1. E NGINEERING D ESIGN L AB II ENGR-102 W EEK 2 L ECTURE – B RIDGE M ODULE Pramod Abichandani, Ph.D. Richard Primerano, Ph.D. O VERVIEW OF W EEK 2 LAB ACTIVITIES T HE M ETHOD OF J OINTS F REE BODY DIAGRAMS

2. T HE T RUSS A structural member composed of triangular sections The members of a truss are connected at nodes Each member is in either tension or compression A technique called the method of joints can be used to determine the forces in all members under a given load. 2

3. E XAMPLE : F IND THE F ORCES IN THIS S TRUCTURE We know the applied loads, the types of supports, and the geometry of the structure From this, we can determine the tension and compression forces in the members AB, BC, and AC 3 100lb A B C pin joint – prevents motion in the X and Y directions. roller joint – prevents motion in the Y direction. 1ft

4. A NALYZING F ORCES IN A T RUSS T HE M ETHOD OF J OINTS 1. Determine the support reactions. 2. Draw a free body diagram for each joint. 3. Write an equilibrium equation for each joint. 4. Solve for the forces at each joint. A good video tutorial on the method of joints http://www.youtube.com/watch?v=ZOtSRbzKCC8 4

5. S TEP 1: D ETERMINE S UPPORT R EACTIONS Employ Newton’s law to find the reaction forces keeping the structure stationary 5 100lb A B C 1ft F cy F ay F cx

6. S TEPS 2&3: D RAW F REE B ODY D IAGRAMS Draw all the forces on node A Each force must be resolved into its X and Y components 6 50 0 F ac F ab 60 º F ab sin 60 º F ab cos 60 º F ab = - 50 / 0.87 = -57.7 lb F ac = 57.7 * 0.5 = 28.9 lb

7. B RIDGE D ESIGN M ODULE W EEK 2 S OFTWARE T OOLS FOR B RIDGE D ESIGN We introduce one of the three design tools (West Point Bridge Designer) that will be used in this module for bridge design and analysis. AutoCAD and Visual Analysis are the others WPBD is an educational package developed and distributed by the United States Military Academy. It allows you to quickly design and simulate bridges using a variety of common construction materials. Using this, you can visualize the forces in a truss for given loads. 7

8. S OME B ASIC P HYSICS S TATIC E QUILIBRIUM When an object is in static equilibrium, all of the particles that make up the object are stationary and there is no net force acting on the object. Basically, the object is at rest and not accelerating All applied forces are balanced by reaction forces. 8 100lb A B C 1ft 50lb 0 lb

9. E XTERNAL F ORCES A CTING ON A S TRUCTURE Applied forces (loads) – these are the forces that the structure is designed to carry: cars on a bridge, shingles on a roof, lift force on a wing Dead loads – loads permanently acting on a structure: the weight of members, road decking, roofing material Live loads – time varying loads acting on a structure: cars on a bridge, people on a floor, snow on a roof The dead loads can usually be calculated with high accuracy. The live loads are generally estimated based on experience of calculation. video video 9

10. E XTERNAL F ORCES A CTING ON A S TRUCTURE Reaction forces – exerted by support structures and responsible for keeping a structure in static equilibrium These forces “adjust themselves” as the applied forces change. 10 100lb A B C 1ft 50lb 0 lb 100lb A B C 1ft 43lb 100 lb

11. E XAMPLE : T UG OF W AR Consider a 3-way tug of war. Force A is exerted along the -X axis Force B is exerted along the -Y axis Force C is exerted at a 45˚ angle w.r.t. the +X axis The knot is in static equilibrium If for force exerted by player A is 150lb, find all remaining forces. 11 +Y +X F A = 150lb FBFB FCFC 45˚

12. S OME INITIAL OBSERVATIONS F A is resisted (balanced) by a component of F C F B plays no role in balancing F A F B is also balanced by a component of F C F A plays no role in balancing F B 12 F A = 150lb FBFB FCFC 45˚ +Y +X

13. R ESOLVING A V ECTOR INTO ITS C OMPONENTS The force F C can be considered as the sum of two component forces, one acting in the X direction and the other acting in the Y direction. 13 FCFC θ+Y +X F CY F CX F CY = F C sin(θ) F CX = F C cos(θ) FCFC F CY F CX θ As F C becomes closer to horizontal: F CX → F C F CY → 0

14. R ESOLVING A V ECTOR INTO ITS C OMPONENTS Note that we usually draw both component vectors with their tails at the origin, not head to tail. 14 FCFC θ +Y +X F CY F CX FCFC θ F CY F CX same thing

15. B ACK TO OUR P ROBLEM All forces are now decomposed into their X and Y components. 15 +Y +X F A = 150lb FBFB 45˚ FCFC F CY F CX

16. T HE C ONDITIONS N EEDED FOR S TATIC E QUILIBRIUM An object in static equilibrium is one that “isn’t moving” Neither translating nor rotating Assuming the ropes can only move in the plane of the screen, this means All forces in the X direction must sum to zero All forces in the Y direction must sum to zero We will come back to the rotation part 16

17. T HE EQUATIONS OF STATIC EQUILIBRIUM The net force acting on the object is zero If it were non-zero, the object would be accelerating 17 +Y +X F A = 150lb FBFB 45˚ FCFC F CY F CX F CY = F C sin(45) = F C / √2 F CX = F C cos(45) = F C / √2 ΣF X = 0 F CX – F A = 0 F A = F CX = F C / √2 F C = 150 √2 lb ΣF Y = 0 F CY – F B = 0 F B = F CY = F C / √2 F B = 150 lb

18. W HAT IS M OMENT OF F ORCE Also called moment or torque. 18 5lb 1ft 5lb 0.707ft 45˚ 5 ft-lbs of torque 3.54 ft-lbs of torque 5lb 0 ft-lbs of torque 1ft

19. C ALCULATING THE R EACTION F ORCES A CTING ON A S TRUCTURE The reaction forces keep the object fixed both transnationally (X and Y directions) and rotationally (about the Z axis) 19 100lb A B C 2ft 1ft F CY F AY F AX

20. T HE R ELEVANT E QUATIONS The object is not translating: ΣF X = 0, ΣF Y = 0 The object is not rotating: Σ torques about any point = 0 Lets choose point A: ΣM A = 0 20 100lb A B C 2ft 1ft F CY F AY F AX ΣF X = 0 F AX – 100 = 0 F AX = 100 ΣF Y = 0 F AY + F CY = 0 F AY = -F CY ΣM A = 0 1ft*100lb + 2ft*F CY = 0 F CY = -50 ft-lb F AY = 50 ft-lb

21. A NALYZING F ORCES IN A T RUSS T HE M ETHOD OF J OINTS 1. Determine the support reactions. 2. Draw a free body diagram for each joint. 3. Write an equilibrium equation for each joint. 4. Solve for the forces at each joint. A good video tutorial on the method of joints http://www.youtube.com/watch?v=ZOtSRbzKCC8 21

22. F INDING THE S UPPORT R EACTIONS Last time we had a symmetrical structure that was symmetrically loaded. What observations can we make Weight distributions between support reactions Sign of each support reaction 22 F CY F AY F AX 100lb 50lb 5 ft A C 10 ft

23. F INDING THE S UPPORT R EACTIONS Last time we had a symmetrical structure that was symmetrically loaded. 23 F CY F AY F AX 100lb 50lb 5 ft ΣF X = 0 F AX = 0 ΣF Y = 0 F AY + F CY = 150lb ΣM A = 0 20F CY – 500 – 250= 0 F CY = F AY = A C 10 ft

24. T HANK YOU ! 24