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STEEL CUTTING CHARGES ACTION: Calculate and place steel cutting charges. CONDITONS: Given a 2 hour block of instruction, students handout, FM 5- 250,

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Presentation on theme: "STEEL CUTTING CHARGES ACTION: Calculate and place steel cutting charges. CONDITONS: Given a 2 hour block of instruction, students handout, FM 5- 250,"— Presentation transcript:

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2 STEEL CUTTING CHARGES ACTION: Calculate and place steel cutting charges. CONDITONS: Given a 2 hour block of instruction, students handout, FM , and Demolition Card GTA STANDARD: Students will correctly calculate and place steel cutting charges. SAFETY: Specific safety considerations will be discussed where appropriate throughout the lesson.

3 SPECIAL CONSIDERATIONS Target Configuration –Structured Steel Target Materials –High-Carbon Steel –Alloy Steel –Cast Iron Type of Explosives Size of Explosives Pg. 3-10

4 PLACEMENT OF CHARGE Split the charge Continuous Contact Width Priming Charge is placed on the side of target Charge is split in half Placed opposite side of each other Offset the web thickness Pg. 3-11

5 BLOCK CHARGES P = 3/8 A P = pounds of TNT 3 / 8 = Constant A = area of cross section of target in square inches Pg. 3-12

6 AREA OF CROSS SECTION FlangesWeb

7 BLOCK CHARGE walk through Cut each I-Beams 1 time, using C-4 Flanges 16 x 1, Web 12 x.5 5 Beams

8 BLOCK SOLUTION Step 1 :Flanges = 16 x 1 x 2,Web = 12 x.5 1 cut, C-4, 5 Beams Step 2 : Flanges = 16 x 1 x 2 = 32 Sq in Web = 12 x.5 = 6 Sq in TOTAL = 38 Sq in P = 3/8 x 38 = lbs TNT Step 3 : = lbs C4 (M112) 1.34 Step 4 : = Step 5 : 5 Beams x 1 cut = 5 Charges Step 6 : 9 x 5 = 45 pkgs C4 (M112) 9 pkgs C4 (M112)

9 15 X 1 12 X 1 Use C4 Total of 4 beams BLOCK CHARGE #1

10 Step 1 :Flanges = 15 x 1 x 2 = 30 Sq in Web = 12 x 1 = 12 Sq in TOTAL = 42 Sq in Step 2 : P = 3/8 A P = 3/8 x 42 = lbs TNT Step 3 : = lbs C4 (M112) 1.34 Step 4 : = pkgs C4 (M112) 1.25 Step 5 : 4 Beams = 4 Charges Step 6 : 10 x 4 = 40 pkgs C4 (M112) BLOCK SOLUTION # 1

11 19.5” x 2” 19.5” x 2.5” 14” x 1.5” How many packages of C-4 are required to cut the 15 beams shown using a block charge

12 BLOCK CHARGE STEP 1 : Given: C4, 15 I-beams Flanges 19.5 x 2 = x 2.5 = x 1.5 = 21 Total Area = STEP 2: P = 3/8A.375 x =40.78 lbs TNT STEP 3: / 1.34 = lbs C4 STEP 4: / 1.25 = Pkg C4 25 pkgs C4 STEP 5: 15 Beams = 15 charges STEP 6: 25 x 15 = 375 pkgs C-4 (M112)

13 Average Thickness of Section (in) Pounds of Explosive (TNT) for Rectangular steel sections of given Dimensions Height of Section (in) /4 1 7/8 3/4 5/8 1/2 3/ NOTE: to use this table – 1.Measure each rectangular section of total member separately. 2.Find the appropriate charge size for the rectangular section from the table. If the section dimension is not listed in the table, use the next-larger dimension. 3. Add the individual charges for each section to obtain the total charge weight. 1/

14 Average Thickness of Section (in) Pounds of Explosive (TNT) for Rectangular steel sections of given Dimensions Height of Section (in) /4 1 7/8 3/4 5/8 1/2 3/ NOTE: to use this table – 1.Measure each rectangular section of total member separately. 2.Find the appropriate charge size for the rectangular section from the table. If the section dimension is not listed in the table, use the next-larger dimension. 3. Add the individual charges for each section to obtain the total charge weight. 1/2 16 3

15 HASTY BLOCK CHARGE Table 3-3, pg 3-12: TNT 1. Measure rectangular sections separately. 2. Use table to find pounds of explosives for each section. Table 3-3, pg 3-12: TNT 3. Add charges from each section to find total charges. 4. If dimensions of sections are not on table, use the larger dimension.

16 How many packages of C4, (M112) are required to cut 5 I-beams with dimensions of 8” x 1/2” for the flanges, 9” x 1” for the web ? HASTY PROBLEM # 1

17 HASTY SOLUTION # 1 Step 1 : [8” x 1/2”] [9” x 1”] [8” x 1/2”] Table Value Step 2 : 8” x 1/2” 1.2 lbs 9” x 1” 2.9 lbs 8” x 1/2” 1.2 lbs Total = 5.3 lbs C4 Step 3 : N/A Step 4 : 5.3 = Pkgs C4 (M112) 1.25 Step 5 : 5 beams = 5 charges Step 6 : 5 x 5 = 25 pkgs C4 (M112)

18 RIBBON CHARGE Step 2. Calculate the volume of explosives. –T x W x L = volume of explosives in cu. in. needed –T = Charge thickness = 1/2 TGT but never less than 1/2 inch –W = Charge width = 3 times the charge thickness –L = Charge length = length of TGT to be cut Step 4. Calculate packages of C4 or M118 –C4: 1” x 2” x 10” = 20 cu inches –M118 sheet:.25” x 3” X 12” = 9 cu inches 4 SHEETS = 1 M118 PACKAGE (use in Step 6) Maximum target thickness = 3 inches Pg. 3-15

19 RIBBON CHARGE PLACEMENT Thickness of charge = 1/2 thickness of TGT Width of charge = 3 times thickness of charge Length of charge = length of TGT Place ½ inch explosive around cap or knot Primed at center with a blasting cap or det cord knot Det Cord or Time Fuse Pg. 3-16

20 WEB < 2 inches offset edge to center > 2 inches offset flange charges edge to edge

21 Beams > 2” thick offset flange charge: Edge to Edge Det Cord branch lines are all of equal lengths forming a (British Junction) Beams < 2” thick offset flange charge: Edge to Center Charge Thickness = 1/2 Thickness of the target never less than.5

22 Charge Thickness (T)Target thickness

23 RIBBON CHARGE Charge thickness (T) Charge thickness (T) Charge width 3 x T Charge width 3 x T Charge width 3 x T Charge length FL - WT =L Charge length FL - WT =L Charge length WL - TCT - BCT = L Required explosive T x W x L=VOL Required explosive T x W x L=VOL Required explosive T x W x L=VOL Charge thickness (T) TOTAL EXPLOSIVES T W L TF WEB BF

24 RIBBON CHARGE WALK THROUGH How much C4 (M112) is required to cut one plate? 24” 2”

25 Step 1. 2” x 24” Step 2. 1” x 3” x 24” = 72 cu. in. Step 3. N/A Step = pkgs C-4 (M112) 20 Step 5. 1 Plate = 1 charge Step 6. 4 x 1 = 4 pkgs C-4 (M112) RIBBON CHARGE WALK THROUGH

26 19.5” x 2” 19.5” x 2.5” 14” x 1.5” How many packages of C-4 are required to cut the 15 beams shown using a ribbon charge

27 RIBBON CHARGE STEP 1 : Given: C4, 15 I-beams Flanges 19.5 x x 2.5 Web 14 x 1.5 STEP 2: T x W x L = VOL cu. in. of explosives TF 1 x 3 x 18 = 54 Web.75 x 2.25 x = BF 1.25 x 3.75 x 18 = TOTAL VOL = STEP 3: N/A STEP 4: = pkgs C-4 (M112) STEP 5: 15 Beams = 15 charges STEP 6: 8 x 15 = 120 pkgs C-4 (M112)

28 19.5” 14” 2” RIBBON CHARGE PROBLEM #2 2” Use (M118 sheet explosive) 15 Beams

29 STEP 1 : Given: M118, 15 I-beams STEP 2: T = 1” W = 3” L = Top Flange = 19.5” - 2” = 17.5” Bottom Flange = 19.5” - 2” = 17.5” Web = 14” - 2” = 12” TOTAL = 47” T x W x L = 1”x 3” x 47” = 141 cu. in. of explosives STEP 3: N/A STEP 4: 141 = sheets (M118) 9 STEP 5: 15 Beams = 15 charges STEP 6: 16 x 15 = 240 sheets (M118) 4 = 60 packages (M118) RIBBON CHARGE PROBLEM #2 SOLUTION

30 8” 16” 1.5” RIBBON CHARGE PROBLEM #3 1.5” Use M Beams 1.5”

31 STEP 1 : Given: M118, 12 I-beams STEP 2: T =.75” W = 2.25” L = Top Flange = 8” - 1.5” = 6.5” Bottom Flange = 8” - 1.5” = 6.5” Web = 16” - 1.5” = 14.5” TOTAL = 27.5” T x W x L =.75” x 2.25” x 27.5” = 46.4 cu. in. of explosives STEP 3: N/A STEP 4: 46.4 = sheets M118 9 STEP 5: 12 Beams = 12 charges STEP 6: 6 x 12 = 72 sheets 4 sheets per pkg = 18 pkgs M118 RIBBON CHARGE PROBLEM #3 SOLUTION

32 SUMMARY v BLOCK CHARGE v Placement and Priming v Formula method v Table method v RIBBON CHARGE v Placement and Priming v Formula v M118 v Round to SHEET in step 4 v Round to PACKAGE in step 6


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