2. THE BASIC FUNDAMENTALS OF STATICS The physical laws used in this study that govern the action and reaction of forces on a body include Sir Isaac Newton’s first and third Laws; First Law – A body at rest will remain at rest and any body in motion will move uniformly in a straight line unless acted upon by a force – which is a condition in structures called EQUILIBRIUM. Third Law – For every force of action, there is a reaction that is equal in magnitude, opposite in direction, and has the same line of action – which is the basic concept of FORCE.

3. Forces acting on a body generally cause two effects, and not necessarily simultaneously: a) It will cause the body to move if it is at rest or change the motion of the body if it is already in motion. b) It will cause a deformation of the body if the body is restrained. a horizontal beam will bend a member in compression will shorten a member in tension will stretch a member in shear will distort

4. A force is a QUANTITY, caused by an object that has mass, or weight – or a force is caused due to a RESISTANCE to a mass or weight. Forces in mathematics are represented by an illustration called a VECTOR, which is simply a graphic representation, a line, that has MAGNITUDE, or value, and a DIRECTION. Use of vectors is a simple way to mathematically represent forces in order to determine the solution to a structural situation. Vectors can be used to represent distance, mass, or area, but their elementary use in structures is to represent forces.

5. CHARACTERIZATION of a vector is evidenced by: 1 MAGNITUDE – the quantity of a force, a numerical measure of its intensity, usually in units of pounds or Newtons. 2 DIRECTION – the line of action of a force, such as gravity pulling downward due to weight, or resistance of a restraint pushing that weight upward to negate actual movement and remain static. 3 POINT OF APPLICATION – a location that generally describes the origin of a force.

6.  A weight supported from a ceiling by a string is represented by a vector, the line with the arrowhead that defines direction  A weight represented by a vector, supported by a post, using a vector to indicate resistance, the direction of each defined by the arrowheads.

7.  The magnitude of the force shown as 100 pounds, and the resistance shown as 100 pounds, both directions defined by the arrowheads. Point of application is the small circle. When the forces are equal and opposite, the object is in EQUILIBRIUM.

8. In analysis of structural elements, it is necessary to add the effects of two or more forces to determine the net effect as though ONLY ONE FORCE produced the same result. This one force is called the RESULTANT of a series of forces. Parallelogram Law: If two forces act concurrently at a point, the resultant force can be represented by the diagonal of the parallelogram formed by the sides, parallel and proportional to the two forces.

9.  Illustration: If an object is under the influence of two forces as shown that tend to drag it along, and the magnitude of the two forces are illustrated by the length of their lines, then; A Which direction will the object be moved? B What is the magnitude of a single force that will move the object.

10.  Consider two forces 1 & 2, each represented by a straight line vector that indicates their magnitudes by the lengths, and directions by the arrowheads.  The two forces, acting together would have an influence on the object.  By the graphic parallelogram method, the RESULTANT, of the two forces can be constructed as shown, its value indicated by the graphic length.  The RESULTANT has the same effect on the object as do forces 1 & 2.

11. Obviously, if the numerical value of the vectors is known, and the directions defined by angles with some reference, the magnitude of the Resultant can be found using trigonometry – by the solution of the sides of a triangle. It can also be found graphically if the diagram is drawn accurately to scale. For instance this illustration was done by a cadd drawing, and the system is accurate enough that the lengths of the lines can simply be measured. Doing this on the drawing, F-1 measures 13.31 units at an angle of 48 degrees, and F-2 measures 8.57 units at an angle of 115 degrees. The Resultant measures 18.45 units at an angle of 78 degrees. The units can be in inches, feet, pounds, Newtons, or any other system of units, and the relationship between the forces and the resultant will remain the same. If the units are in pounds, realize that the single 18.45 pound force has the same influence on the object as do the two forces of 13.31 and 8.57 pound forces.

12.  On a cadd system, the angles are measured from the right side X axis of the Cartesian Coordinates, in a counter – clockwise direction.

13.  In the study of statics, the relationship of forces will be significant in determining structural qualities required to resist forces that occur in the built environment.  The X and Y axes of the Cartesian Coordinates system will become the reference for forces that have directions upward, downward, or at any angle.  Forces that are not parallel to either the X and Y axes, are the RESULTANT of component forces that are parallel and perpendicular to those axes. Structural elements, whether rigid or flexible, have volume and mass, and are subject to the forces of gravity.

14. Most all structural elements are necessary, primarily because they are affected by the forces of gravity, and must exist in order to restrain themselves, and possibly some other structural element against such affects of gravity in order to remain at rest – to maintain equilibrium. A body in equilibrium acted upon by another body reacts with an equal magnitude and opposite action, called a reaction.

15. During equilibrium, a person sitting at rest has weight due to gravity, which pushes down on the chair in which that person is at rest. The chair pushes upward an equal and opposite amount. The person and the chair have weight, which pushes down on the floor, which in turn pushes upward the same amount of weight of the chair and the person. So it is in structure; Roofing pushes down on roof deck, which pushes down on joists, which push down on beams, which push down on columns, which push down on foundations, which push down on earth - - - each because of gravity, and each successive element pushes upward the same amount of the sum of the weights above.

16. We think in terms of these items pushing downward, while actually, those items of structure don’t DO anything. They simply have weight that reacts to the force of gravity... and gravity has a natural tendency to pull them downward toward the earth. So man, places these items in opposition to gravity in order to create those things we call enclosed space, which we as architects refer to as The Built Environment And those elements are successful in maintaining that space according to the integrity of their strength.

17. OBJECTS IN EQUILIBRIUM

18. In order to simplify the graphical analysis of structural members, it may be necessary to illustrate the subject as a FREE – BODY DIAGRAM which means that any body can be illustrated as an object suspended in space as long as the forces that act to hold it in equilibrium are shown. And further, any object can be divided into parts, whether separated at connections, or whether members are shown to be cut at some point or plane - - - and each part can be illustrated as an object suspended in space if the forces that hold it in equilibrium are shown. In the analysis of structure, after all acting forces are calculated, a Free-Body Diagram simplifies the calculation of the resisting forces that hold it in equilibrium.

19. A free body diagram is a graphic of an object under the influence of forces with all restraints removed and replaced by forces that hold the body in equilibrium. Free body diagrams are useful in isolating a structural member for analysis.

20. Consider this free body diagram of a beam with a load located directly at its center. It is easy to realize that the sum of the upward forces at A and B must equal the 50 lbs. And it also may be easy to realize, that if the load on the beam is at the center, that A and B are the same, 25 lbs. each.

21. If the diagram is inverted, nothing changes in the amount of the loads, and it is easy to see that A and B must be equal if the beam is to remain balanced at its center. Imagine the beam is 10’ long so the distance from A to the 50 lb load is 5’, and is the same for B. The point at the 50 lb load is a fulcrum, or a point where the beam could rotate. The force at A tends to rotate downward, or counterclockwise about the 50 lb load. Force B tends to rotate downward, or clockwise about the 50 lb load. In other words A and B oppose each other in opposite directions of rotation, and must be equal if the distances are the same.

22. Forces that tend to rotate about a certain point create a MOMENT with respect to that center point, equal in magnitude to the force times the perpendicular distance from the center point to the line of action of the force. MOMENT equals Force times perpendicular distance.

23. Consider that the action of tightening or loosening a bolt with a wrench, the force applied at the handle of the wrench creates a MOMENT about the pivot point – the head of the bolt – and it is this MOMENT that acts to turn the bolt to make it tight or loose.

24. FORCE DISTANCE PERPENDICULAR TO LINE OF ACTION OF FORCE pounds feet In this case, the amount of moment created by the FORCE is with respect to the PIVOT CENTER of the bolt. The FORCE has a TENDENCY to rotate about the PIVOT CENTER. Moment equals FORCE times the perpendicular DISTANCE of the line of action of the force to the pivot center Pivot center

25. Forces that tend to rotate about a certain point create a MOMENT with respect to that center point, equal in magnitude to the force times the perpendicular distance from the center point to the line of action of the force. MOMENT equals Force x perpendicular distance. So, for the beam as loaded, at its center, the force at A and at B must be 25 pounds each. The force at A creates 25lbs x 5ft = 1255 ft-lbs of moment with respect to the point of the 50 lb load, which is equal and opposite the moment created by the load at B - and since the forces create equal and opposite moments, they have a tendency to BEND the beam.

26. For any body to remain in equilibrium, two things must be realized in this demonstration of forces on a free body diagram; ONE, vertical forces in the upward direction must equal vertical forces in the downward direction, or The sum of all vertical forces must = zero; If vertical forces are represented by “y”, then Summation of F y = 0 Moment caused by Forces that tend to cause clockwise rotation about any point must be equal and opposite to moment caused by Forces that tend to cause counter- clockwise rotation about that same point, or The sum of ALL moments with respect to the same point must = zero Summation of M 0 must equal 0

27. The same is true that for horizontal forces to the right must be equal to the horizontal forces to the left, or if horizontal forces are represented by “x”, then The sum of F x = 0 Mathematically, it is convenient to adopt a sign convention for forces and moment, and the standard is set thus: Vertical forces upward are + (positive) Vertical forces downward are - (negative) Moments counter-clockwise are + (positive) Moments clockwise are - (negative) Horizontal forces to the right are + (positive) Horizontal forces to the left are - (negative)

28. To demonstrate the sum of moments, assume that Both A and B equal 25 lbs. Then pick a point of rotation, a point that would represent a pivot if all the forces had a tendency to rotate about that point - say at A, and sum the moments created about that point by the force at B and the 50 lb force. Realize that the force of 25 lbs at A will not create any moment about A, since the distance equals zero. Use A as a point of rotation: - (50 x 5) + (25 x 10) = 0 - 250 + 250 = 0, so the amount of moment created by the two forces cancel to zero.

29. Next, pick B as the point rotation, and sum the moments created by the 50 lb force and the force of 25 lb at A: Realize also that the force of 25 lbs at B will not create any moment about B, since the distance equals zero. Use B as a point of rotation: + (50 x 5) - (25 x 10) = 0 + 250 - 250 = 0, so the amount of moment created by the two forces cancel to zero. Note the results are identical, except the signs are opposite.

30. Next, arbitrarily pick a point that is 3’ to the left of B, that is 3’ to the left of B, and call it point “C” Sum the moments of all the forces about point C... * (A) (B) - 25 x 7 + 50 x 2 + 25 x 3 = 0 - 175 + 100 + 75 = 0 so, - 175 + 175 = 0 and no matter where rotational points are selected on the beam, the MOMENTS will sum to 0. And that is because the beam, loaded, with reactions as shown remains in equilibrium. C

31. But what if the load on a beam is not at its center? How can the reactions at A and B be found? A and B be found? When there are two unknowns, such as the reactions at A and B, pick a point of one of the unknowns and sum the moments about that point, so there will be only one unknown. Choose the point at A: write the equation, - (50 x 4) + (By x 10) = 0, and – 200 + 10By = 0, then solving for By ; 10 By = 200, and By = 20 lbs And since the sum of the forces in the vertical direction must equal zero, then Ay = 50 – 20 = 30 lbs

32. And to demonstrate that the 30 lb load at A is accurate, Sum the moments about Point B: There is no moment created at B by the load at B, since the distance equals zero. Choose the point at B, and solve for the unknown at A: write the equation, + (50 x 6) - (A y x 10) = 0, and + 300 – 10A y = 0, then solving for A y ; 10 A y = 300, and A y = 30 lbs Consequently, since the sum of the forces in the vertical direction must equal zero, then B y = 50 – 30 = 20 lbs

33. NOW CONSIDER THIS EXERCISE For the beam loaded as shown, find the value of the reactions at A and B

34. The process of finding the reactions to a beam is the same for one with multiple loads as in the previous examples. Consider a beam loaded as shown and calculate the reactions. Since there are two unknowns, sum the moments created by the 3 loads shown about point A ;

35. The force at A will not create a moment at A because the dimension equals zero... - (200 x 5) – (300 x 10) – (400 x 18) + 24B y = 0 - 1000 – 3000 – 7200 + 24B y = 0 24By = 7200 + 3000 + 1000 24By = 7200 + 3000 + 1000 24By = 11,200; By = 11,200 24By = 11,200; By = 11,200 24 24 B y = 466.67 lbs Then sum the vertical loads to find A y ; Then sum the vertical loads to find A y ; A y = - 200 – 300 – 400 + 466.67 = 433.33 lbs A y = - 200 – 300 – 400 + 466.67 = 433.33 lbs

36. NOW CONSIDER THIS EXERCISE TWO: For the beam loaded as shown, calculate the magnitude of the reactions at points A and B. Take moments about point A: -200x7 – 500x14 + By x 20 = 0 So By = 1400+7000/20 and By = 420; Ay = 280

37. Consider a beam loaded as shown, with one end that projects over its support. Calculate the magnitude of the reactions. The procedure is the same as previous, paying careful attention to sign convention.

38. First, consider that there are two unknowns, the reactions at A and B. Select one of the points and sum the moments about that point that produce rotation. Select A, and remember the reaction force at A will not produce any moment at A, because the distance is zero. Begin at the left - - - +(400 x 5’) –(600 x 13’) –(200 x 19’) + 25 x B y = 0 + 2000 – 7800 – 3800 + 25 B y = 0 ; 25 B y = 9600 25 B y = 9600` B y = 9600 = 384 lb B y = 9600 = 384 lb 25 25

39. To find the reaction at A, find the algebraic sum of all known vertical loads: + 384 – 400 – 600 – 200 + A y = 0, and A y = 816 lb Or, take moments about B to find A y : +(400 x 30’) + (600 x 12’) + (200 x 6’) – 25 A y = 0 +12,000 + 7,200 + 1,200 - 25 A y = 0 25 A y = 20,400 ; A y = 20,400 = 816 lb 25 25 384 lb

40. Remember, the cantilever will not present a problem of confusion IF you remember the proper sign convention as the forces have a tendency to rotate about the point you choose. Sometimes it helps to keep the sign convention in order if you place a fingertip at the point of rotation, and then visualize how the forces would have a TENDENCY to rotate about your finger.  Consider finally, if you don’t know the direction of an unknown force, use judgment and assume a direction - - - If you assume the wrong direction, it will be evident if your answer turns out to be a negative number.