Unit 1: Bending and Stresses in Beams

SM Bending & Stresses in Beams INTRODUCTION Many structural members can be classified as Thin walled Open or closed cross-section The structural members are subjected to axial forces, shear force & bendang Require checks with von Mises' yield criterion to ensure that local yielding of the material does not Occur MOMENTS OF AREA CENTROID Longitudinal stress due to a bending moment M/I = E/R = 8/Y Need to determine I, which is the moment of the first moment of area of the cross-section about an axis called it second moment of area The x, y & & arxes pass through the centred of the cross-sectron C 0 The x' * y' * z' axes pass through an arbitrary origin O & are parallel to the x, y & u. V k w denote displacements in the x, y & & directions w Area A= iint^ rho dA= iint^ rho d * y' * d * z' First moment of area M y' = iint z^ prime dA= iint z^ prime dy^ prime dz^ prime M 0 * y' = iint y^ prime dA= iint y^ prime dy^ prime dz^ prime y.v B delta*A = (delta * y')' * delta * x' = delta_{y}*delta*x C Coordinates of controld y' c = M * y^ prime A B_{e}' = (Mo * g')/A Second moments of arex I o * y' = iint z^ prime ^ 2 d lambda= iint z^ prime ^ 2 dy^ prime dz^ prime 18 I 0 * y' = iint^ beta y^ prime ^ 2 dA= iint y^ prime2 ^ 2 dy^ prime dx^ prime Product second M of A I a * y' * y' = iint^ theta y^ prime y^ prime dA= iint y^ prime z^ prime dy^ prime dz^ prime Three parallel axis theorems are useful in determining the section properties of profiles made up of on assemblage of discrete areas 0 y^ prime =y^ prime 2 + y z' = 3' * c + z Lay SS 3dASS(z²+2zz+zé²) dA = I + Αγό Iog' SSy" dASS(y+zyyé+ y²²) dA= Icy + Aye Joy's Sly's'da = SJ (9) + yz + zyć + yézé) da = Ioys + Ayézé EXAMPLE Rectangular area Determine the second moments of area Area = bd Second moments of area at C (centroid) (y z axes) I by = iint z^ 2 dA =b[ gamma^ 3 3 ] (- d) / 2 ^ (d / 2) = (b * d ^ 3)/24 - (- (b * d ^ 3)/(2 ^ 4)) = int -d/2 ^ +d/2 overline y ^ 2 * bdz = (6d ^ 3)/12 I cy = iint^ 1 Y^ 1 dA = int -b/x ^ +b/z y^ z .d d y =d[ y^ 3 3 ] (- b) / 2 ^ (4b / 2) = (d * b ^ 3)/24 - (- (d * b ^ 3)/24) = (d * b ^ 3)/12 d/2 b/2 d y C Product second moment of aren I ayj = iint y* y* dA=0 because of symmetry zoro with respect to any p of aves in hitich at least one axis is an arts of sytometry. (tue) & (-ve) components cancel Second moments of area I o * y' =I ey +A gamma c ^ prime ^ 2 at O (arbitory axes origen 0) (y' z' aves) = (b * d ^ 3)/12 + bd * (d/2) ^ 2 = (b * d ^ 3)/12 + (b * d ^ 3)/4 = (b * d ^ 3)/3 I 0 * q' =I Cq +Ay c ^ 2 = (d * b ^ 3)/12 + (b * d ^ 3)/4 = (d * b ^ 3)/12 + bd * (b/2) ^ 2 = (d * k ^ 3)/12 Ιουζ' = Ιηζ + Αγόζί = 0 + b * d(d/2)(b/2) = (b ^ 2 * d ^ 2)/4 EXAMPLE 2 T-section Determine second moments of area about the centroid al axes Area = (100×10) x 2 = 2000 mm² Moy' = JS z'dA M 0 * y' = boxed 100*10*50 + boxed 100*10*50 1mn mm^ 3 aligned = sum 1 dA\\ =[100*10*5] [100 * 10 * 5] * 65ooom * m ^ 2 + boxed 100*10*6 \\ aligned gamma^ prime epsilon = M ny ^ prime A = 65000 2000 =32.5mm matrix prime prime&y epsilon ^ prime =& (M_{0} * y')/A &=&10000 2500 &50 mm 100x10mm² 100x10mm Second moments of anex I_{y} = Sigma((b * d ^ 3)/12 + A * gamma_{e}' ^ 2) flange W I 4f = Sigma( (b * d ^ 3)/12 +A* y 4 ^ prime2 ) Icygo (due to axis of symmetry) = boxed 100*10 ^ 3 12 +100*10*(32.5-5)^ 3 + boxed (10 * 100 ^ 3)/12 + 100 * 10 * (50 + 10 - 31.5) ^ 2 = boxed (10 * 100 ^ 3)/12 + 10 * 100 * 0 + (100 * 10 ^ 3)/12 + 100 * 10 * 0 = 841 * 667m * m ^ 4 = 2354167mi * n ^ 7 PRINCIPAL AXES The orthogonal axes passing through the centroid about which the product second moment of area T cgY =0 3 is the clockwise angle from the arbitary y' - y' axes to the principal y-z axes β C The y y coordmates of a point on the anss section are related to its y' * z' coordinates by the equation y = y' * cos beta + gamma * sin beta z = z' * cos beta - y' * cos beta Arbitary axes Principle cours delta*A = delta * y' * delta * z' = delta*y*delta*z The product second moment of area relative to the the y-z axes I yy = iint y y dA = iint^ rho (y' * cos beta + z' * sin beta)(z' * cos beta - y' * sin beta) * dA =: iint\ y^ prime y^ prime \ cos^2 beta - sin^2 beta )+\ y^ prime2 -y^ prime2 )sinp cos beta\ dA =I y' * gamma' cos 2 beta-(Ig^ prime -Ig^ prime ) sin 2 beta 2 = iint\ y' * y' * cos 2beta -(y^ prime i -y^ prime a ) sin 1 beta 2 \ dA Equating I cyl =0 tan 2 beta= 2I*epsilon * y' * y' I cy^ prime -I epsilon y^ prime can be wed to Transform section pap ang yg anes to any other y- ares (net neorssarily principal axes) Second moments of area about the principal y & J doves I oy = iint gamma^ 2 dA = iint^ rho (y' * cos beta - y' * sin beta) ^ 2 * dA = iint(y^ prime2 cos^2 beta - 2 * y' * y' * sin beta * cos beta +y^ prime2 sin^ 4 beta)dA I_{y} * cos^2 beta - 2Icyy * sin y * cos beta + Icy * sin^2 beta I q = iint y ^ 2 * dA = iint (y' * cos beta + y' * sin beta) ^ 2 * dA = iint( y' ^ 2 * cos^2 beta + 2 * y' * y' * sin beta * cos beta +y^ prime2 sin^ 2 beta)dA = Ic * y' * cos^2 beta + 2Io * y' * gamma' * sh*beta * cos beta + I * y' * s * h ^ 2 * beta or asymmetry, then the centriod lies on that ans If a section possesses an axis of symmetry If a section has two orthogonal aves of symmetry then the principal axes coincide with these axes EXAMPLE Determine the orientation of the principal axes including Ing, Icy k Icyz for the thon walled 2. section of thickness tmm, assuming t<<6. t<< 6 mam 6mm 6mm 6mm 6 mm 5-22.5 6mm :: Ignore terms containing t Icy' = web tx123 12 - 576t + 3 Range x2 b+bxt6 12 144t |--Sin=-0.3/26 Iug y^ prime = boxed (12xtxD)/(sin t) + boxed 6* t*(-3)(-6) + boxed 6* t*(+3)( pi6) Iq^ prime = boxed 12* q^ 3 12 + boxed ((4 * 6 ^ 3)/12 + 6t * 3 ^ 2) * 2 tan 2 beta= 2 I cy^ prime y^ prime I c * y' - Ic * y' beta = - 22.5 deg cos beta = 0.92388 2beta =-45^ + sin 2beta = - 0.7671 = (2(216t))/(144t - 578t)