Bending of straight beams • In mechanics of materials we cover symmetrical cross sections and bending in one plane. Now we will consider the more general case • Moment perpendicular to a plane at an angle phi from x-z plane (plane of loads). Centroidal axes. Cantilever beam with an arbitrary cross section subjected to pure bending Shear loading • We assume for now pure bending with no twist. This implies shear forces passing through shear center Cantilever beam with an arbitrary cross section subjected to shear loading Symmetrical bending • Moments of inertia I = y 2 dA x ∫ I = x2 dA y ∫ J = ∫ r 2 dA I = xy dA xy ∫ • Moments of inertia also a tensor so has principal axes Symmetrical and anti-symmetrical cross sections Equilateral triangle Open channel section Angle section Z- section • Are these also principal axes? Symmetrical bending • Euler-Bernoulli beam theory, Leonhard Euler (1707-1783) and Daniel Bernoulli (1700-1782) • What are the assumptions? • For symmetrical cross section M Y x M X y σ zz = − + IY I X • Neutral axis σ zz = 0 Rectangular cross section • Maximum bending stress 6 | M | σ = X max bh2 Cantilever beam with rectangular cross section Unsymmetrical bending • Equations of equilibrium 0 = σ dA ∫ zz M = yσ dA x ∫ zz M = − xσ dA y ∫ zz • Plane sections remain plane ∈zz = a′+ b′x + c′y σ zz = E ∈zz σ zz = a + bx + cy • Combining it all ⎛ M I + M I ⎞ ⎛ M I + M I ⎞ σ = − ⎜ y x x xy ⎟ x + ⎜ x y y xy ⎟y zz ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ I x I y− I xy ⎠ ⎝ I x I y− I xy ⎠ Moments • Moment is perpendicular to the plane of the loads. If the plane of the loads makes an angle φ with the x-axis, M x = M sinφ and M y = − M cosφ M y cotφ = − ⇒ M y = − M x cotφ M x Neutral axis • For bending stress to be zero ⎛ M I + M I ⎞ σ =⎜ x xy y x ⎟ x = x tanα zz ⎜ ⎟ ⎝ M x I y+ M y I xy ⎠ ⎛ M I + M I ⎞ tanα =⎜ x xy y x ⎟ ⎜ ⎟ ⎝ M x I y+ M y I xy ⎠ I − I cotφ tanα = xy x I y − I xy cotφ Pure bending of a nonsymmetrically loaded cantilever beam Example 7.3 • A cantilever beam of length 3m as shown in the figure has a channel section. A concentrated load P=12.0 kN lies in the plane with an angle φ = π/3 with the x-axis. The plane of the loads passes through the shear center C. Locate points of maximum tensile and compressive stresses and find the magnitude of stresses. Location of max stresses Section properties : 2 6 4 A=10000mm I x =36.69 x10 mm I xy =0 6 4 y0 =82.0mm I y =30.73x10 mm Locate Neutral Axis : I − I cotφ I tanα = xy x = − x cotφ I y − I xy cotφ I y ⎛ π ⎞ cos⎜ ⎟ π ⎛ π ⎞ ⎝ 3 ⎠ φ = ⇒ cot⎜ ⎟ = = 0.5774 3 ⎝ 3 ⎠ ⎛ π ⎞ sin⎜ ⎟ ⎝ 3 ⎠ tanα = − 0.7457 ⇒ α = − 0.6407 rad Since moment is negative, the part above N.A is in tension and the bottom Moments : part is in compression. Therefore M = − 3.00P = − 36.0kN.m maximum tensile stress occurs at M = M sinφ = − 31.18kN.m point A and maximum compressive x stress at point B. Maximum stresses Stress at A(-70,-118) . Mtx ()yx− anα σ zz = IIxx− ytanα Mx ()−−118 ()−70 tanα σ A ==133.7 MPa IIxx− ytanα Stress at B(70,82) M8x ()2− ()70tanα σ B ==−105.4MPa IIxx− ytanα Deflections • Determine separately x and y components of displacement. Here we show y component •Curvature 2 1 ∈zz 1 d v − = where ≈ 2 Ry Ry Ry dz 2 d v M x M x I y + M y I xy 2 = − = − 2 dz E(I x − I xy tanα) E(I x I y − I xy ) Total displacement u = − v tanα v δ = u 2 + v2 = cosα Components of deflection of a nonsymmetrically loaded beam. Example 7.6 • A simply supported beam of length 3m has a channel section. A concentrated load P= 35.0 kN applied at the center of the beam, lies in a plane with an angle φ = 5π/9 with the x-axis. Locate points of maximum tensile and compressive stresses and magnitude of stresses. Find the maximum deflection. E=72 GPa. Section properties : 2 6 4 A=10000mm I x =36.69 x10 mm I xy =0 6 4 y0 =82.0mm I y =30.73x10 mm Solution Stress : M (y − x tanα) Locate Neutral Axis: σ = x zz I − I tanα I x xy tanαφ=− x cot ⇒φ=5π9 M x (−118−(−70) tanα) I y σ = = 63.8MPa A I − I tanα tanαα=⇒0.2277 =0.2239 rad x xy M x (82−(70) tanα) σ B = = −87.2 MPa I x − I xy tanα Deflection Simply supported beam : PL3 ∆ = Moment : 48EI PL3 sinφ PL v = = 6.78 mm M = = 26.25kN.m 48EI x 4 u = − v tanα = −1.54 mm M x = M sinφ = 25.85kN.m v δ = u 2 + v2 = =6.95 mm cosα Reading assignment Sections 7.3-5: Question: Consider a horizontal cantilever beam under a tip vertical load. What condition is required so that tip will move both down and sideways? Source: www.library.veryhelpful.co.uk/ Page11.htm.