WLD 204 Non Destructive Testing I Math Review

Transcription

1 WLD 0 Non Destructive Testing I Math Review NSF-ATE Project

2 Index Math Review - Inspection Gages 6- This project was supported, in part, by the National Science Foundation Opinions expressed are those of the authors And not necessarily those of the Foundation NSF-ATE Project 1

3 READING NUMBERS The most important part of any number is the decimal point. Every number is written around a decimal point. Whole units are located to the left of it, and anything less than a whole unit is located to the right of it. The decimal point may be considered as a point of reference, identifying each digit by its relative position. For example, the following number (1,.67) is read: one thousand, five hundred thirty-four and three hundred sixty-seven thousandths. This means there are 1, whole units, plus 67/1000 of one unit. The following are examples of numbers and how they are read: 1,7,.6 Twelve million, nine hundred seventy-eight thousand, five hundred forty-three and eight hundred ninety-six thousandths. 1,,601.7 One million, four hundred twenty-three thousand, six hundred one and seventy-eight hundredths. 670,0. Six hundred seventy thousand, eight hundred nine and nine tenths. 6,06 Fifty-six thousand, two hundred six. 7,000 Seven thousand.,0 Three thousand, nine hundred eighty. ADDITION Addition is the process of combining two or more numbers so as to obtain a number called their sum or total. The numbers being added are Addends. The result is the Sum.. Addend 17. Addend 61.7 Sum To prove the accuracy of your addition, you merely reverse the order and add again. SUBTRACTION Subtraction is the process of finding the difference between two numbers. The number from which another is to be subtracted is the Minuend. The number to be subtracted from another is the Subtrahend. The result is the Difference. 0 Minuend - 7 Subtrahend 1 Difference or Remainder To prove the accuracy of your subtraction, you add the Difference to your Subtrahend and the result should be the same as your Minuend. NSF-ATE Project

4 CREDIT OR NEGATIVE BALANCE When subtracting a larger number from a smaller number, invert and subtract. The result is called a Credit or Negative Balance. $1. $ $1. Credit Balance MULTIPLICATION Multiplication is repeated addition. The number to be multiplied is the Multiplicand. The number by which another is multiplied is the Multiplier. The result of the multiplication is the Product. Although the multiplicand and multiplier are interchangeable, the product is always the same. 1 Multiplicand x 1 Multiplier 6 Product If one number is larger than the other, the larger number is usually used as the multiplicand. MULTIPLICATION WITH DECIMALS When multiplying with decimals, the product is pointed off from right to left as many decimal places as there are in the multiplicand and multiplier. 7.6 Multiplicand has decimal places x1. Multiplier has 1 decimal place 10.6 Product has decimal places To multiply a decimal by any multiple of ten, move the decimal point as many places to the right as there are zeros in the multiplier..6 x 10 = 6..7 x 100 = 7.. x 1000 = 0 NSF-ATE Project

5 DIVISION Division is repeated subtraction. The number to be divided by another is the Dividend. The number by which another is divided is the Divisor. The result of the division is the Quotient. Any part of the dividend left over when the quotient is not exact is the Remainder. Quotient /--- Remainder Divisor---- /--Dividend The division sign (/) means "divided by." However, a division problem may be set up in several acceptable ways. For example, / / To prove the accuracy of your division, multiply the Quotient by the Divisor and add the Remainder, (if there is one) to the result. The final product should be the same figure as your Dividend. DIVISION OF DECIMALS Proof 1 Quotient x Divisor + Remainder Dividend figure To divide a decimal by a whole number, place the decimal point in the quotient directly above the decimal point in the dividend and proceed as with whole numbers. / 1. = 1.7 To divide a decimal by a decimal, move the decimal point in the divisor to the right until it becomes a whole number. Then move the decimal point in the dividend the same number of places to the right. (Add zeros if necessary.) Then place the decimal point in the quotient directly above the decimal point in the dividend..6 / 1. =. To divide a smaller dividend by a larger divisor or to extend a quotient to a given number of decimal places, add enough zeros to the right of the dividend until the dividend contains the desired number of decimal places. /.0 =. 1 / = 1.0 Answer rounded to nearest hundredth (1.0) NSF-ATE Project

6 As a rule, carry out the division problem one more decimal place than is needed, before rounding off. If the last figure is or more, drop it and add 1 to the figure in the pre-ceding place. If the last figure is less than, just drop it entirely, (as above). To divide a decimal or a whole number by 10 or by a multiple of 10, move the decimal point as many places to the left as there are zeros in the divisor. / 10 =. 16 / 100 = / 1000 =.067 FRACTIONS A fraction is a number that is less than a whole number (1/). The number below the line is the Denominator and indicates the complete number of equal parts into which the whole quantity has now been subdivided. The number above the line is the Numerator and indicates how many of these equal parts are being considered. Numerator The line dividing the two numbers means divided by. Denominator Common fractions may appear in various forms such as proper or improper, simple or complex. Proper fractions, like 1/, 1/, / are called Proper fractions because their value is less than one. Improper fractions, like /, /, 17/ consist of a numerator larger than the denominator and are generally reduced to mixed numbers, e.g. / = 1 1/ Simple fractions, like those above consist of whole numbers in the numerator and denominator. Complex fractions have mixed numbers in the numerator and/or denominator, e.g., / 1 1 / 1/16 NSF-ATE Project

7 DECIMALS Decimals are another way of expressing Parts of a whole. Any proper fraction can be converted to its Decimal Equivalent by dividing the numerator by the denominator and then writing the quotient in decimal form Decimal Equivalent of / / = /.0000 READING DECIMALS First decimal place Tenths.7 7/10 Second decimal place Hundredths.07 7/100 Third decimal place Thousandths.007 7/1000 Fourth decimal place Ten Thousandths /10000 Fifth decimal place Hundred Thousandths / Read the number to the right of the decimal Point as a whole number and give it the name of the last decimal place.. is read as four hundred eighty-nine thousandths. PERCENTAGE Percentage and percents are used to show the relationship between two numbers in terms which provide more meaningful information. Percentage is a quantity and refers to a part of a whole quantity expressed in relation to 100. (The whole quantity is considered 100%.) Percent or % is a rate and refers to a fractional or decimal part of one hundred. For example, 0 % means 0 parts of 100 or 0/100 or.0. (Notice that 0/100 can be reduced to 1/.) In working with a problem the % is changed to a decimal. To change a Percent to a decimal, remove the Percent sign and move the decimal point two places to the left. 0% =.0 1.1% =.011 % =. To change a decimal to a percent, move the decimal point two places to the right and add a percent sign..7 = 7 % 1.1 = 110%.6 = 6.% NSF-ATE Project 6

8 RECIPROCALS The Reciprocal of a number is the quotient obtained by dividing 1 by that number. The reciprocal of is1/ or.. The reciprocal of is 1/ or.0. Reciprocals are used to simplify some arithmetic problems by substituting multiplication for division. For example, 6 / = 16. Using the reciprocal method this problem becomes 6 x 1/ or. (Reciprocal of Divisor) = 16. Reciprocal Method: Dividend x Reciprocal of Divisor = Quotient COMPLEMENTS The Complement of a number is the difference between that number and the next higher power of 10. For example, the complement of is (10 - ). The complement of is 6 (100 -). The complement of 00 is 600 ( ). Complements are used to simplify some arithmetic problems by eliminating one or more steps. A Complementary Percentage is the difference between a percent and 100 %. For example, the complementary percentage of % is 7%. PERCENTAGE BASE - The number on which the percent is to be calculated. RATE - The number of hundredths taken (Rate is always a %). PERCENTAGE - The number obtained by multiplying the Base by the Rate. THREE TYPES OF PERCENTAGE PROBLEMS (1) To find a certain percent of a number, multiply the number by the percent expressed decimally. Example: Find 1% of 0. Percentage = Rate X Base.1 X 0 = 7. NSF-ATE Project 7

9 PRACTICE PROBLEMS 1. What is %of.00?. What is 0% of $.0?. What is 0% of.00?. What is % of 117.7?. Whatis% of 1.? NSF-ATE Project

10 () To find what percent one number is of another, the amount to which the comparison is made is used as the divisor... (the number preceded by the word "of" is always the divisor). Example: 7 is what % of 100? Percentage Rate = Base 7 / 100 =. or % PRACTIC E PROBLEMS 1. 0 is what % of 00?..00 is what % of.00?. 6. is what % of 7.6?..00 is what % of 10.00?. is what % of.? NSF-ATE Project

11 Math for Welding A) Arithmetic 1) Manipulate fractions: Simplify, add, subtract, multiply, divide, and convert from proper fractions to improper fractions and vice versa: Simplify: Fractions can be simplified by dividing out the common fractions: = =, = =, = =, Add and subtract: To add or subtract find a common denominator, multiply numerator and denominator by a number to make the denominators equal to this common denominator: 7 + least common denominator is, multiply numerator and denominator of by = + = = 7 least common denominator is, multiply numerator and denominator of by = = = 7 7 least common denominator is 1, multiply numerator and denominator of by, and the numerator and the denominator of by 1 1 = = = NSF-ATE Project 10

12 7 Two pieces of steel one 1 inches long, the other 1 inches long are to be welded together. Find total length Solution: A number with a whole number and a fraction is called a mixed number. To add or subtract mixed numbers, add or subtract the whole numbers, and add or subtract the fractions: = 1 7 = + 1 = 1 Proper and improper fractions: An improper fractions is a fractions whose numerator is bigger that its denominator, like 1 or. To turn a improper fraction into a proper one divide the denominator into the numerator, the whole number from the division is written in front, the remainder is the numerator, and the denominator stays as is: 1 = 1 ( goes into 1 one times, remainder is ) 1 = 7 ( goes into seven times, remainder is 1) To turn a mixed number into an improper fraction multiply the whole number by the denominator and add the result to the numerator. This will be the numerator of the improper fraction. The denominator stays as is: = = = = NSF-ATE Project 11

13 Multiply and divide fractions: To multiply fractions multiply the numerators and multiply the denominators: 7 1 = 0 To divide fractions invert the denominator and multiply: 7 7 = = To multiply a whole number by a fraction, multiply the whole number by the numerator: 1 6 = 1 10 = 1 A ft long pieces of quarter inch steel weighs lb/ft. Find total weight. 6 6 lb/ft ft NSF-ATE Project 1

14 Solution: To multiply mixed numbers first turn them into improper fractions then multiply: 6 ft = ft/lb = Total weight ft 17 6 lb/ft 17 = ft 6 66 = lb 1 = 7 lb lb/ft A steal rod weighs pounds, and is feet long. Find the weight per foot. lb Solution: ft Weight per foot is weight divided by length. To divide mixed numbers turn them into improper fractions, then divide: lb ft = = 7 1 = 1 7 = 0 7 lb ft NSF-ATE Project 1

15 Other solved exercises: Simplify the fractions: 1 =, = 1 Add or subtract fractions and mixed numbers: =, 6 + = 11, 6 = Turn an improper fraction to a mixed number, and a mixed number to an improper fraction: = 1, = Multiply or divide whole numbers and fractions and mixed numbers: 6 =, = =, = = = = = = 6 7 = = = 1 NSF-ATE Project 1

16 Exercises: 7 1)Two pieces of steel one 1 inches long, the other together. Find the total length. 7 inches long are to be welded 7 )A 1 ft long pieces of quarter inch steel weighs lb/ft Find total weight. 7 )A steal rod weighs 7 pounds, and is 1 feet long. Find the weight per foot. )Simplify the following fractions: 7, 1, 1, 6 1, 1 6 )Add or subtract fractions and mixed numbers: , + 1 6, 1 NSF-ATE Project 1

17 6)Turn improper fractions to mixed numbers: 1, 1 16, 7, 16, 7)Turn mixed numbers to improper fractions: 7,, 16,, )Multiply or divide the following: 7,, ,, 7, 7 7, 1 7 ) Powers Find area of square, and volume of cube: side = a side = a Area of square : Solved exercises: A = a Volume of cube : V = a Find the area of a square with sides equal to in. Area of square is the square of one side: A = = in NSF-ATE Project 16

18 Find the volume of a cube with sides equal to in. Volume of cube is the cube of one side: V = = 6 in Find the area of a square with sides equal to inches: Solutions: Turn the mixed number into an improper fraction first, then square. To square fractions square the numerator and square the denominator. Turn the answer into a proper fraction. A = ( ) = ( ) = ( ) 16 1 = 16 squareinches Convert mixednumber toimproper fraction. Squarenumerator and denominator. Convert back tomixednumber. Find the area of a square with sides equal to.6 feet and the volume of a cube with sides equal to 7. inches: Solution: A =.6 V = 7. = 1.6 = 7. square feet (ft cubic inches (in ) ) NSF-ATE Project 17

19 Exercises: 1) Find the area of a square sheet metal with sides equal to 6 inches. ) Find the area of a square sheet metal with sides equal to 6 inches. ) Find the area of a square sheet metal with sides equal to 6.7 inches. ) Find the volume of a cube with sides equal to 6 inches. ) Find the volume of a cube with sides equal to 6.7 inches. B) Algebra: 1) Manipulate equations with whole numbers and fractions: The equation : C = ( F ) converts degrees Fahrenheit to degrees Centigrade. The equation : F = C + converts degrees Centigrade to degrees Fahrenheit. NSF-ATE Project 1

20 Solved Exercises: Given the temperature 0 in degrees Fahrenheit ( o F ), use the conversion equation to find the temperature in degrees Centigrade (Celsius) ( o C ) : C = ( F ) C = (0 ) C = (1 ) o C = C Fahrenheit plug in 0 for F to Centigrade simplify inside the paretheses multiply 1 by conversion equation Given the temperature 70 in degrees Centigrade (Celsius) ( o C ), use the conversion equation to find the temperature in degrees Fahrenheit ( o F ): F F F F = C + = (70) + = = 17 o F Centigrade to Fahrenheit conversion equation plug in 70 for C multiply before adding add Exercises: 1) Given the temperature 7 in degrees Fahrenheit ( o F ), use the conversion equation to find the temperature in degrees Centigrade (Celsius) ( o C ) ) Given the temperature 0 in degrees Centigrade (Celsius) ( o C ), use the conversion equation to find the temperature in degrees Fahrenheit ( o F ) ) Create tables of values for linear algebraic equations: Solved exercises: Make a table of values for Celsius (Centigrade) to Fahrenheit conversion equation F = C + when C varies from 0 o to 0 o in 10 o intervals. NSF-ATE Project 1

21 Solution: plug in 0, 0, and 0 in the above equation for C and calculate F: F = (0) + F = (0) + F = (0) + = 6 + = + = 7 + = 6 C F = 6 = 10 Make a table of values for Fahrenheit to Centigrade conversion equation for F ranging from 00 to 700 o F in 0 degree steps: C = ( F ) Solution: Plug in 00, 0, 600, 60, and 700 in the above equation for F and calculate C: C = (00 ) = (6) = 6 C = (0 ) = (1) = 7. C = (600 ) = (6) = 1.6 C = (60 ) = (61) =. F C Find the equation for total weight and make a table for weight vs. L for L varying from feet to feet in increments of 1 foot: lb. lb/ft L NSF-ATE Project 0

22 Solution: The equation for total weight is. times length plus pounds (.L + ): W =.L + lb. Make a table of W vs. L : W =.() + = 7 + = L (ft) W (lb.) W =.() + = = 11. W =.() + = 1 + = 16 Exercises: 1) Make a table of values for F = C + when C varies from 7 o to 100 o in o intervals. ) Make a table of values for C = ( F ) for F ranging from 100 to 00 o F in degree steps ) Find equation for total weight and make a table for weight vs. L for L varying from feet to feet in increments of 1 foot: lb 7. lb/ft L NSF-ATE Project 1

23 ) Electric Power: Given voltage (V) in volts, current(i) in amperes or amps, and resistance (R) in ohms, power (P) in watts is: P = I R or P = IV Solved Exercises: Find power if current is amps, and resistance is 1 ohms. Solution: P = I R write the equation P = 1 plug in the values for I and R = 1 = 00 watts square first before multiplying Find current if power is 1 kw (1 kilowatts or 1,000 watts) and voltage is 10 volts. Solution: P = IV 1000 = I = I 10 I = 100 amps write the equation plug in the values for P and V divide both sides by 10 to solve for I Find current if power is kw and resistance is 10 ohms. Solution: P = I R 000 = I 000 = I 10 I = I = I =.16 amps write the equation for power plug in values for P and R divide both sides by 10 solve for take the square root of both sides to solve for I I NSF-ATE Project

24 Exercises: 1)Find power if current is set at 10 amps, and resistance of welding rod is 1 ohms. )What should the current be set to if welding power is specified to be 1 kw (1 kilowatts or 1,000 watts) and voltage is 10 volts? )What should the current be set to if power is specified to be kw and welding rod resistance is 0 ohms? ) Heat input Given Current (I) in amperes or amps (A), voltage (V) in volts (V), and welding travel speed (S) in inches per seconds (in/s), heat input H in Joules per inch (J/in) is given by: H = I V S Solved exercises: Calculate the heat input in kj/in and convert to metric (kj/mm) when automatic GMAW is being performed using the following settings: Current (I) = 00 A Voltage (V) = 0 V Travel Speed (S) = in / min NSF-ATE Project

25 Solution: Convert speed to inches per second (in/s): in 1 min Speed = min 60 sec in = 60 s in = 1 s Plug values into the heat input equation: I V Heat input = S 00A 0V = in 1 sec 00A 0V 1s = in = 6700 J/in = 67. kj/in Convert to metric units: Heat input kj 1 = 67. in. kj =.7 mm in mm NSF-ATE Project

26 Exercises: 1)Calculate the heat input in kj/in and convert to metric (kj/mm) when automatic GMAW is being performed using the following settings: Current (I) = 0 A Voltage (V) = V Travel Speed (S) = 1 in / min )Calculate the heat input in kj/in and convert to metric (kj/mm) when automatic GMAW is being performed using the following settings: Current (I) = 0 A Voltage (V) = V Travel Speed (S) = 6 in / min C) Geometry 1) Two dimensional forms: Rectangle: b a Area = a b, Circumference = a + b Triangle: b h c Area =(1/)a h, Circumference = a + b + c a NSF-ATE Project

27 Circle: Radius = r, Diameter = D = r Area = πr, Circumference = πr = πd Equilateral, isosceles, and right triangles. Equilateral Isosceles Right triangle All sides are equal Two sides are equal One angle is 0 degrees Solved Exercises: Find circumference of a triangle with sides equal to, 7., and 10 inches. Solution: C = a + b + c C = = 0. in. Find the area of a triangle with base equal to ft and height equal to. ft. Solution: 1 A = ( )()(.) A =. sq. ft. (ft ) NSF-ATE Project 6

28 Find the area of a circle with radius equal to 1. ft: Solution: A = π r A = π (1.) A =.1. A = 7.07 sq. ft. (ft ) Find the circumference of a circle with radius equal to 1. ft.: C = π r C = π 1. C =.1 1. C =. ft. Find the radius of a circle with area equal to 1 ft Solution: A = π r 1 = π r 1 = r π r =. r =. r = 1. ft equation of the area of a circle plug in1 ft for area to solve for r divide both sides by π 1 use calculator to find π take the square root of both sides to find r use calculator to find square root of. A NSF-ATE Project 7

29 Exercises: 1) Find circumference of a triangle with sides equal to., 7., and 1.6 cm. ) Find the area of a triangle with base equal to. ft and height equal to 1.7 ft. ) Find the area of a circle with radius equal to. ft. ) Find the circumference of a circle with radius equal to. ft ) Find the radius of a circle with area equal to 1 ft NSF-ATE Project

30 ) Pythagorean Theorem: a c b c = a + b c is called hypotenuse, a and b are called legs or sides. Solved exercises: Given a =. in. and b = 6.7 in., calculate c. Solution: c c = a + b = = = 67.1 c = 67.1 =.1 in. Pythagorean equation plug in numbers for a and b square numbers take the square root of both sides to solve for c If the hypotenuse of a right triangle is ft, and one leg is feet, find the other leg. Solution: c = a = 6 = + b 6 = b 0 = b + b + b b = 0 b = 6. ft. Pythagorean equation plug in numbers for c and a square and subtract from both sides to solve for b take the square root of both sides use calculator to find 0 NSF-ATE Project

31 A steel bar (C) is to be welded to the shelf shown to strengthen it. Find the length of this bar. A=1 in. B= in. C Solution: C C C = A = 1 + B + = C = 0 C = 0 C = 1. in. Pythagorean equation plug in numbers for sides A and B square 1 and add take the square root of both sides to solve for C use calculator to find 0 Exercises: 1) Given the two sides of a right triangle equal to.6 in. and. in., calculate the hypotenuse. ) If the hypotenuse of a right triangle is 10 ft, and one leg is 7 feet, find the other leg. NSF-ATE Project 0

32 ) Find the length of the steel bar (C) to be welded to the shelf shown. 1 in 1 in C ) Three dimensional forms: Volume of three-dimensional forms: Solid Rectangle Solid parallelogram Cylinder Solid Trapezoid Solid triangle Volume of solid rectangle = length width height Volume of solid rectangle = area of base height Volume of solid triangle = area of cross section length Volume of cylinder = π r height Volume of cylinder = area of base height NSF-ATE Project 1

33 Solved exercises: Find volume of the cylinder shown: h = 1 in Diameter = in Solution: r = D / = 1. V = Area of base height = ( π r = (.1 1. ) 1 = 106 ) h in radius is1/ of diameter equation of volume of cylinder plug in numbers and use calculator to find volume Find the weight of the above cylinder if the weight density is 0. lb/in. Solution: Weight = volume weight density = 106 in 0. lb/in = 7.1lb. Find weld weight of the following welded groove. The groove width is 1. inch (in.), the groove height is 1. inch (in.) and the length of the groove is feet (ft). The weight density of the welding metal is 0. pound per cubic inch (lb/in ). 1. NSF-ATE Project 1. in. ft.

34 Solutions: Volume of grove = area of cross section length of groove Weight = volume weight density = area of cross section length weight density 1 in. lb = 1. in 1. in ft 1 0. ft. in = 6.11 lb. Exercises: 1) Find volume of a cylinder with radius equal to in. and height equal to. feet. ) Find the weight of the above cylinder if the weight density is 0. lb/in. ) Find the weight of a weld in a groove with width of 0.7 in., height of 1.0 in., and weld weight density of 0. lb/in. NSF-ATE Project

35 D) Right Triangle Trigonometry 1) Given sides, find cosine and sine of angles Angles in triangles are written using upper-case letters like A, B, or C and their opposite sides are written by lower-case letters like a, b, or c. c hypotenuse a Given angle b Opposite side Adjacent The cosine of the angle A, cos(a ), and the sine of the angle A, sin(a ), are given by: Adjacent cos( A ) =, sin( A) = Hypotenuse Opposite Hypotenuse Solved exercises: a=1 in c angle A b=1 in Find sine and cosine of the angle A shown: NSF-ATE Project

36 Solution: Find c (the hypotenuse) first: a 1 + b + 1 = c c = 0 c = 0 c = 1. = c Pythagorean Theorem plug in numbers for the two sides a and b square 1 and 1 and add and change sides take the square root of both sides t solve for c use calculator to find 0 Now find the cosine and the sine of angle A: cos( A) = sin( A) = = = Exercises: Adjacent Hypotenuse Opposite Hypotenuse 1 = 1. = 0.6 angle B a= in c angle A b=11 in 1) For the above triangle find cos(a), sin(a). ) For the above triangle find cos(b), sin(b). NSF-ATE Project

37 ) Given the hypotenuse and angles, find the sides Given the hypotenuse of a right triangle and one of the angles, we can find the sides by the following formulas: Adjacent = Hypotenuse cos(angle) Opposite = Hypotenuse sin(angle) a = L sin A L A b = L cos A a =L cos B B L b = L sin B Solved Exercises: Given the following right triangle, find all the side and angles: a B c = A=6 o b NSF-ATE Project 6

38 Solution: b = c cos(6 o ) = cos (6 o ) = 0.0 =. in. a = c sin(6 o ) = sin(6 o ) = 0. =. in. B = 0 o 6 o = o Given the following right triangle, find all the side and angles: b B c A=6 o a= Solution: a = c cos(6 ) = c 0.0 c 0.0 = c = 0. adjacent = hypotenuse cos (A) plug in " for a, use calculator divide both sides by 0.0 to solve for to find cos(6) c b = c sin(6 ) b = b = 1.17 opposite = hypotenuse sin(a) plug in 0. for c, use calculator to find sin(6) NSF-ATE Project 7

39 The shelf shown is to be constructed with the steel bar c equal to 16 inches, and the angle B equal to 0. Find the lengths a and b. a b B c=16 in. Solution: a = c cos( B) = 16 cos(0 ) = = 1. in. b = c sin( B) = 16 sin(0 ) = = in. adjacent = hypotenuse cos(b) plug in16 for c and 0 for B use calculator to find cos(0) opposite = hypotenuse sin(b) plug in16 for c and 0 for B use calculator to find sin(0) Exercises: a B c = b B c A= 0 o A= o b a=1 Fig. 1 Fig. NSF-ATE Project

40 1) Given the right triangle in Fig. 1, find sides a and b and angle B: ) Given the right triangle in Fig, find sides b and c and angle B: ) The shelf shown is to be constructed with the steel bar c equal to 0 inches, and the angle B equal to. Find the lengths a and b. a b B c=0 in. NSF-ATE Project

41 E) Units and Unit Conversions 1) Units Property Unit Symbol length centimeter, meter, inch, foot cm, m, in.,ft area square centimeter, square cm,m, in,ft meter, square inch, square foot volume cubic centimeter, cubic cm, m, in,ft, L meter, cubic inch, cubic foot, cubic liter mass gram, kilogram, pound g, kg, lb temperature degree centigrade degree Fahrenheit o C, F time second, minute, hour sec, min, hr pressure pascal (Newtons/sq. meter) pounds/square inch Pa (N/m ) psi speed meter/second, foot/second m/s, ft/s electric resistance ohm Ω electric current ampere (amp) A electric voltage volt V power watt (Joule/secon) W (J/s) energy Joule J frequency Hertz (cycles/s) Hz Heat input Joule/inch, kilo Joules/inch, Joule/millimeter J/in, kj/in, J/mm deposition rate kilogram/hour kg/h electrode force Newton N wire feed speed millimeters/second mm/s flow rate liter/minute L/m NSF-ATE Project 0

42 )Unit Conversions: There are: 1 in/ft 1 in /ft. cm/in 6. cm /in. mm/in 100 cm/m 10,000 cm /m 1000 mm/m. ft/m 0.0 m/ft 6 Pa/psi 1000 g/kg. lb/kg 1000 J/kJ 1000 kw/w 1000 Hz/kHz Solved Exercises: The length of a piece of steel is feet. Find the length in meters and centimeters. Solution: 0.0 m ft = ft ft =. m 100 cm. m =. m m = cm A piece of sheet metal is 16 square feet. Convert the area to square inches: Solution: in 16 ft = 16 ft 1 ft =,6 in Heat input is 6700 J/in. Convert to kj/in and metric units: Solution: Heat input = 6700 J/in J 1kJ = 6700 in 1000 J = 67. kj/in NSF-ATE Project 1

43 Convert to Metric Units : kj Heat Input = 67. in kj =.7 mm 1. in mm 1 in multiply by,. mm so inches cancel Convert 70 psi to Pascals: Solution: 6 Pa 70 psi = 70 psi psi =,60 Pa 6 Pa multiply by psi, so psi's cancel Welding speed is specified to be inches per minute. Convert to centimeters per seconds: in cm 1 min. = min in 60 sec cm 1.0 sec cm (multiply by. in and 1 60 min, sec so inches and minutes cancel) Exercises: 1)The length of a piece of steel is 1 feet. Find the length in inches, meters and centimeters. )A piece of sheet metal is 0 square feet. Convert the area to square inches. )Heat input is,000 J/in. Convert to kj/in and metric units. ) Welding speed is inches per minute. Convert to cm/sec. NSF-ATE Project

44 F) Reading Graphs: To read graphs, locate the value on the horizontal axis, draw a vertical line (up or down) to meet the graph, then draw a horizontal line (left or right) to meat the vertical axis. Read the number on the vertical axis. Solved Exercises: 1) Shear strength in ksi is given as a function of joint thickness in inches for pure silver joints for 0. inch diameter steel drill rod by the following graph. Find shear strength when joint thickness is and 0.00 inches. Solution: Vertical lines at joint thickness of and 0.00 are drawn up to meet the graph and horizontal lines are drawn from those points to the left to meet the vertical axis. Shear strength values are read as respectively 6 and 1 ksi. Shear Strength vs. Joint Thickness Shear Strength (ksi) Joint Thickness (in.) ) The following figure gives deposition rate in pounds per hour (lb/h) versus arc energy in kilowatts (kw) for gas tungsten arc welding with cold and hot steel filler wires. Find deposition rates for cold and hot filler wires when arc energy is kw. Solution: Vertical lines at arc energy of kw are drawn up to meet the graphs and horizontal lines are drawn from those points to the left to meet the vertical axis. Deposition rate values are read as respectively and 16 lb/h. NSF-ATE Project

45 Deposition Rate for GTAW Deposition Rate, lb/h Cold W ire Hot W ire Arc Energy, kw Exercises: 1) The following graph shows tensile shear strength in 1000 s of lbs. vs. weld nugget diameter in inches for three different sheet metal thickness. a)find tensile shear strength for a 0.07 in. thick sheet steel when weld nugget diameter is 0. in. b)find tensile shear strength for a 0.1 in. thick sheet steel when weld nugget diameter is 0. in. Tensile Shear Strength X1000 LBS Tensile Shear Strength vs. Nugget Size Steel thickness =.0 in. Steel thickness =0.07 in. Steel thickness = 0.1 in. Nugget Size (in.) NSF-ATE Project

46 ) Voltage versus current for aluminum tungsten arc welding with gas shielding is given in the graphs below. a) Find voltage when current is set at 0 Amps with argon gas shielding. b) Find voltage when current is set at 100 Amps with helium gas shielding. Voltage vs Current with Gas Shielding 0 Arc Voltage (V) Argon Helium Arc Current (A) NSF-ATE Project

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48 NSF-ATE Project 7

49 NSF-ATE Project

50 NSF-ATE Project