CH. 4 BEAMS & COLUMNS

Size: px

Start display at page:

Download "CH. 4 BEAMS & COLUMNS"

Stuart Booth
6 years ago
Views:

CH. 4 BEAMS & COLUMNS BEAMS Beams Basic theory of bending: internal resisting moment at any point in a beam must equal the bending moments produced by the external loads on the beam Rx = Cc + Tt - If

1 CH. 4 BEAMS & COLUMNS BEAMS Beams Basic theory of bending: internal resisting moment at any point in a beam must equal the bending moments produced by the external loads on the beam Rx = Cc + Tt - If the beam is in equilibrium then all moment forces must cancel out - Positive: clockwise rotation - Negative: counterclockwise rotation - Moment increases as the distance from the reaction increases or as the distance from the neutral axis increases, therefore in a simply supported beam: - Maximum moment occurs at the center span - Beam is subject to highest bending stress at extreme top and bottom fibers - In order for a beam to support loads, material, size and shape of beam must be selected to sustain the resisting moments at the point on the beam where the moment is greatest Section modulus: ratio of the beams moment of inertia to the distance from the neutral axis to the outermost part of the section (extreme fibers) S = I/c Shear: tendency of two adjacent portions of the beam to slide past each other in a vertical direction Horizontal shear: two adjacent portions of a beam to slide past each other in the direction parallel to the length of the beam Horizontal shear stress is: V h = VQ/Ib Rectangular sections horizontal shear stress at mid-depth of the beam (where stress is a maximum) V h = 3V/2bd (@ neutral axis of beam) Types of Beams Basic beam types: all have ends that are free to rotate as the load is applied - Simply supported - Overhanging - Continuous beam held up by more than two supports Other beam types: one or both sides restrained against rotation - Cantilever beam - Fixed-end beam

2 Typical load types - Concentrated load (P) - Uniformly distributed load (W) - Resultant of uniformly distributed loads is at the center of the load - Useful when summing moments of partial uniform loads Statically determinate: reaction can be found using the equations of equilibrium OR summation of horizontal, vertical and moment forces equals zero - Simply supported beams - Overhang beams - Cantilever beams Statically indeterminate: more complex calculations required - Continuous beams - Fixed-end beams Basic requirements for structural design of a beam - Determine stresses due to bending moment and vertical shear caused by particular loading conditions - However, before these are determined reactions of supports must be calculated - Principles of equilibrium - Sum of all vertical forces acting on body must equal zero - Sum of all horizontal forces acting on body must equal zero - Sum of all moments acting on body OR the moment of all forces about a point on the body must equal zero Example 4.1: Find the reactions of the beam shown W P 1 P 2 R 2 + 3(500 x 6 ) + (2500 x 10 ) + (7000 x 17 ) 14R 2 = ,000-14R 2 = ,000-14R 2 = 0 (solve for R2) -153,000 + (+ 153,000-14R 2 ) = 0 153,000-14R 2 = -153,000 (1/14) 14R 2 = 153,000 (1/14) (negatives cancel out, Isolate R2) R 2 = 153,000/14 R 2 = 10, lbf - Since summation of all vertical forces must be equal reaction R 1 can be found by subtracting 10,929 from the total of all loads acting on the beam. W P 1 P 2. [(500 x 6 ) ] = 1,571 R 1 = 1,571 lbf - Once all loads and reactions are known, shear and moment diagrams can be drawn - Critical to know values at points where shear and moment are at their maximum and where they are zero

3 Shear Diagrams Graphic representation of the values of vertical shear anywhere along a beam To find values: - Take a section at any point and algebraically sum the reactions and loads to the left of the section - Same can be done by taking values on right but convention is to work from left to right Example 4.2: What is the vertical shear at points 4 and 10 to the right of reaction R 1? - First: compute reactions using summation of moments - Solve for R 2 about R 1 : P 1 R 2. + (8000 x 8 ) 20R 2 = 0 +64,000-20R 2 = 0-64,000 + (+64,000-20R 2 ) = 0 64,000-20R 2 = -64,000 (1/20) 20R 2 = -64,000 (1/20) (negatives cancel out, Isolate R2) R 2 = 64,000/20 R 2 = 3,200 lbf - Solve for R 1 : Total sum of forces R 2 P 1 R ,200 = R 1 R 1 = 4,800 lbf - Second: consider the point 4 from the left reaction A-A - At this point there is only the upward reaction of 4800lbf - At the point 10 from the reaction there are two forces to the left - Reaction R 1 of 4800lbf - Load of 8000lbf - This net shear is 3200lbf which is exactly the same as the reaction R 2 b/c no other loads or reactions so the summation of all vertical loads equals zero as dictated by the Principles of Equilibrium - Draw the shear diagram of the beam - Draw the beam and loads to scale (see a ) - Draw horizontal line length of beam and decide on vertical scale to represent the loading - Beginning from the left, positive values plotted above the line and negative values below - Notice: - When no intervening loads between reaction and load (or between two loads) the portion of the shear diagram is a horizontal line - Concentrated load or reaction causes shear diagram to change abruptly in the vertical direction

4 - Uniform loading of the simply supported 1500plf - Calculate total load 1500plf x 20 = 30,000lbf - Total load is equally distributed between two reactions R 1 the only force is the reaction of lbf - Long version for clarity solving for reaction R 2 about R 1 W R 2. + (1,500 x 20 ) 20R 2 = ,000 20R 2 = 0-300,000 + (+300,000 20R 2 ) = 0 300,000-20R 2 = -300,000 (1/20) 20R 2 = 300,000 (1/20) (negatives cancel out, Isolate R2) R 2 = 300,000/20 R 2 = 15,000lbf - Beginning at R 1 for the remainder of the beam the uniform load acts in a downward direction from reaction, shear = 15,000 (1 )(1500) = 13,500lbf from reaction, shear = 15,000 (3 )(1500) = 10,500lbf mid-span, shear is zero and begins to be negative value b/c accumulating load is now larger than the reaction R 1 - At reaction R 2, the positive value of the reaction brings the shear back to zero which is consistent with the Principles of Equilibrium Moment Diagrams Graphic representation of the moment at all points along a beam To find the moment at any point remember: - Bending moment is the algebraic sum of the moments of the forces to the left of the section under consideration - Moment is the value of the force times distance: M = PL Example 4.3: Find the moments at sections A-A and B-B

5 - To visualize, draw free-body diagrams of the two conditions - Solving for moment about A-A, moment at the reaction is zero since the moment arm distance is zero - At point A-A, 4-0 from the left reaction the moment is: R 1 _ + (4 x 4,800lbf) = 19,200 ft-lbf - At point B-B, 10-0 from the left reaction the moment is: R 1 P _ +(10 x 4,800) (8000 x 2 ) = 32,000ft-lbf - B/c maximum moment occurs where shear diagram crosses zero, calculate moment at this point also. This is the point where the concentrated load (P) occurs - The moment just to the left of the point load (P) is: R 1 _ +(8 x 4,800) = 38,400ft-lbf - Draw the moment diagram - Calculate moments at several points (all reactions & forces) and connect them with line - Notice: - First: maximum moment does occur where shear diagram passes through zero and is indicated by the highest point of the moment diagram - Second: when the shear diagram is a constant horizontal line between two concentrated loads or reactions, the moment diagram between these two points is a straight, constant sloped line - Third: where shear changes abruptly, the slope of the moment line also changes abruptly

6 Example 4.4: Draw the moment diagram for the uniform load of 15000plf over the simply supported beam. (from example 4.2) - Since the distance of the moment arm at each reaction is zero, the moment at these two points will also be zero - Maximum moment will occur where shear diagram passes through zero so start here - When calculating moments due to uniform loads, moments resultant is at the center of the uniform loads and therefore moment of those loads is calculated as if the total load were concentrated at the midpoint of the loads - This is why the uniform load on the beam is multiplied by 5 - It is recorded as a negative number b/c it tends to cause a counter clockwise rotation about the center of the beam where moments are taken W (total) w (midpoint of uniform load bet. R1 & W) M = (15,000 x 10 ) [(1,500 x 10 )5] M = 150,000 75,000 M 75,000ft-lbf - Note: - Area of the shear diagram between two points along the beam is numerically equal to the change in moment of the beam between the same two points. Thus at the midpoint of the beam as shown the area of the triangle is [(15,000 x 10 )/2] = 75,000ft-lbf Example 4.5: find the reactions and draw the shear and moment diagrams of the beam

7 - First: find reactions sum of moments about one reaction and set them equal to zero - Solve for R 2 about R 1 : W P 1 P 2 R 2. [(1000 x 10 )5] + (3000 x 6 ) + (5000 x 15 ) 12R 2 = , , ,000 12R 2 = ,000 12R 2 = 0-143,000 + (+143,000) 12R2 = 0 143,000-12R 2 = - 143,000 (1/12) 12R 2 = 143,000 (1/12) (negatives cancel out, Isolate R2) R 2 = 143,000/12 R 2 = Solve for R 1 : Total sum of forces R 2 W P 1 P 2 R 2. 10, , ,000 11,916.7 = 6, Second: Find vertical shears and draw the shear diagram starting with left reaction - Construct free-body diagrams - Starting with the left reaction R 1, the shear is lbf in the upward (+) direction - Shear just to the left of the 3000lbf load is the sum of the loads and reactions so at this point it is: V = (1000 x 6 ) V = 83.3lbf - At P lbf load, the shear is: V = (1000 x 6 ) 3000 V = lbf - Since the load between R 1 and P 1, 3000lbf is uniformly distributed, the two points connect with a sloped line - Shear at the end of the uniform load is found in the same way. From this point to the reaction R 2 there are no loads or reactions so the line is horizontal - At reaction R 2 the load changes abruptly in the magnitude of the reaction or 11,916.7lbf which is added to the negative shear of lbf given a net value of +5000lbf - No other loads are encountered until the P lbf load at the end of the overhang. Since it is downward direction (-) it brings the shear to zero at the end of the beam - Expected result b/c of the Principles of equilibrium - Third: Calculate and draw the moment diagram - In this example, there are two points where the shear diagram crosses zero will be two maximum moments - One positive similar to simply supported beam - One negative means the beam is bending upward above a support instead of downward b/c the way the loads are applied - Both moments need to be calculated to determine which is the greather - Highest value is one used to design beam

8 - Two ways to find the moments: - Draw free-body diagrams at each point of interest and take the moment about that point - Find the area of the shear diagram between points of interest (simpler) - To find the maximum moment at the P1 3000lbf load, find the area of the trapezoid with the bases of lbf and 83.3lbf and the height of 6-0 A = 1/2(b1 + b2)h A = 1/2 ( ) x 6 A = 18,500ft-lbf - To find the moment 10 to the right of the reaction R1 find the area of the trapezoid from the P1 3000lbf load to the end of the uniform load and subtract it from the previous moment. - This is because this area is below the baseline oof the shear diagram and is negative A = 1/2 ( ) x 4 A = 19,667ft-lbf - The moment is therefore: 18,500 19,667 = -1167ft lbf - Next calculate the area of the rectangle from the end of the uniform load to the reaction R2 (13,833ft-lbff) and subtract this from the last value COLUMNS Resist axial compressive forces Other considerations: - Tendency of a long slender column to buckle even though the columns material and size can withstand the load - Column will fail in buckling under a much smaller load F a = P/A - Combined loading: due to the normal compressive force plus lateral load (wind) - Combined loading can also be induced by an eccentric compressive load applied off the centroidal axis of the column - In case column acts like a beam standing on end with one face in compression & other in tension - Compressive forces due to the eccentric load add to the normal compressive stress on one side and subtract from the normal compressive stresses on the other - Flexural stress caused by eccentricity is given by F = M c /I Radius of Gyration Ability of a column to withstand a load is dependent on - Length - Cross sectional shape - Area - Moment of inertia Combined properties of area and moment of inertia is called radius of gyration R = I/A

9 Slenderness Ratio Most important factor in column design Slenderness ratio = l/r

Chapter 4.1: Shear and Moment Diagram

Chapter 4.1: Shear and Moment Diagram Chapter 5: Stresses in Beams Chapter 6: Classical Methods Beam Types Generally, beams are classified according to how the beam is supported and according to crosssection