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ARCHITECTURE 2351 INTRODUCTION TO STRUCTURE CONCEPTS

ARCHITECTURE 2351 INTRODUCTION TO STRUCTURE CONCEPTS. Structural elements, whether rigid or flexible, have volume and mass, and are subject to the forces of gravity.

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ARCHITECTURE 2351 INTRODUCTION TO STRUCTURE CONCEPTS

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  1. ARCHITECTURE 2351 INTRODUCTION TO STRUCTURE CONCEPTS

  2. Structural elements, whether rigid or flexible, have volume and mass, and are subject to the forces of gravity. 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. A body at rest is said to be in equilibrium. A body in equilibrium acted upon by another body reacts with an equal magnitude and opposite action, called a reaction.

  3. 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.

  4. 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.

  5. OBJECTS IN EQUILIBRIUM

  6. 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 equilibrium 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 equilibrium 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.

  7. 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.

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. Moment equals FORCE times the perpendicular DISTANCE of the line of action of the force to the pivot center FORCE pounds Pivot center DISTANCE PERPENDICULAR TO LINE OF ACTION OF FORCE 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.

  13. 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.

  14. 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 Fy = 0 Moment caused by Forces that tend to cause clockwise rotation about any point must be equal and oppositeto 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 M0 must equal 0

  15. 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 Fx = 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)

  16. 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.

  17. 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.

  18. Next, arbitrarily pick a point that is 3’ to the left of B, and call it point “C” Sum the moments of all the forces about point C . . . * 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.

  19. But what if the load on a beam is not at its center? How can the reactions at 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

  20. 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) - (Ay x 10) = 0, and + 300 - 10By = 0 , then solving for Ay ; 10 Ay = 300, and Ay = 30 lbs Consequently, since the sum of the forces in the vertical direction must equal zero, then By = 50 – 30 = 20 lbs

  21. NOW TAKE A SHEET OF PAPER AND DO THIS EXERCISE ONE For the beam loaded as shown, find the value of the reactions at A and B

  22. 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 ;

  23. 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) + 24By = 0 - 1000 – 3000 – 7200 + 24By = 0 24By = 7200 + 3000 + 1000 24By = 11,200; By = 11,200 24 By = 466.67 lbs Then sum the vertical loads to find Ay ; Ay = - 200 – 300 – 400 + 466.67 = 433.33 lbs

  24. NOW ON THE SAME SHEET OF PAPER DO THIS EXERCISE TWO: For the beam loaded as shown, calculate the magnitude of the reactions at points A and B.

  25. 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.

  26. 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 By = 0 + 2000 – 7800 – 3800 + 25 By = 0 ; 25 By = 9600 ` By = 9600 = 384 lb 25

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

  28. 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.

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