# M5 Simple Beam Theory (continued)

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1 M5 Simple Beam Theory (continued) Reading: Crandall, Dahl and Lardner In the previous lecture we had reached the point of obtaining 5 equations, 5 unknowns by application of equations of elasticity and modeling assumptions for beams (plane sections remain plane and perpendicular to the mid plane): u( x,y,z) = -z dw dx (1) w( x,y,z) = w( x) () e xx = -z d w dx (3) e xx = s xx E x (4) s xx x + s zx z = 0 (5) 5 eqns, 5 unknowns : w,u,e xx,s xx,s xz Solutions: Stresses & deflections First consider relationship between stresses and internal force and moment resultants for beams (axial force, F, shear force, S, and bending moment M). Consider rectangular cross section (will generalize for arbitrary cross-section shortly)

2 Resultant forces and moments are related to the stresses via considerations of equipollence: F = h Ú s xx bdz (6) -h h S = - Ú bdz (7) -h s xz h M = - Ú s xx bzdz -h (8) Now we combine and substitute between the equations derived above. combine (3) and (4) to obtain: s xx = E x e xx = -E x z d w dx Now substitute this in (7) F = -E x d w dx h h È d w z Ú zbdz = Í -E x dx b -h Î Í -h = 0 Which is correct, since there is no axial force in pure beam case. Similarly in (8) M = E x d w dx +h Ú. -h z bdz We can write this more succinctly as: M = E x I d w dx curvature this is the moment curvature relation - positive moment results in positive curvature (upward). The quantity EI defines the stiffness of beam sometimes called flexural rigidity note that it depends on both the shape and the material property, E.

3 h Where I = Ú z bdz is the second moment of area (similar to moment of inertia for -h dynamics see 8.01) Recall that s xx = -E x z d w dx = -E x z M E x I I yy = z h +y(z) Ú da = Ú Ú z dydz I yy is second moment of area about y axis. A -h -y(z) Note that there is also a moment of inertia about the z axis I zz. In general we will be concerned about the possibility of bending about multiple axes, and the bending of nons xx = - Mz I This is the moment - stress relationship Implies a linear variation of stress, Maximum stress at edges (top and bottom) of beam More general definition of I : See Crandall, Dahl and Lardner 7.5 For general, symmetric, cross-section

4 symmetric cross-section. However, we will not consider these cases in Unified (see 16.0) Note, for particular case of a rectangular section I = 1 1 bh3 [L 4 ] We need to define centroid: center of area (analogous to center of mass). Position calculated by taking moments of area about arbitrary axis. Hence z position given by: z Centroid = Ú A zda Ú da A No net moment about centroid In bending centroid is unstrained also known as the neutral axis or neutral plane Can also use parallel axis theorem to calculate second moments of area I total = Â I i + ÂA i d i Second moment of area about centroid of area area Distance of centroid from total section centroid Useful to simplify problems, e.g. for I beam cross-section (structurally efficient form, effectively used as spars in many aircraft). is equivalent to:

5 1 1 bh3 I total = I y1 y 1 + A 1 d 1 + Iy y + I y3 y 3 + A 3 d 3 Changing area distribution by moving material from the web to the flanges increases section efficiency measured as I A - resistance to bending for a given amount of material used.

6 M6 Shear Stresses in Simple Beam Theory Reading: Crandall, Dahl and Lardner 7.6 Returning to the derivations of simple beam theory, the one issue remaining is to calculate the shear stresses in the beam. We would like to obtain an expression for s zx ( z). Recall: Shear stresses linked to axial (bending) stresses via: s xx x and + s zx z = 0 fi s zx z = - s xx x Shear force (for case of rectangular cross-section, width b) linked to shear stress via: (5) S = - h s xz -h Ú bdz (7) Multiply both sides of (5) by b and integrate from z to h/ (we want to know s zx ( z) Ê and we know that s zx ± h ˆ Á = 0 ) Ë

7 h b s zx s Ú dz = - Ú xx bdz z z z x h [ bs xz (z)] z h h Ê -dm ˆ zb = - Ú Á z Ë dx I dz s xx = -Mz I yy recall b s xz h dm dx = S h Ê ˆ Ë I dz yy ( ( ) - s xz ( z) ) = SÁ zb Ú Note that at z = h (top surface) s zx = 0 Also define z Q = h Ú zbdz z Hence: s xz (z) = - SQ I yy b first moment of area above the center shear stress -force relation For a rectangular section: Q = h Ú z È zbdz = z Í b Î Í h z = b È h Í 4 - z Î Í

8 Parabolic, maximum at z = 0. The centerline. Minimum = 0 at top and bottom "ligaments" For non-rectangular sections this can be extended further if b is allowed to vary as a function of z. I.e. can write: s xz (z) = - SQ I yy b z ( ) But need to beware cases in which we have thin walled open or closed sections where shear is not confined to s xz. Will also have s xy (on flanges). E.g.

9 or We will defer discussion of these cases until So this is the model for a simple beam. What can we do with it?

10 M7 Examples of Application of Beam Theory Reading: Crandall, Dahl and Lardner Example: Calculate deflected shape and distribution of stresses in beam. Refer back to example problem in lecture M3. Example: Uniform distributed load, q (per unit length, applied to simply supported beam. Free Body diagram: Apply method of sections to obtain bending moments and shear forces: S(x) = ql - qx:

11 M (x) = qlx - qx : Let beam have rectangular cross-section, h thick, b wide. Material, Young s modulus E: Deflected shape of the beam: obtain via moment curvature relationship: w EI d w dx = M fi d dx = M (x) EI Integrate to obtain deflected shape w(x) w( x) = M (x) Ú Ú dx dx EI and for EI constant over length of the beam: fi w( x) = 1 EI For our case: Ú Ú M (x)

12 M (x) = -qx (x - L) dw dx = - 1 qx Ú (x - L)dx EI = -q Ê È x 3 EI 6 - Lx ˆ Í Á Ë Î Í 4 + C 1 So w(x) = 1 EI ( q È Ú Í Î Í x Lx 4 + C 1 )dx w(x) = -q È x 4 EI 4 - Lx3 Í 1 + C 1 x + C Î Í C 1 and C are calculated from values of w(x) and w (x) at boundary conditions. Two constants of integration, need two boundary conditions to solve. x = 0, w = 0 fi C = 0 x = L, w = 0 Hence: 0 = -L4 4 + L4 1 + C 1L = 0 fi C 1 = -L3 4 w(x) = - q [ 48EI x4 - Lx 3 + L 3 x] Maximum deflection occurs where: dw(x) dx = 0 dw dx fi 4x3 - blx + L 3 = 0 at x = L w = ql 4 EI yy

13 Find Stresses: s xx = - Mz I yy substitute for M(x) Max tensile (bending) stress at x = L, s xxmax = 3qL 4bh M max = ql 8, z = ± h Shear stresses: s xz = -SQ Ib for rectangular cross-section occurs where max shear force S = x=0, x=l h Q(z) = Ú z bd z z = b z h È Í = b È h Í Î Í z 4 - z z Î Í s xzmax = bh 8 ql 1 1 bh 3 b = 3 ql 4 bh Note that this is much smaller than s xxmax if L/h is large (i.e slender beam).

14 M8 Application of Simple Beam Theory: Discontinuous Loading and Statically Indeterminate Structures See Crandall, Dahl and Lardner 8.3, 3.6, 8.4 Example Free body diagram Shear forces:

15 Bending moments: Note in passing M(x) and hence s xx = -Mz = constant I for L x L - useful for mechanical testing of materials four point bending Could solve for deflections by writing separate equations for: M(x) for 0 < x < L, L < x < L, L < x < 3L and integrating each equation separately and matching slope and displacements at interfaces between sections. This is a tedious process. Or we can use:.

16 Macaulay s Method Relies on the knowledge that the integral of the moment (i.e. the slope, and deflection) are continuous, smooth functions. Beams do not develop kinks. This is a particular example of a class of mathematical functions known as singularity functions which are discontinuous functions, defined by their integrals being continuously defined. Apply method of sections to the right of all the applied loads Write down equilibrium equation for section, leaving quantities such as {x-l} grouped within curly brackets { } M x=x = 0 M + P{x - L} + P{x - L} - Px = 0 {x-a} is treated as a single variable, only takes a value for x > a otherwise { } = 0 Proceed as before - integrate moment-curvature relationship twice, applying boundary conditions M = Px - P(x - L) - P(x - L) = EI d w dx EI dw dx = Px - P { x - L} - P { x - L} + A EIw = Px3 6 - P 6 x - L { }3 - p 6 x - L { }3 + Ax + B apply boundary conditions: w = x = 0 for x = 0 only P(x)3 6 w = x = 3L is evaluated = 0, therefore B = 0

17 for x = 3L fi PL3 6 - PL3 6 - P 6 ( 8L3 ) + A ( 3L ) = 0 fi A = 4 9 PL w(x) = 1 È Px 3 EI 6 - P 6 { x - L}3 - P 6 { x - L}3 + 4 Í 9 PL x Î Í Provides complete description of deflected shape of beam in one equation. Can also use for distributed loads. (Several ways to approach.) Statically Indeterminate Beams Reading: Crandall, Dahl and Lardner, ) By simultaneous application of three great principles i.e., Set up: equilibrium Moment - curvature Geometric Constraints Solve simultaneously (reactions remain as unknowns in momentcurvature equations).) Superposition Decompose problem into several statically determinate structures. Replace constraints by applied loads Solve for loads required to achieve geometric constraints. Example: cantilever beam under uniform distributed load, with pin at right hand end.

18 R B =? Equivalent to: (1) + () From standard solutions or calculation obtain expressions for deflections in case (1) and case ():

19 d 1 = - ql4 8EI, d = + R B L3 3EI fi d 1 + d = 0 ql4 8EI = R B L3 3EI fi R B = 3qL 8 This is a powerful technique for solving beam problems statically indeterminate, or otherwise complex loading. Note. For superposition to work we must be superimposing the loads and boundary conditions on the same beam. Only change one variable in each loading case.

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