is weight of the fabricated steel column that is shown below? (2 x 36 x 2 ) + (2 x 12 x 2 ) x 3.4 ( 144 sq in + 48 sq in) x 3.4 = 653 lbs per ft Col is 36 long & weighs 653 plf x 36 = 23,500 lbs = 12T Pl 36 x 2 36 Pl 12 x 2 ea end Pl 36 x 2 15

16 Compare to Area Method Example What is weight of this 36ft long steel section? 2 Steel = 2 x 40 psf = 80 psf Area per ft = 2 x 3 sq ft + 2 x 1 sq ft = 8 sq ft Weight per ft = 8 x 80 = 640 plf Total weight = 640 x 36 = 23,040 lbs Exact weight = plf (only 2% off) Pl 36"x 2" Pl 12"x 2" ea end Pl 36"x 2" 16

17 Steel Column at WTC 17

18 Weight of Odd Shaped Objects L or other Odd Shaped Slabs Divide into individual rectangles & add-up. 8 /12 (10 x x 7 ) x 150 pcf = Weight 8/12 (102 sq-ft) x 150 pcf = 68 cu-ft x 150 = 10,200 lbs 12 ft 10 ft 8 thick 6 ft 7 ft 6 ft 18

19 2ft National Urban Search & Rescue Response System Properties of Triangle (with 90 deg corner) Area is H x W/2 Volume is H x W/2 x Thickness 6 x 3 /2 x 8 /12 = 6 cu-ft Weight = 6 cu-ft x 150 pcf = 900lb Location of C.G. is at 1/3 Point 1/3 the length each way from the 90 deg corner 8 thick 3ft 1ft 6ft 19

20 Weight of Odd Shaped Objects Tapered Slabs Divide into Uniform Thickness Slab + Triangular Cross-section Area of Triangle is Length x Width divided by 2 (8 /12 x 4 x /12 x 4 /2 x 7 ) x 150 pcf = Wt (18.67 cu-ft cu-ft) x 150 pcf = 4,550 lbs 8 thick 4 ft 10 thick 7 ft 8 thick 20

21 Finding Center of Gravity 21

22 Center of Gravity X Y Z X Z Point on an object around which object s mass is evenly distributed Center is at the junction of three axes X-axis = Horizontal, side to side Y-axis = Vertical Z-axis = Horizontal, front to back Y 22

23 Effects of Center of Gravity Lifting point CG CG Object will rotate until CG in under lift point 23

24 Center of Gravity and Load Stability Stable Unstable CG CG Connection point below CG makes object unstable 24

25 Estimating Center of Gravity Not difficult for simple shapes = Middle Beams, Columns, Rectangular Slabs Single & Dbl Tees, Slabs w/ Holes Most Steel and Wood Members What about Odd Shaped Slabs? See Crosby Method, next 3 slides What about Tapered Sections? Divide into Rectangle + Triangle 25

26 2ft National Urban Search & Rescue Response System Finding Center of Gravity Crosby Method C.G. is Located Toward Heavier Piece Heavy Piece Total Weight = C.G. of each prism = C.G. of group 900lb 1ft 2/3 8 thick 3ft 1800lb 6ft 1/3 3ft Total Wt = 2700lb 29

27 Center of Gravity Calculation by StS (2400 x x 6) H = = (2400 x x 1.5) V = = = C.G. of each prism = C.G. of group 8 thick 1200lb 3ft 4ft 4ft 8 thick H = 3.33ft V = 2400lb 2.5ft Total Wt = 3600lb 6ft 30

28 2ft National Urban Search & Rescue Response System Center of Gravity Calculation by StS (1800 x x 4) H = = (1800 x x 2) V = = lb 1ft 8 thick 3ft H = 2.33ft 1800lb 6ft = C.G. of each prism V = 2.67ft = C.G. of group 3ft Total Wt = 2700lb 31

29 4.44 ft National Urban Search & Rescue Response System Center of Gravity Hole in Slab (can t use Crosby Method) V = (4000 x x 2.5) (16, ) V = = = C.G. of each prism = C.G. of group 5ft 8 thick 4,000lb -900lb 3ft hole 3ft 1ft 1ft 8ft Wt = 4, lb = 3,100 32

30 Summary - Calculating Weights & C.G. Use Volume x Unit Weight Method for calculating weight of simple objects Use Weight per Foot x Length Method for calculating weight of more complex objects May use Area Method for Steel Shapes Use Crosby Method for finding C.G. of non rectangular slabs Use StS to calculate C.G. of complicated Shapes If no StS is available, make best estimate, then Rig it and do Trial Lift 33

31 Conclusion of Calculating Weights & Center of Gravity Questions? Discussion? Next: Team Exercise Each Team work together this is a contest Find the weight and CG of the following 6 shapes 34

32 Problem 1 10 ft 10 ft 6 thick 4 ft 6 ft 6 ft Problem 2 12 ft 1 ft 5 ft 8 thick 5 ft 5 ft 5 ft 6 ft 35

33 Problem 3 Problem 4 4ft 8 thick 8ft 7ft 1ft 8 thick Hole 4ft 8ft 4ft 4ft 4ft 1ft 36

34 Problem 5 3 ft dia. 4 thick 20 ft long Problem 6 Single Tee x 90ft long thick 37

35 Team Debriefing Instructor s Solution 38

36 Problem 1 Wt = ft ft L = 6 x 10 x 75psf = 4500lb L ft.71 4 ft R 6 ft 6 thick R = 6 x 4 x 75psf = 1800lb 1800lb lb = 6300lb 1800lb / 6300lb = lb / 6300lb =.71 39

37 Problem 2 Wt = ft 5 ft 6 ft F B = 5 x 12 x 100psf = 6000lb F = 5 x 6 x 100psf = 3000lb 6000lb lb = 9000lb 3000lb / 9000lb = lb / 9000lb = ft B 5 ft 5 ft 5 ft 8 thick 40

38 3.56 = V National Urban Search & Rescue Response System Problem 3 Wt = 3200lb +1600lb = 4800lb, 1600lb / 4800lb =.33 (3200 x x 5.33) H = ft 3200lb 3.11 = H ( ) H = = ft (3200 x x 2.67) V = 1600lb ft 4ft (12, ) V = = thick

39 4.4 = V National Urban Search & Rescue Response System Problem 4 Wt = (7 x 8 4 x 4 ) x 100psf = 4000lb (5600 x x 4) H = (19, ) H = = (5600 x x 3) V = (22, ) V = = ft 7ft Hole 4ft 4ft 1ft 3.3 = H 8 thick 8ft 42

40 Problem 5 Wt=8570 CG at mid length 3 ft 4 thick 20 ft long 0.8 (3 x x 2.33 ) x 20 x 150pcf 57.1 cu-ft x 150pcf = 8,570 lbs. 43

41 Problem 6 Wt = 560plf = 50.4k CG at mid length Single Tee x 90ft long Single Tee x 90 ft Long (2 x x 46 ) x 1.0 (PC Conc) (192 sq-in sq-in) x 1.0 = 560 plf For 90ft long = 560 plf x 90ft = 50,400 lbs 2 thick 44

42 Evaluation Please complete the evaluation form for Module 2 Unit 2: Calculating Weights & Center of Gravity 45