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_14_EGR1301_Fall2014_Statics_141006.pptx Newtonβs Laws Calibri 20 I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it. Arial 18 I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it. Sans Serif 18 I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it. Times New I. Every object in a state of uniform motion tends to remain in Roman 20 that state of motion unless an external force is applied to it. Calibri 20 II. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector. Calibri 20 III. For every action there is an equal and opposite reaction. 1 _14_EGR1301_Fall2014_Statics_141006.pptx Force Terminology I. Forces acting along the same line of action are called collinear forces II. Forces whose lines of action pass through the same point in space are called concurrent forces III. Forces that lie in the same plane are called coplanar forces Moments at a Point If only one force can cause rotation at a point, the moment is the product of that force and the perpendicular distance from the line of action of the force to the point. If more than once force is involved, then superimpose their individual moments on the point. Counterclockwise is the reference (positive) direction of rotation. Equilibrium A body is in equilibrium when the sum of all external forces and moments acting on the body is zero. This requires the body to be at rest or moving with a constant velocity. ππ = π ππ = π π΄ = π for any point that can rotate. Use counterclockwise as positive. 2 _14_EGR1301_Fall2014_Statics_141006.pptx Forces Possible from Supports, Surfaces, and Connectors 3 _14_EGR1301_Fall2014_Statics_141006.pptx Draw the Free-Body Diagram Cut the body free from all others, all supports, and all connectors. They are replaced with external forces acting upon the body under consideration. 4 _14_EGR1301_Fall2014_Statics_141006.pptx Write the Equations 4.0 m 6.5 m 8.5 m Unknowns π΄π₯ , π΄π¦ , π΅π¦ 1 0 0 π΄π₯ β47 0 1 1 π΄π¦ = 81.41 0 0 6.5 π΅π¦ 880.0 47 πΉπ₯ = π΄π₯ + 94 cos 60° = 0 81.41 πΉπ¦ = π΄π¦ + π΅π¦ β 94 sin 60° = 0 All points are free to rotate. Usually the only points not free are a beam imbedded into a wall, or a clamp. A good rule is to chose a point that has the most unknowns. In this case, point A is the best choice. β188.0 β188 -692.0 ππ΄ = β94 cos 60° β 4.0 + β94 sin 60° β 8.5 + π΅π¦ β 6.5 = 0 5 _14_EGR1301_Fall2014_Statics_141006.pptx Draw the Free-Body Diagram 6 _14_EGR1301_Fall2014_Statics_141006.pptx Some trig, then write the equations tππ π = 6 , 8 P π = 36.9° Pang=90°+β +π Py P Py Px Px ππ₯ = ππππ (ππππ ) ππ¦ = ππ ππ(ππππ ) = 196.14 N Slope of GH is given as 1: 1, π π β +π = 45°, β = 45° β 36.9° = 8.1° ππππ = 90° + 45° = 135°, ππ₯ = π πππ (135°), ππ¦ = π π ππ (135°) πΉπ₯ = πΊπ₯ + ππ₯ = 0, πΊπ₯ β0.7071π = 0 Unknowns πΊπ₯ , πΊπ¦ , π πΉπ¦ = πΊπ¦ + ππ¦ β 196.14, πΊπ¦ + 0.7071π β 196.14 = 0 ππΊ = ππ¦ β 10πππ 8.1° β ππ₯ β 10π ππ8.1° β 196.14 β 12πππ 8.1° = 0, π π ππ(135°) β 10πππ 8.1° β π πππ (135°) β 10π ππ8.1° β 196.14 β 12πππ 8.1° = 0 7 _14_EGR1301_Fall2014_Statics_141006.pptx πΉπ₯ = πΊπ₯ + ππ₯ = 0, πΊπ₯ β0.7071π = 0 Unknowns πΊπ₯ , πΊπ¦ , π πΉπ¦ = πΊπ¦ + ππ¦ β 196.14, πΊπ¦ + 0.7071π β 196.14 = 0 ππΊ = ππ¦ β 10πππ 8.1° β ππ₯ β 10π ππ8.1° β 196.14 β 12πππ 8.1° = 0, π π ππ(135°) β 10πππ 8.1° β π πππ (135°) β 10π ππ8.1° β 196.14 β 12πππ 8.1° = 0, π β 0.7071 β 10 β 0.9900 β π β (β0.7071) β 10 β 0.1409 β 196.14 β 12 β 0.9900 = 0, π β 7.000 β π β (β0.9963) β 2330 = 0, 1 0 β0.7071 πΊπ₯ 0 0 1 0.7071 πΊπ¦ = 196.14 0 0 7.996 2330 π 8 _14_EGR1301_Fall2014_Statics_141006.pptx Draw the Free-Body Diagram 9 _14_EGR1301_Fall2014_Statics_141006.pptx Write the Equations π΄π₯ π΄π¦ πΉπ₯ = π΄π₯ β 270 = 0 πΉπ¦ = π΄π¦ β 250 β 180 + π΅π¦ = 0 π΅π¦ ππ΄ = β250 β 1.3 β 180 β 3.0 β 270 β 1.0 + π΅π¦ β (1.3 + 1.7 + 1.0 β 1.5) = 0, β1135 + 2.5π΅π¦ = 0, 1 0 0 0 0 π΄π₯ 270 1 1 π΄π¦ = 430 0 2.5 π΅π¦ 1135 10 _14_EGR1301_Fall2014_Statics_141006.pptx Draw the Free-Body Diagram 11 _14_EGR1301_Fall2014_Statics_141006.pptx Some Trig to Determine the Cable Angle Tang E 42β° D π΄π΅πππ 42° = 3.1β² , π΄π΅ = 4.171β² π·π΅ = π΄π΅π ππ42° = 2.791β² πΆπΈ = πΆπ΄ β π·π΅ = 3.9 β 2.791 = 1.109β² πΆπΈ 1.109 ππππ = 180° β ππ‘ππ = 180° β ππ‘ππ = 180° β 19.7° = 160.3° πΈπ΅ 3.1 12 _14_EGR1301_Fall2014_Statics_141006.pptx Freebody Diagram and Equations Tang = πππ. π° E T Ty Tx π. πππβ² π. πππβ² 42β° Ax D Ay πΉπ₯ = π΄π₯ + ππ₯ = 0, π΄π₯ + ππππ 160.3° = 0 1 0 0 1 0 0 0 β0.9415 π΄π₯ 0.3371 π΄π¦ = 1200 7680 3.673 π πΉπ¦ = π΄π¦ + ππ¦ β 1200 = 0, π΄π¦ + ππ ππ 160.3° β 1200 = 0 ππ΄ = ππ¦ β 3.1 β ππ₯ β 2.791 β 1200 β 6.4 = 0 = ππ ππ 160.3° β 3.1 β ππππ 160.3° β 2.791 β 7680 = 0 1.045 -2.628 13 _14_EGR1301_Fall2014_Statics_141006.pptx Draw the Free-Body Diagram 14 _14_EGR1301_Fall2014_Statics_141006.pptx Draw the Free-Body Diagram 15 _14_EGR1301_Fall2014_Statics_141006.pptx Some trig, then write the equations 8.0 β 6.0 π πππ πΆ πͺ Angle of πͺ β πΆπ¦ πΆπ₯ β 6.0 πππ π If π is known (0° to 75° in increments of 1°), β’ Then angle β of the top triangle can be found, π΄π₯ β’ then the angle of C is (180 β β), β’ then πΆπ₯ = πππ of C, and πΆπ¦ = π ππ of C πΉπ₯ = π΄π₯ + πΆπ₯ = 0, π΄π₯ + πΆπππ (ππππ) = 0 π΄π¦ πΉπ¦ = π΄π¦ + πΆπ¦ β 1200 = 0, π΄π¦ + πΆπ ππ ππππ β 1200 = 0 ππΆ = βπΆπ₯ β 6.0π πππ + πΆπ¦ β 6.0 πππ π β 1200 β 8.0πππ π = 0, βπΆπππ (ππππ) β 6.0π πππ + πΆπ ππ ππππ β 6.0 πππ π β 1200 β 8.0πππ π = 0 1 0 0 0 πππ (ππππ) 1 π ππ(ππππ) 0 β6 cos ππππ β π πππ + 6 π ππ ππππ β πππ π π΄π₯ 0 π΄π¦ = 1200 9600πππ π πΆ 16 _14_EGR1301_Fall2014_Statics_141006.pptx Two-Dimensional Problem: Loosening a Bolt with a Wrench, When the Force is Not Perpendicular to the Wrench y Bolt Head P π« ππ· ππΉ x π Perpendicular distance π = π· π ππ(ππ· β ππΉ ), π = F ππΉ = F cos ππΉ πx + sin ππΉ πy = πΉπ₯ πx + πΉπ¦ πy π« = D ππ· = D cos ππ· πx + sin ππ· πy = π·π₯ πx + π·π¦ π y Moment π = πΉ π 17 _14_EGR1301_Fall2014_Statics_141006.pptx Useful trig identities for this problem are sin π΄ ± π΅ = sin π΄ cos π΅ ± cos π΄ sin π΅ cos π΄ ± π΅ = cos π΄ cos π΅ β sin π΄ sin π΅. Thus, P can be expressed as π = π· π ππ(ππ· β ππΉ ) = π· sin ππ· cos ππΉ β cos ππ· sin ππΉ . Now, substitute P into the M equation, Moment π = πΉ π = πΉ π· sin ππ· cos ππΉ β cos ππ· sin ππΉ . In determinant form, Moment π = πΉ π· cos ππΉ cos ππ· sin ππΉ sin ππ· π = F cos ππΉ πx + sin ππΉ πy = πΉπ₯ πx + πΉπ¦ πy π« = D cos ππ· πx + sin ππ· πy = π·π₯ πx + π·π¦ πy πx πy π΄ = π × π« = πΉπ₯ πΉπ¦ π·π₯ π·π¦ Same πz 0 πΉπ§ 0 π·π§ = πΉπ₯ π·π¦ β πΉπ¦ π·π₯ πz 18 _14_EGR1301_Fall2014_Statics_141006.pptx Chapter 13. Strength of Materials Stress π is force per unit area. Unit is Pascal ( π ) Newtons per square meter. The conversion to U.S. Customary units is 1 MPa = 145.05 psi (pounds force per square inch) Example, 310 πππ ππ π 145.05 πππ = 44,966 ππ π = 44.966 πππ π In reverse, 1 kpsi = 6.894 MPa Example, 44.966 πππ π πππ 6.894 πππ π = 45,966 ππ π = 310 πππ 19 _14_EGR1301_Fall2014_Statics_141006.pptx Three Types of Stress ( π ) β’ Tensile. Axial load such as a cable. Tends to pull the atoms apart. β’ Compressive. Axial load such as driving a fence post into the ground. Tends to push atoms together β’ Shear. Tends to slide atoms across each other. Such as sliding the top half of a stack of plywood across the bottom half, without lifting. 20 _14_EGR1301_Fall2014_Statics_141006.pptx Strain ( πΊ ) β’ Dimensionless ratio, change in length to the original length Stress ( π ) βπ³ πΊ= π³ β’ Proportional Limit β linear region β’ Elastic Limit β Material will still return to original length when load is removed Strain ( πΊ ) β’ Yield Strength β permanently deformed β’ Ultimate Strength β Fails 21 _14_EGR1301_Fall2014_Statics_141006.pptx Some trig, then write the equations acd T cda 8.0 m dac ead Tang sin(πππ) 9.2 = sin(50°) , 8.0 so πππ = 61.8°. πππ = 180° β πππ β πππ, so πππ = 68.2°. πππ = 90° β πππ = 21.8°. πππ = 90° β πππ = 28.2°. dea E Tang = 180° β πππ = 151.8°. Material Yield Strength MPa Iron 210 Structural Steel 240 Stainless Steel 210 Aluminum 85 Copper, hard drawn 415 22 _14_EGR1301_Fall2014_Statics_141006.pptx Draw the Free-Body Diagram Beam weight is modeled as a force at the center of mass π»π = π»πππ(π»πππ ) 12.4 m 6.2 m π»π = π»πππ(π»πππ) 8.2 kg/m β’ 12.4 m = 101.7 kg, corresponding to 997 N 6.2 m 21.8β° π»π π»π Ay 21.8β° πΉπ¦ = π΄π¦ + ππ¦ β 1471 β 997 = 0 ππ΄ = β997 (6.2cos 21.8°) β π»π (9.2sin 21.8°) + π»π (9.2cos 21.8°) β 1471(12.4cos 21.8°) = 0 Write the Equations π» πΉπ₯ = π΄π₯ + ππ₯ = 0 150 kg β’ 9.807 = 1471 N Beam 101.7 kg, 997 N Tang = 151.8° Ax 23 _14_EGR1301_Fall2014_Statics_141006.pptx ππ₯ = ππππ (ππππ ), ππ¦ = ππ ππ(ππππ ), ππππ = 151.8° -0.8813 πΉπ₯ = π΄π₯ + ππ₯ = 0, π΄π₯ + ππππ (151.8° ) = 0 0.4726 πΉπ¦ = π΄π¦ + ππ¦ β 1471 β 997 = 0, π΄π¦ + ππ ππ(151.8° ) β 2468 = 0 5739 -0.8813 3.417 0.4726 8.542 ππ΄ = β997 (6.2cos 21.8°) β ππππ (151.8° ) β (9.2sin 21.8°) + ππ ππ(151.8° ) β (9.2cos 21.8°) β 1471(12.4cos 21.8°) = 0 16936 π΄π₯ β0.8813 0 π΄π¦ = 0.4726 2468 β3.011 + 4.037 π 5739 + 16936 1 0 0 0 1 0 1 0 0 0 β0.8813 π΄π₯ 0 1 0.4726 π΄π¦ = 2468 0 1.026 22675 π 24 _14_EGR1301_Fall2014_Statics_141006.pptx 25 _14_EGR1301_Fall2014_Statics_141006.pptx Cubic Meter of Air, Moving at V m/s Air density ππ π = 1.225πΎπ πΎπ΄ 3 π where πΎπ and πΎπ΄ are correction factors for altitude and pressure, and equal 1 at standard 15β°C, 1 atmosphere (sea level). Air mass passing through A square meters of surface normal to V is ππ π ππ 2 π 3 π΄π π = ππ΄π π π π Kinetic energy of the mass passing through surface A per second (i.e., power) is 1 ππ ππ΄π 2 π π2 π2 π 2 1 ππ β π2 1 1 π½ππ’πππ 1 3 3 = ππ΄π β = ππ΄π = ππ΄π 3 πππ‘π‘π 2 2 π π 2 π 2 Joules/sec Force and pressure of the air are 1 ππ β π2 1 π 1 1 3 2 πππ€π‘πππ , 2 πππ πππs πΉ = ππ΄π β ÷ π = ππ΄π π = ππ 2 2 2 π π π 2 Joules/sec ÷ meters/sec 26 _14_EGR1301_Fall2014_Statics_141006.pptx 27 _14_EGR1301_Fall2014_Statics_141006.pptx 28 _14_EGR1301_Fall2014_Statics_141006.pptx 29 _14_EGR1301_Fall2014_Statics_141006.pptx 30 _14_EGR1301_Fall2014_Statics_141006.pptx 31 _14_EGR1301_Fall2014_Statics_141006.pptx 32 _14_EGR1301_Fall2014_Statics_141006.pptx 33 _14_EGR1301_Fall2014_Statics_141006.pptx Cubic meter of air, moving at V m/s Air density ππ π = 1.225πΎπ πΎπ΄ 3 π where πΎπ and πΎπ΄ are correction factors for altitude and pressure, and equal 1 at standard 15β°C, 1 atmosphere (sea level). 34 _14_EGR1301_Fall2014_Statics_141006.pptx 35 _14_EGR1301_Fall2014_Statics_141006.pptx 36 _14_EGR1301_Fall2014_Statics_141006.pptx 37 _14_EGR1301_Fall2014_Statics_141006.pptx 38 Prob. 1. A playground see-saw has a 10 kg child on one end, and an 80 kg adult on the other end. Write the equations to find the distance x for the center of mass of 100 kg to achieve balance. 80 kg m 100 kg m 10 kg m x 2.5 m Pivot 2.5 m πΉπ₯ = 0 (there are no forces in the x direction) πΉπ¦ = β10 β 100 β 80 9.807 + πΉπππ£ππ‘ = 0, πΉπππ£ππ‘ = 190 β 9.807 π ππππ£ππ‘ = 10 9.807 β 2.5 + 100(9.807)π₯ β 80(9.807) β 2.5 = 0 In this problem, only the ππππ£ππ‘ = 9.807(10 β 2.5 + 100π₯ β 80 β 2.5) = 0 moment equation is needed to answer the question. 175 25 + 100π₯ β 200 = 0, π₯= = 1.75 π 100 39 EGR1301, Test 2, Oct. 08, 2014. Show all work on these sheets. Name. __________________________ y-axis 3β3πππππ° π Prob. 2. A 100 kg mass is suspended by a cable, twothirds the way up on a 3 meter long, weightless pipe. The pipe is cabled to the wall. Tension T 3πππππ° 3πππππ° ο· 1m wall ο· 3m 2m 3πππππ° 2πππππ° 100 kg ππ° Fy Write the two force equations, and the moment equation. Use the joint at bottom of the beam for the moment equation. Put the equations into standard matrix form, combining terms where possible. Do not solve. πΌ = ππ‘ππ 3 β 3πππ 30° 3πππ 60° 60β° Fx x-axis πΉπππππ π₯ = πcos(180° β πΌ) + πΉπ₯ = 0 2πππππ° 3πππππ° πΉπππππ π¦ = π sin 180° β πΌ + πΉπ¦ β 100(9.807) = 0 π = βππππ (180° β πΌ) β 3πππ 30° + π π ππ 180° β πΌ β 3πππ 60° β 100(9.807) β 2πππ 60° = 0 1 0 0 0 1 0 cos(180° β πΌ) sin(180° β πΌ) βππππ (180° β πΌ) β 3πππ 30° + π π ππ 180° β πΌ β 3πππ 60° πΉπ₯ 0 πΉπ¦ = 980.7 980.7 β 2πππ 60° π 40 Prob. 3. The Sky Cam at the football stadium is 60 kg and supported by four cables. In this problem, you will consider a two-dimensional version of the Sky Cam. Assume that the Sky Cam assembly is a weightless rigid frame, and the entire 60 kg mass is in the large ball at the bottom. ο· Write the two force equations, and the moment equation. Use the left-hand ring for the moment equation. ο· Put the equations into standard matrix form, combining terms where possible. Do not solve. y-axis Tension TL 120β° Tension TR 1m 1m πΉπ₯ = ππΏ πππ 120° + ππ πππ 30° = 0 30β° x-axis 2m πΉπ¦ = ππΏ π ππ120° + ππ π ππ30° β 60(9.807) = 0 60 kg π = ππ π ππ30° β 2π β 60(9.807) β 1π = 0 πππ 120° π ππ120° 0 πππ 30° π ππ30° 2π ππ30° 0 0 0 0 ππΏ ππ = 60(9.807) 60(9.807) 0 In this problem, the first two equations were sufficient. The third equation can be used as a check. 41
