Outline: Centres of Mass and Centroids Centre of Mass Centroids of Lines, Areas and Volumes Composite Bodies Beams External Effects Beams Internal Effects 1
Up to now all forces have been concentrated forces (force applied at single point) But forces are actually applied over some contact area If contact area is significant compared to the other dimensions we must account for it Sum the effects of the distributed force over the contact region using mathematical integration 2
Three types of Force Distribution (see fig. 5/2): Line Distribution - vertical load supported along a cable Intensity is in N/m Area Distribution - hydraulic water pressure against dam face Intensity is in N/m 2 or Pascal (Pa) Called pressure for the action or fluid forces Called stress for internal distribution in solids Volume Distribution - gravitational attraction Intensity is in N/m 3 For gravity the intensity is the specific weight (density times gravitational field strength ρg) 3
For external forces a concentrated load can be used to replace a distributed load Need to find the position for a concentrated load that will give an equivalent system Typically, we are dealing with weight Need to find Centre of Gravity/Mass for rigid bodies (for location of load) Also, we need to be able to analyze the internal forces caused by distributed loads 4
Integration is the inverse of differentiation The integral of a function is a measure of the area between it and the x-axis The symbol of integration is, an elongated S (stands for "sum"). The integral is written as and is read "the integral of f-of-x with respect to x." b a f ( x) dx 5
Example: If we know that (dy/dx)=3x 2 and we need to know the function this derivative came from, then we "undo" the differentiation process. y = x 3 is ONE antiderivative of 3x 2 (since the derivative of any constant is zero) In general: y = x 3 + C, is the indefinite integral, C is called the constant of integration Indefinite integrals deal with algebraic quantities (define an entire function) 6
Integrals with a definite bound are definite integrals The definite integral is written as and is read "the integral from a to b of f-of-x with respect to x." b a f ( x) dx a and b are the lower limit and upper limit, respectively, of integration, defining the domain of integration; f is the integrand, to be evaluated as x varies over the interval [a,b]; and dx is the variable of integration. 7
Back to the example: to evaluate the integral of 3x 2 between x=1 and x=5: 5 1 3x 2 dx x 3 5 1 (5) 3 (1) 3 124 dx x C Some integration rules and properties adx dx a dv dx dx dv x n dx n1 x n 1 C 8
How does it balance? Image from http://24hours7days.com/toys/circus_sam_balancing_man_1972.html Accessed March 2009 9
Rigid Body The center of mass is a unique point whose location is a function of the distribution of mass throughout the body It is the point through which the total weight vector acts B A B A G G 10
Consider a 2D body: Divide the plate into n small elements each with weight DW and coordinates x & y 11
co-ordinates for the center of gravity: W x dw xdw W y ydw z W zdw W 12
For a Wire: W gal DW gadl x y z xdw W ydw W zdw W ρ (density), g, and A (the crosssectional area) are constant over the length (L) ga x y z xdl L ydl L zdl L 13
For an Area: W gta DW gtda x y z xdw W ydw W zdw W ρ (density), g, and t (the thickness) are constant over the surface area (A) gt x y z xda A yda A zda A 14
For a Volume: W gv DW gdv x y z xdw W ydw W zdw W ρ (density), and g are constant over the volume (V) g x y z xdv V ydv V zdv V 15
Steps to solve: Choose a differential element for integration Express da, dl or dv in terms of 1 variable Express x, y or z (location of centroid of element) in terms of 1 variable Substitute into equations and integrate Lx Ly Lz xdl ydl zdl Ax Ay Az Use axes of symmetry whenever possible xda yda zda Vx Vy Vz xdv ydv zdv 16
Find the centroid of the area under y=x 2? 17
Expression for differential element (tiny rectangle) Integrate to get expression for A 18
ANSWER: The centroid of the area is 3 4, 3 10 m 19
Find the centroid of the cone, with radius a and height h? 20
An irregularly shaped area may be divided into the smaller shapes with known G s Summing moments as before: For plates and wires with uniform thickness and homogeneous material: ENGR 1205 Chapter 5 21 i i i i i i A z A Z A y A Y A x A X i i i i i i L z L Z L y L Y L x L X 3 3 2 2 1 1 3 2 1 ) ( x W W x W x X W W W
An irregularly shaped body may be divided into smaller shapes with known G s (see Appendix D for centroids of common shapes) As before: X W x W i i X V xv i i Y W y W i i Y V y V i i Z W z W i i Z V z V i i 22
Steps to solve a composite body: Choose a reference frame Divide the composite area/volume into parts whose centroids you know (APPENDIX D) or can easily determine Determine the coordinates of the centroids and the volume/ area/ length for each part Watch for cases of symmetry that can simplify calculations 23
Find the centroid of the L-shaped bar of negligible thickness. 1 m 1 m 24
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Find the centroid of the irregularly-shaped plate below. All dimensions in cm. y 50 40 20 r=10 30 40 80 x 27
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PART A x Y xa ya 1 2 3 4 Sum 29
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Locate the center of mass of the composite body made of a conical frustrum and a hemisphere, shown below. 31
Beams are bars of material that support lateral loads (perpendicular to the axis of the beam) Beams are probably the most important structural member Beams can support both concentrated and distributed loads Analyze load carrying capacity of beams for: Equilibrium and external reactions Internal Resistance (strength characteristics) 32
We will only analyze statically determinate beams in this class 33
Distributed loads: intensity w of a distributed load is expressed as force per unit length of the beam (can be constant or variable) The resultant of the distributed load is equal to the area formed by the intensity w and the length The resultant of the distributed load passes through the centroid of the area 34
For a more general load distribution: R = wdx Rx = xwdx 35
To Determine External Reactions for Beams: Reduce each distributed load to an equivalent concentrated load (find total force applied and location) Use given concentrated loads and equivalent concentrated loads with 2-D equations of equilibrium 36
A beam is subjected to the loads shown. Determine the reactions at the supports A & B. 37
Given: See FBD Find: Ax, Ay, By TOTAL AREA (RESULTANT FORCE) CENTROID (POINT OF ACTION) 1 1 2 2 m 300 N/m = 300 N 2 3 2 m = 4 3 m = 1.33 m 2 6 m 300 N/m = 1800 N 1 2 6 m + 2 m = 5 m 3 1 2 (4 m) (300 N/m) = 600 N 1 3 4 m + 8 m = 9.33 m 38
For Equilibrium ΣM = 0 & ΣF = 0 ΣM A = 300 1. 33 1800 5 600 9. 33 + B Y 12 B Y = 1250 N ΣF X = 0 A X = 0 N ΣF Y = 0 A Y 300 1800 600 + B Y = 0 A Y = 1450 N = 0 ANSWER: The reactions at the supports are A Y = 1450 N [up] & B Y = 1250 N [up] 39
Beams can resist: Tension/ Compression Shear force (V) Bending moment (M) Torsional moment (T) The amount of each kind of internal load can change throughout the length of the beam, depending on external loads 40
We can cut the beam at any point along its length to analyze the internal forces at that point It is often difficult to tell the direction of the shear and moment without calculations so represent V and M in their positive directions and let the result tell you (+/-) whether you drew it the right way CONVENTION 41
The maximum bending moment is often the primary consideration in the design of a beam Variations in shear and moment are best shown graphically 42
Use equilibrium of whole beam to find reaction forces Then cut beam concentrated loads, replace any distributed loads with and solve for unknown shear and moment at. Plot V (shear vs x) and M (moment vs x) along the beam 43
Solve for R 1 and R 2 Isolate left section & solve for V and M between the left support & 4kN load Isolate right section and solve for V and M between the right support & 4kN load Isolate a section between every change in an external load, but DO NOT CUT the section at the concentrated load The area under a shear curve between two points is equal to the change in bending moment between the same two points. 44
Given: See Diagram Find: Maximum bending moment Draw shear force and bending moment diagrams Solve: For Equilibrium F = 0 and M = 0 45
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Draw the shear and bending moment diagrams for the beam and loading shown and determine the location and magnitude of the maximum bending moment. 48
Draw the shear and bending moment diagrams for the beam and loading shown. 49