ME 108 Statics Ahmet Erkliğ erklig@gantep.edu.tr Office No: 315 4th floor of Mechanical Engineering Departmen
What is Engineering Mechanics? Engineering Mechanics is a discipline that studies the response of solids, structures, fluids and devices made of these materials to mechanical, thermal, and other types of load events. It emphasizes a basic science approach that draws heavily on physics, mathematics and computer science. Mechanics is crucial to addressing modern problems in: aerospace engineering mechanical engineering civil engineering nuclear engineering materials science geological engineering agricultural engineering biomedical engineering electrical engineering (electronics packaging) chemical engineering (polymeric material modeling)
Particles vs Objects Particles: Point mass No geometry Rotation is not important Objects Contain mass Have geometry Rotation is important
What is Statics? Statics is the study of particle and rigid body structure equilibrium. Static means unchanging position in time. The basic principle we will use is Newton s 2nd law of motion F ma We will spend considerable time becoming proficient in applying this law to engineering problems.
The Basic Concept If an object is at rest (or moving with a constant velocity), we can say quite a lot about the forces acting on it and within it!
Importance of Statics
What You Will Learn in Statics? How to determine the resultant of multiple forces and distributed forces. How to determine the moment of forces about points and axes. How to draw free body diagrams and identify all of the forces acting on a body. How to use equilibrium equations to determine unknown or unspecified forces and moments acting on (and within) bodies, including structures and machines. How to identify and quantify friction forces. How to determine centroids, centers of mass/gravity, and moments of inertia for shapes and objects.
And what will you do with these competencies??? Use information about forces and moments to analyze and design parts, assemblies, mechanisms, and structures that are appropriate to their intended (and sometimes unintended) functions. Use information about centroids, centers of mass/gravity, and moments of inertia in support of the above.
Why is Statics Important? ME 108 is the fundamental building block for all of engineering courses. Good habits developed in Statics will help you in subsequent courses. Performance in Statics is factored into your Math/Science GPA for admission to a department
ME 108 Statics ME 223 mech of materials 1 ME 208 dynamics ME 223, 307-308 advanced dynamics robotics satellite mechanics vibrations others advanced mech of materials stress analysis structural engineering machine design finite element analysis others
Tips for Success in Statics (i.e., what is expected!) Do all assigned reading before class. Solve extra problems Don t spend more than 20 minutes/problem. Get help! Attend lecture regularly. Ask questions!!!
Courtesy During Lectures Arrive for lecture on time. No sleeping, reading newspapers, conversations, etc. during lecture. I expect your attention to be focused on statics for 50 minutes. Please wait until class is dismissed before leaving.
Text Book Text Book: Engineering Mechanics Vol-1;Statics By J. L. Meriam and L. G. Kraige Auxiliary Book: Vector Mechanics for Engineers; Statics By Ferdinand P. Beer and E. R. Johnston Engineering Mechanics Statics By R.C. Hibbeler Many books related with the Statics in the Library
Grading 2 midterm 2 x 20% Homeworks+Quiz 20% final 40% 100% Exams 1st midterm 26-07-2011 2nd midterm 09-08-2011 final??-08-2011
Course Overview Basic principles Force systems Equilibrium Structural analysis and machines Centroids, distributed load systems Beams Area and mass moments of inertia
Fundamentals, units, calculations Units Length need to know position and geometry of objects Time need to determine succession of events Mass related to amount of stuff in a body, found using gravitational attraction Weight force due to gravity acting on a mass, W=mg, where g=9.8m/s2
Fundamentals, units, calculations Basic Quantities Force push or pull on a body, can be direct (contact) or indirect (no contact) Moment turning effect caused by a force applied at some distance away from the axis of rotation
Fundamentals, units, calculations Engineering Concepts Idealizations all real problems are simplified to some degree Particle mass acting is if it were concentrated at a singe point Rigid Body particle collection in a shape that doesn t change with applied force Concentrated Force force acting as if it were at a single point
Newton s laws of motion 1st law: A particle remains at rest, or continues to move in a straight line with uniform velocity, if there is no unbalanced force acting on it. 2nd law: The acceleration of a particle is proportional to the resultant force acting on the particle, and is in the direction of this force. F ma F ma 0 for staticcase 3rd law: The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and colinear.
Newton s law of gravitational attraction Consider two particles of mass m 1 and m 2. Newton stated: F G m 1m 2 r 2 F = force of attraction between the two particles. G = universal gravitational constant, experimentally found to be = 66.73 (10-12 ) m 3 /(kg-s 2 ). r = distance between particles.
weight of a particle Consider a particle with mass m 1 =m, and let the earth s mass be denoted by m 2. Letting W=F where we call W the weight, we can write W=mg where g = acceleration due to gravity = Gm 2 /r 2 Observation: We commonly think of g as a constant. However, in our application of F=Gm 1 m 2 /r 2, the earth is not a particle, it has nonuniform density, and r depends on position. Nonetheless, commonly accepted values for g are: g = 32.2 ft/s 2 = 386.4 in/s 2 = 9.81 m/s 2
Units used in Engineering Statics SI Units Dimension Unit SI Symbol Length (l) meter m Mass (m) kilogram kg Time (t) second s Derived Unit Basic Units Symbol Area (A) m2 - Volume (V) m3 - Density (ρ) kg/m3 - Velocity (v) m/s - Acceleration (a) m/s2 - Force (F) kgm/s2 N Energy (E, U, K, etc.) kgm2/s2 J Be consistent when using units!
Units System length mass force time English foot, ft slug==lb-sec 2 /ft pound, lb second, s metric (SI) meter, m kilogram, kg Newton==kg-m/s 2, N second, s Remark: the relationship among these units is governed by f=ma. By selecting 3 fundamental units, a 4th derived unit can be obtained. The derived units are: English system: 1 slug f a 1 lb 1 ft/s 2 or...1 lb (1 slug)(1 ft/s 2 ) Suggestion: Avoid slugs and simply work with fundamental units. metric (SI): 1 Newton ma 1 kg 1 m/s 2
Unit Conversions If p=expression, to change the units of p, multiply the r.h.s. by dimensionless factors of unity until the desired units are obtained. e.g., Convert the velocity v = 10 m/s to English units. Since 1 ft 0.3048 m, 1 ft 0.3048 m 0.3048 m 0.3048 m 1 v 10 m s 1 10 m s 1 ft 0.3048 m 1 32.808 ft/s
Unit Conversions Length Mass 1 km = 0.62137 mi. 1 kg = 2.2 lbs 1 mile = 5,280 ft = 1.6093 km 1 lb = 453.59 g = 16 oz. 1 meter = 1.0936 yds 1 metric ton = 1000kg 1 inch = 2.54 cm (exact) 1 slug = 32 lb = 14.594 kg Volume Pressure 1 ml = 1 cm 3 1 atm = 760 torr (mm Hg) 1 L = 1 dm 3 = 1.0567 qts. 1 atm = 14.70 lb/in 2 1 gal = 4 qts = 8 pts = 3.7854 L 1 bar = 100,000 Pa 1 in 3 = 16.4 cm 3 1atm = 101325 Pa 1 qt = 0.946 L Force Energy 1N = 0.2248lb f 1 J = 1 kg-m 2 /s 2 1N = 2.248x10-4 kip 1 cal. = 4.184 J 1 kip = 1000lb f 1N = 7.233 poundal (lb m ft/s 2 )
Prefixes Useful to reduce numbers to manageable size, especially SI units. factor prefix symbol 10 9 giga G 10 6 mega M 10 3 kilo k 10-3 milli m 10-6 micro 10-9 nano n
Numerical calculations Dimensional Homogeneity: In addition to being numerically correct, an equation must also be dimensionally correct. Always carry units along with your calculations. Accuracy: You cannot create accuracy with your calculations. e.g., 12.34 * 12.3 = 151.782 In lengthy calculations, retain one or two extra significant digits*. Your final results should be reported with the same accuracy as your data (in this case... 152.) * Occasional exceptions to this.
example 1 convert 175 lb/ft 3 to SI units: convert 6 ft/hr to SI units: convert 1.13 kn-m to English units: A: a) 27.5 kn/m 3 b) 0.508 mm/s c) 833 ft-lb
example 2 Two steel spheres with 100 mm diameter are placed beside one another at the earth s surface. Compute the gravitational force W (weight) and the force of mutual attraction F in Newtons. The unit weight of steel is = 0.284 lb/in 3 A: W = 40.3 N, F = 1.13 (10-7 ) N
Problem Solving Strategy 1. Interpret: Read carefully and determine what is given and what is to be found/ delivered. Ask, if not clear. If necessary, make assumptions and indicate them. 2. Plan: Think about major steps (or a road map) that you will take to solve a given problem. Think of alternative/creative solutions and choose the best one. 3. Execute: Carry out your steps. Use appropriate diagrams and equations. Estimate your answers. Avoid simple calculation mistakes. Reflect on / revise your work.
Definitions: FORCE SYSTEMS - Force A scalar is a quantity that is completely characterized by a single number. Mass, volume, and length are examples of scalars. A vector is a quantity that has both magnitude (or size) and direction. Velocity, force (e.g., weight) and position are examples of vectors. Common notation for vector quantities: F or F
Vector Notation To report a vector, its magnitude and direction need to be stated. We will discuss a few convenient methods for doing this including: Simple statement of magnitude and direction relative to some arbitrary reference direction (effective for 2-D). e.g., Statement in terms of rectangular Cartesian components (effective in both 2-D and 3-D... more on this shortly)
Vector Nomenclature: Head or Tip Tail 30 Line of Action
Basic Vector Operations Addition: There is more than one way of adding vectors: parallelogram addition head-to-tail: note that this can be done in any order Multiplication: multiplication of a vector by a scalar changes its magnitude but NOT its direction. Subtraction: equivalent to multiplying the vector to be subtracted by -1 and adding it to the vector from which it is being subtracted.
Summary: Scalar multiplication: s(a + B) = sa + sb, s is a scalar Vector addition is commutative: A + B = B + A A A R = A+B B B A B R = B+A R = A+B B A
Resolution of a vector into components a Imagine two directions, a and b in 2-dimensional space. Provided a and b are not parallel, any vector can be resolved into components in the a and b directions. a R b A R B b R = A + B Vectors A and B are often called the projections of R in the a and b directions, respectively. more...
more on resolution of vectors... When the a and b directions are orthogonal (i.e., a and b intersect at a right angle) it is usually straightforward to determine the vector components. When the a and b directions are non-orthogonal, it is usually more work to determine the vector components. The law of sines and law of cosines may be useful. a A A a R B R B b b more...
A c B b a C law of sines A sin a B sin b C sin c law of cosines C = A2 B 2 2AB cosc
Rectangular Components: A force F is said to have been resolved into two rectangular components if its components are directed along the coordinate axes. Introducing the unit vectors i and j along the x and y axes, y F x = F cos q F = F x i + F y j F y = F sin q j F y = F y j i q F F x = F x i x tan q = F = F y F x 2 2 F x + F y
When three or more coplanar forces act on a particle, the rectangular components of their resultant R can be obtained by adding algebraically the corresponding components of the given forces. R x = S R x R y = S R y The magnitude and direction of R can be determined from tan q = R y 2 2 R R = R x + R y x
Addition of Several Vectors Step 1: resolve each force into its components Step 2: add all the x components together and add all the y components together. These two totals become the resultant vector. Step 3: find the magnitude and angle of the resultant vector.
Illustration
Magnitude and Direction
Example 1 Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force R. P = 500 N T = 200 N P T B 5m A C 75 o D 3m
P = 500 T = 200 q q R tan BD AD 3 5sin 75 5cos 75 48.4
Law of cosines: c 2 a 2 b 2 2abcos( c) R 2 200 2 500 2 2(200)(500) cos(48.4 ) R 396.5 N 200 sinq Law of sines: 396.5 sin 48.4 q 22.2 R 396. 5 N 22.2
Example 2 A barge is pulled by 2 tugboats. The resultant of the forces exerted by the tugboats is a 5000 N force directed along the center axis of the barge. Determine tension in each rope if α =45 degrees Barge 30 α A B
T sin 45 T 1 2 sin30 5000 sin105 T T 1 2 3660 N 2590 N
Example 3 Given: Three concurrent forces acting on a bracket. Find: The magnitude and angle of the resultant force.
Solution
Solution Summing up all the i and j components respectively, we get,
Example 4 A ring supports three forces. What is the resultant force applied to the ring? A: R = 998 N @ 134.9 more...
Hand Solution
This may be simplified to: more...
Add force vectors to get the resultant. Because vector addition is commutative, the order of addition is not important.
Example 5 A sail boat tacks into the wind such that the force perpendicular to the sail is 110 N. Resolve this force into two components, one parallel to the keel and one perpendicular to the keel. 25 F = 110 N Wind A: F = 46.5 N F = 99.7 N
Hand Solution