Tutorial #1 - CivE. 205 Name: I.D:
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1 Tutorial # - CivE. 0 Name: I.D: Eercise : For the Beam below: - Calculate the reactions at the supports and check the equilibrium of point a - Define the points at which there is change in load or beam shape - Draw the Shear Force Diagram (S.F.D.) and Bending Moment Diagram (B.M.D.). Drawing the diagrams to-scale and show the compression side. 0 KN m a 3 KN / m m 4 m S.F.D B.M.D.
2 Eercise : - Draw the (N.F.D.), (S.F.D.), and (B.M.D.) for the following frame. 0 KN 0 KN KN / m KN.m m KN m m c a 4 m 4 m. m b m N.F.D S.F..D. 3 The column has zero shear C sum of moments = B.M.D. 0
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5 Tutorial 3: S.F.D. 0. m B.M.D 0.3 m Beam cross section 4 M =0 3 M =0 V = Properties of area: I = b.t 3 / = 0.3*0. 3 / = m 4 Q 6, 4 = A. = 0.*0.3*0.87 = m 3 Q 3 = A. = 0.*0.3*0. = m 3 Q,, = 0 m 3 Forces: M = 0 KN.m V = KN Stresses:, = M. / I = -0 * 0. / = -800 KN/m = M. / I = 0 * 0. / = 800 KN/m 6 = M. / I = -0 * 0. / = -400 KN/m 4 = M. / I = 0 * 0. / = 400 KN/m 3 = M. / I = 0 * 0 / = 0 KN/m,, = V. Q / t. I = * 0 / 0.3*0.003 = 0 KN/m 6, 4 = V. Q 6 / t. I = * / 0.3*0.003 = -0 KN/m 3 = V. Q 3 / t. I = * / 0.3*0.003 = KN/m Element, Element3 Element6 Element4 Element
6 Element & 800 = - 800, = 0, = 0 tan θ p = / ( ) θ p = 0 o, min = ( + ) / +/- (( ) / ) +, min = 0, Element = 0, = 0, = tan θ p = / ( ) θ p = 4 o 4 o, min = ( + ) / +/- (( ) / ) +, min = +/ Element o 6.6 = - 400, = 0, = + 0 tan θ p = / ( ) θ p = o 406.6, min = ( + ) / +/- (( ) / ) +, min = 6.6, Element = 400, = 0, = + 0 tan θ p = / ( ) θ p = 7.0 o, min = ( + ) / +/- (( ) / ) o, min = 406.6, Element = 800, = 0, = 0 tan θ p = / ( ) θ p = 0 o, min = ( + ) / +/- (( ) / ) +, min = 800, 0
7 Tutorial 4 - E.: Element Mohr s Circle principle plane ma. shear = (, -) R = min min = 0 = 0 θ p = 90 ο 4 ο From point to ( av, - ma ) = (, -) ( min, 0) (, 0) ma = (0, ) θ s from point to av & ma R =.6 θ p from point to θ p = 9.3 ο = 8. min = -3. θ s = 3.3 ο ma =.6 av =. ( min, 0) ( av, - ma ) (, 0) θ from point to θ p = 3.7 ο θ s = 76.7 ο ( av, ma ) θ s from point to av & ma R =.6 = 8. min = -3. ma ma =.6 av =. 4 ο min θ p = 90 DFrom point to min = - = R = = min = -
8 Element Mohr s Circle principle plane ma. shear min min = = ο R = 7.07 θ p = 4 ο From point to θ s from point to av & ma. ο = 7.07 min = ma = 7.07 ma 3 ( av, - ma ) ( min, 0) (, 0) 4 min = -. = 8. θ s =.9 ο ma R =.0 ( av, ma ) θ s from point to av & ma θ p from point to θ p= 4. ο = 8. min = -. ma =.0 av = 3. = = (, 0) R= 0 θ p = θ s = 0 (one point) 4 ο min R = θ s from point to av & ma ma = av = 0 ma
9 Eercise : 4 kips I. kips 6 kips 7.8 kips I S.F.D. 4 in 39.4 kip.ft 43 kips 3 in Beam cross section B.M.D 40 kip.ft Properties of area: I = b.t 3 / = * 3 / = ft 4 Q = A. = **0. = 0. ft 3 Q =Q 3 = 0 Forces: M = 39.4 kip.ft V =. kips Element 39. Stresses: = M. / I = * / = -39. kip/ft = M. / I = * 0 / = 0 kip/ft 3 = M. / I = 39.4 * / = 39. kip/ft = V. Q / t. I =. * 0 / *0.667 = 0 kip/ft = V. Q / t. I =. * 0./ *0.667 = 9. kip/ft 3 = V. Q 3 / t. I =. * 0 / * = 0 kip/ft Element Element = (0, -9.) 9. 4 ο min = -39. = 0 min = -9. = 9. min = 0 = 39. θ p from point to Element, R = 79. Element, R = 9. Element3, R = 79.
10 Tutorial # The state of strain at a point on a wrench has components ε = 0 (0-6 ), ε = 00 (0-6 ), and γ = -700 (0-6 ). - Use Mohr s circle to determine the equivalent in-plane strains on an element oriented at an angle of θ = 30 o clockwise from the original position. - Sketch the deformed elements at the original and the new orientation. - Sketch the elements at the principal plane and the maimum shear plane. - Determine the absolute maimum shear strain.
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12 Tutorial # 6. From the given strains, calculate the state of plain stress (draw the element). Strain Values ε = + 30 (0-6 ), ε = (0-6 ), γ = (0-6 ) = G.γ =.8*0 6 * (-400)* 0-6 = - 4 psi E. ε = - υ. 9*30 = -.8 () 08. 9*600 = -.8 () 4 B solving Eq. () and () = psi & = 08. psi Draw Mohr s circles for stress and for strain. min = R= 3 = Calculate the required ield stress ield for the material so that to prevent failure with respect to both Von Mises and TRESCA criteria, with a factor of safet of.. Von Mises: For -D: -. + < ield -. + < ( ield / F.S ) * = = ( ield / F.S ) ield = psi TRESCA: ma (3-D) = / = 9.87 psi ma = ield / ma = ield / F.S ield = *.*9.87 ield = psi = 48.7 Select a material with ield = psi answer
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