Bending BADI 1 John Errington MSc Loaded Beams • When a beam is supported at both ends and loaded it will bend. • The amount by which it bends depends on – the load, and the way it is distributed – the distance between supports – the elastic modulus for the material, and – the distribution of material in the beam Loaded beam F 2x For our design we need to know how much the beam will bend, and the maximum stress in the beam to determine whether it will break. Here we have a solid square section beam, centrally loaded with a force F and unsupported distance 2x The beam bends into an arc, so the top is shorter than the bottom. This means the top is under compression, and the bottom under tension. The material in the middle (the neutral axis) contributes less to the strength of the beam than the material in the top and bottom quarters of the beam. Figure 1. A side view of a simply supported beam (top) bending under a distributed lateral load (bottom). Figure 2. The internal forces and the cross-sectional stress distribution of a beam in bending. Note the position of the neutral axis Joists and floorboards Floor loading • The timber floor of a property obviously weighs something. This weight consists of the timber joists, the plasterboard ceiling underneath it, and the floorboards. All of these are known as "the dead load" and the joists themselves must be able to support this dead load without sagging. For a normal property, with chipboard or timber floors and plasterboard ceilings, this dead load is generally taken to be no more than 0.50 kiloNewtons per square metre. (kN/sq.m) • The weight we place upon a floor by way of bathroom suites, beds, wardrobes etc, is known as the "imposed load". It is again accepted that, for normal household requirements, the imposed load will not exceed 1.5kN/sq.m. • The building regulations tables A1 and A2 list the size of joist necessary to support this weight, over a maximum span. Table A2 uses timbers known as "SC4" which are high strength timbers containing very few, if any, knots. They are not common in modern day construction unless specified and we will deal with the more generally used timbers, dealt with in table A1. These timbers are known as SC3 and will have C16 stamped on them. This table is for a dead load of more than 0.25 but not more than 0.50 and allows for an imposed loading of no more than 1.5 kN/sq.m. This is an abridged representation of Building regulations Table A1 Spacing (distance apart) of joists. 400m m Spacing (distance apart) of joists. 450m m 600m m Size of joist 38 x 97 38 x 122 38 x 140 38 x 147 38 x 170 38 x 184 38 x 195 38 x 220 38 x 235 mm mm mm mm mm mm mm mm mm 1.72 2.37 2.72 2.85 3.28 3.53 3.72 4.16 4.43 m m m m m m m m m 1.56 2.22 2.59 2.71 3.1 3.33 3.52 3.93 4.18 m m m m m m m m m 1.21 1.76 2.17 2.33 2.69 2.9 3.06 3.42 3.64 m m m m m m m m m 47 x 97 47 x 122 47 x 147 47 x 170 47 x 195 47 x 220 mm mm mm mm mm mm 1.92 2.55 3.06 3.53 4.04 4.55 m m m m m m 1.82 2.45 2.95 3.4 3.89 4.35 m m m m m m 1.46 2.09 2.61 2.99 3.39 3.79 m m m m m m 400m m 450m m 600m m Size of joist 50 x 170 mm 50 x 195 mm 50 x 220 mm 3.61 m 4.13 m 4.64 m 3.47 m 3.97 m 4.47 m 3.08 m 3.5 m 3.91 m 63 x 97 63 x 122 63 x 147 63 x 170 63 x 195 63 x 220 mm mm mm mm mm mm 2.19 2.81 3.37 3.89 4.44 4.91 m m m m m m 2.08 2.7 3.24 3.74 4.28 4.77 m m m m m m 1.82 2.45 2.95 3.4 3.9 4.37 m m m m m m 75 x 122 75 x 147 75 x 170 75 x 195 75 x 220 mm mm mm mm mm 2.97 3.56 4.11 4.68 5.11 m m m m m 2.86 3.43 3.96 4.52 4.97 m m m m m 2.6 3.13 3.61 4.13 4.64 m m m m m Steel and concrete • Steel or concrete are often used to support loads, in large buildings and for bridges. • Steel is good under tension and compression • Concrete is excellent under compression, and is lighter than steel, but has low tensile strength Steel sections Rounds Hexagons Squares Channels "H" Beams "I" Beams Flats Equal Leg Angles Unequal Leg Angles Tee Bars Diamond Floor Plate Rectangular Tubing Seamless and Welded Pipe Seamless and Welded Tubing Half Rounds Bending moment for a point load A uniform beam 5m long is supported at its ends and has a load of 6kN at 3m from one end. 1. What is the reaction in the supports? 2. What is the bending moment? 3m 6kN 5m Calculate reaction in supports 3m 6kN A B 5m Ra Rb CW = clockwise CCW = counter clockwise Taking moments about A we find a CW moment of 3m * 6kN and a CCW moment of -5m * Rb As the system is in equilibrium CW = CCW so 3m * 6kN = -5m * Rb Rb = - 18/5 kN = -3.6kN Also Ra + Rb = - 6kN Ra = - 6kN – (-3.6kN) = -2.4kN Ra = - 2.4kN Rb = - 3.6kN Bending moment for beam with point load 6kN 3m Xm A B To find the bending moment at a point p, x metres from a we take moments about P. To the left of P: Reaction Ra causes a CW moment of 2.4x p 5m Ra 2.4kN Rb 3.6kN To the right of P: The load causes a CW moment of 6 * (3-x) Reaction Rb causes a CCW moment of 3.6*(5-x) Total moment = (18 -6x) – (18 -3.6x) = -2.4x i.e. 2.4x CCW So the bending moment for this loaded beam at a distance x from A is Mx = 2.4x. Bending moment and shear force for point load Plotting BM vs x we get this graph, showing BM is a maximum of 7.2 at x =3. We can also plot the shear force acting on the beam. Imagine the beam cut at point p and work out the force acting on Bending moment and shear force for point load 8 Bending moment 7 Shear force 6 5 4 the left side of the cut. 3 2 1 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -1 -2 A -3 6kN 3m Xm 5 B p -4 distance from A 5m Ra 2.4kN 3.6kN Bending moment for a distributed load A beam 3m long is supported at its ends and has a distributed load of 2 tonnes/m. 1. What is the reaction in the supports? 2. What is the bending moment? A 2 tonnes/m 3m B Calculate reaction in supports A 2 tonnes/m B 3m Ra Rb Distributed load of 2 tonnes / m = 2000kg * 9.81 = 19620N / m The total weight of the load is 19620N * 3m = 58860N By symmetry this is evenly split between the ends, so Ra = Rb = 588600/2 = 29430N Bending moment for beam with distributed load A 19620N/m B p x 3m Ra 29430N Rb 29430N CW = clockwise CCW = counter clockwise To find the bending moment at a point p, x metres from a we take moments about P. To the left of P: Reaction Ra causes a CW moment of 29430x The load causes a CCW moment of 19620 * x *x/2 = 9810 x2 Bending moment Mx= 29430x - 9810 x2 We can show (by plotting, or by differentiating this expression and looking for stationary turning points, then differentiating again) that the maximum bending moment M! for this loaded beam occurs at the mid point, where M! = 29430* 1.5 - 9810 *1.5*1.5 M! = 44145 – 22072 = 22073 Nm Bending moment and shear force for distributed load Bending moment and shear force for distributed load 30000 BM SF 20000 10000 0 0 0.5 1 1.5 -10000 -20000 -30000 Distance from A 2 2.5 3 Here you can see that the bending moment is a maximum at the centre, while the shear force is a maximum at the ends. TO BE CONTINUED • Enough for one week. • Next week we will use this information to evaluate how much stress we will get in a loaded beam, and how much it will sag. Calculate maximum stress in beam and radius of curvature of neutral axis 0.2m 0.1m Thickness t = 0.02m Suppose our beam from the previous example was a steel box girder with dimensions as shown here. • How much would it bend? • Would it break? To answer these questions we need to calculate the radius of curvature , and the maximum stress in the beam. HOW? /y=M/I=E/R Where: is the strain at y metres from the neutral axis M is the bending moment I is the Moment of Inertia for the beam E is Young’s Modulus of elasticity for the material of the beam, and R is the radius of curvature at the neutral axis Maximum stress The stress is a maximum at the outer faces (top and bottom) of the beam, at a distance D/2 from the neutral axis, where D is the depth of the beam. Using / y = M / I = E / R ! = M! D / 2 I Remember we are using ! to signify a maximum value For our beam M! is 22073 Nm and I = 0.1 * 0.23 /12 – 0.06 *0.163 / 12 I = 6.667*10-5 -2.048*10-5 I = 4.619*10-5 m4 (Formula for calculation of I shown later) ! = 22073 * 0.2 / 2 * 4.619*10-5 ! = 47.8 * 106 Pa (1Pa = 1N/m2) Steel has a tensile strength of 1.5*109 N/m2 so no problem Radius of curvature For our steel beam M! is 22073 Nm E = 2.1 * 1011 N/m2 I = 4.619*10-5 m4 Using / y = M / I = E / R R=IE/M The radius of curvature is given by R = (4.619*10-5 )* (2.1 * 1011) / 22073 R = 439m So how much will our beam sag? (calculated) (tables) calculated) How far does the beam sag? The amount by which the beam sags, x is found using Pythagoras theorem. R2 = a2 + (L/2)2 a2 = R2 - (L/2)2 a = sqrt(R2 - (L/2)2) x = R - sqrt(R2 - (L/2)2) R a x L/2 Substituting values for R=439m and L=3m x = 439 - sqrt(4392 - (3/2)2) x = 439 - sqrt(192721 - 2.25) x = 439 – 438.9974 = 2.56 *10-3 m The beam sags by 2.56mm in the centre Moment of Inertia Defined • The moment of inertia measures the resistance to a change in rotation. – Change in rotation from torque – Moment of inertia I = m * r2 for a point mass • The total moment of inertia I is due to the sum of masses at a distance from the axis of rotation. N I mr i i i 1 2 Moment of Inertia More importantly for our purposes the moment of inertia describes the way in which the mass of the body is distributed. As we have already seen we can use this parameter to evaluate how a regular isotropic beam will respond to a load. Mass at a Radius • Extended objects can be treated as a sum of small masses. • A straight rod (M) is a set of identical masses Dm. • The total moment of inertia is I (Dm )r 2 • Each mass element contributes Dm (M / L )Dr I ( M / L ) r D r 2 distance r to r+Dr • The sum becomes an integral L length L I ( M / L ) r dr 2 0 I ( M / L )( L / 3 ) (1 / 3 ) ML 3 axis 2 Rigid Body Rotation • The moments of inertia for many shapes can found by integration. – Ring or hollow cylinder: I = MR2 – Solid cylinder: I = (1/2) MR2 – Hollow sphere: I = (2/3) MR2 – Solid sphere: I = (2/5) MR2 Point and Ring • The point mass, ring and hollow cylinder all have the same moment of inertia. • The rod and rectangular plate also have the same moment of inertia. I = (1/3) MR2 I = MR2 • The distribution of mass from the axis is the same. • All the mass is equally far away from the axis. M M R M R length R M axis length R Moment of Inertia of common shapes y is the distance of the NA from the top of the shape. ** indicates raised to power e.g. D**3 = D cubed Moment of Inertia of a Cross Section Box Rectangle h h2 b I rect h1 b2 1 b1 bh 12 3 I box 1 12 3 1 b1 h 1 12 3 b2 h2 A box girder of dimension 0.2m * 0.1m and 0.02m thickness has a csa of 0.012m This amount of material would make a solid bar of 0.12m * 0.1m I(bar) = 1.44 *10-5 m4 I(box) = 6.66 *10-5 – 1.36 *10-5 = 5.3 *10-5 m4 Other shapes • We can calculate the M of I for other shapes just by adding or subtracting, as we did for the box girder. See also next slide. • We can evaluate the M of I at a different point within the shape, using the parallel axis theorem • We can evaluate the M of I about an axis in another plane, using the perpendicular axis theorem Comparison of moments of inertia for different shapes all having the same cross-sectional area 1m 1m a: Ia =bh**3 / 12 = 1*(1*1*1)/12 = 1/12 Ia = 1/12 a b: Ib = I(b outer) - I(a) = (1.414 * 1.414**3) /12 – 1/12 1.41m = 4/12 – 1/12 = 3/12 Ib = 3/12 b c: Ic = I c 2m -I -I +I Ic = { (2*(2**3)) - (1*(2**3)) - (2*(1**3)) + (1*(1**3)) } / 12 Ic = { (2*(8)) - (1*(8)) - (2*(1)) + (1*(1)) } / 12 Ic = { 16 - 8 - 2 + 1} / 12 = 7 / 12 Ic = 7/12 Parallel Axis Theorem • Some objects don’t rotate about an end, but rather about the axis at the center of mass. • The moment of inertia for a rod about its center of mass: h = R/2 M • The moment of inertia depends on the distance between axes. axis (1 / 3 ) MR 2 I CM M ( R / 2 ) I CM (1 / 3 ) MR I I CM Mh 2 2 I CM (1 / 12 ) MR (1 / 4 ) MR 2 2 2 Perpendicular Axis Theorem Top view Iy = (1/12) Ma2 b Ix = (1/12) Mb2 M a Side view For flat objects the rotational moment of inertia of the axes in the plane is related to the moment of inertia perpendicular to the plane. Iz = (1/12) M(a2 + b2) Iz Ix Iy To Increase the Moment of Inertia • Increase the size: – But as you increase the size, you increase the weight and cost • Use a denser material – But it may not have the properties you need and it will be heavier • Change the cross-sectional shape: – A hollow cross-section is stronger for the amount of material used Summary We can now work out the stress in a loaded beam, such as a table top or beam bridge. We can find out whether the structure will be strong enough, and how much it will deform under load. Question. A bridge is made of reinforced concrete using pre-stressed reinforcing rods. Should the rods be placed at the top, middle or bottom of the section? WHY?