of 39 /39
Institut de Technology de Cambodge Département de Gènie Civil Professeur: VONG Seng Etudiant : IM David Groupe : I3-A GCI ID : e2008130 Annéé Scolaire: 2010~2011

Rapport Rdm

Embed Size (px)

Text of Rapport Rdm

  • Institut de Technology de Cambodge Dpartement de Gnie Civil

    Professeur: VONG Seng

    Etudiant : IM David

    Groupe : I3-A GCI

    ID : e2008130

    Ann Scolaire: 2010~2011

  • Insitute de Technologie du Cambodge Rapport RDM

    Page2 Professeur: Vong Seng tudiant: Im David

    RAPPORT

    3. Contrainte

    1. Au point dun material de deux dimension de la contrainte est difini par x = 60.0 MN/m2, traction y = 45.0 MN/m2, compression Txy = 37.5 MN/m2, contrainte cissaillement:

    Claculer x, y, xy. Evaluer la valeur et la direction de la contrainte principal et la valeur maximum de la contrainte cissaillement.

    Rsolution:

    On a :

    1/2(x + y ) = 1/2 (60.0 - 45.0) = 7.5 MN/m2

    1/2(x + y ) = 1/2 (60.0 +45.0) = 52.5 MN/m2

    Alors,

    Et

    Donce :

  • Insitute de Technologie du Cambodge Rapport RDM

    Page3 Professeur: Vong Seng tudiant: Im David

  • Insitute de Technologie du Cambodge Rapport RDM

    Page4 Professeur: Vong Seng tudiant: Im David

    3. Cercle de Mohr

    On donne:

    avec et

    1. Tracer le cercle de Mohr 2. Calculer les contraintes principales et indiquer les orientation des directions

    principales de lacontrainte. 3. Calculer la valeur de la contrainte de cisaillement maximale et l'orientation de la

    facette correspondante. 4. Trouver la contrainte normale et la contrainte tangentielle pour une facette oriente

    par le vecteur

    Rsolution:

    1. Le centre du cercle est un point d'abscisse . Pour une facette oriente par le vecteur on a : et

    On peut donc tracer le cercle:

    Figure 2: Cercle de Mohr 2. Le rayon du cercle est et

    . Sur le cercle de Mohr, l'angle entre et est L'angle entre et est dans la ralit :

    3. La contrainte de cisaillement maximale est , elle est obtenue pour une facette oriente par le vecteur tel que l'angle entre et soit sur le cercle de Mohr avec soit

    l'angle entre et est dans la ralit : 4. L'angle entre et est dans la ralit: , sur le cercle de Mohr, on a

    Si on pose: , on a :

    et

  • Insitute de Technologie du Cambodge Rapport RDM

    Page5 Professeur: Vong Seng tudiant: Im David

    4.Dformation;

    1.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page6 Professeur: Vong Seng tudiant: Im David

    2.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page7 Professeur: Vong Seng tudiant: Im David

    3.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page8 Professeur: Vong Seng tudiant: Im David

    5.propit mcanique des Materiaux

    1,

    2.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page9 Professeur: Vong Seng tudiant: Im David

    3. An element of material in plane strain (see figure) is subjected to strains 6480 10 ,x = 670 10 ,x

    = 6420 10xy= .Determine the following quantities: (a) the strains for

    an element oriented at an angle= 75, (b) the principal strains, and (c) the maximum shear strains. Show the results on sketches of properly oriented elements. Solution:

  • Insitute de Technologie du Cambodge Rapport RDM

    Page10 Professeur: Vong Seng tudiant: Im David

    6.Caractristiques Gomtriques dun Section

    1.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page11 Professeur: Vong Seng tudiant: Im David

    2.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page12 Professeur: Vong Seng tudiant: Im David

    3.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page13 Professeur: Vong Seng tudiant: Im David

    7.Type dAppuis.

    1.

    Find the reactions at the supports for a simple beam as shown in the diagram. Weight of the beam is negligible.

    Apply equilibrium equations

    In X direction

    FX

    = 0 RAX

    = 0

    In Y Direction

    FY = 0 ; R

    AY+R

    BY 100 160 = 0

    RAY

    +RBY

    = 260

    Moment about Z axis (Taking moment about axis pasing through A)

    MZ

    = 0 ; We get,

    MA

    = 0

    0 + 250 N.m + 100*0.3 N.m + 120*0.4 N.m - RBY

    *0.5 N.m = 0

    RBY

    = 656 N (Upward)

    Substituting in Eq 5.1 we get MB = 0 R

    AY * 0.5 + 250 - 100 * 0.2 120 * 0.1 = 0 R

    AY = -436 (downwards)

    2. A beam with a hinge is loaded as above. Draw the shear force and bending moment diagram.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page14 Professeur: Vong Seng tudiant: Im David

    Solution:

    A hinge can transfer axial force and shear force but not bending moment. So, bending moment at the hinge location is zero. Also, without the hinge, the system is statically indeterminate. The hinge imposes an extra condition thus rendering the system determinate. We first find the support reactions. Fx = 0 Rax = 0 MB = 0 RAY 2 = 0 RAy =0 MD = 0 10 5+ 42= Rcy 2 Rcy = 22.5kN RDy = 10+54 - 22.5 = 7.5 kN 3.

    Determine the reactions at A and B for the beam shown due to the applied force. At A, the reaction components is x and y directions are RAx and RAy. The reaction RB acting at B and inclined force F can be resolved into two components along x and y directions. This will simplify the problem.

    Calculation:

    Fy = 12, Fx = 9; (By resolving the applied force) MA = 0 12 X 3 RBy * 9 = 0 RBy = 4 kN = RBx MB = 0 12 * 6 RAy * 9 = 0 RAy= 8kN Fx = 0 RAx 9 4 = 0 RAx = 13kN So,

    RA= (82 + 132 )1/2= 233 kN RB = (42 + 42 )1/2= 4 2 kN

  • Insitute de Technologie du Cambodge Rapport RDM

    Page15 Professeur: Vong Seng tudiant: Im David

    8.Elments de Rduction

    1. AB is a vertical post of a crane; the sockets at A and B offer no constraint against flexure. The horizontal arm CD is hinged to AB at C and supported by

    the strut FE which is freely hinged at its two extremities to AB and CD. Construct the bending moment diagrams for AB and CD.

    It is clear from considering the equilibrium of the whole crane that the horizontal reactions at A and B must be equal and opposite, and that the couple due to them must equal the moment of the 20 kN force. Let R be the magnitude of the horizontal reactions at A and B, then

    7R = 7(20000) and therefore R = 20000 N Let P = the pull in CE, and Q = the thrust in FE. Then taking moments about C for the rod CD. we have: And, Q = 58300 N Resolving horizontally for AB we have The vertical reaction at E = Q sine = 35 000 N. We can now draw the bending moment diagrams for AB and CD, considering only the forces at right-angles to each beam; let us take CD first. CD is a beam freely supported at C and E and loaded at D. The bending moment at E = 3 x 20 000 = 60 000 Nm, to which value it rises uniformly from zero at D;fr om E to C the bending moment decreases uniformly to zero. AB is supported at A and B and loaded with equal and opposite loads at C and F. The bending moment at C is Bending moments and shearing forces (2) (20 000) = 40 000 Nm.

    The bending moment at F is (2) (-20 000) = -40 000 Nm.

    At any point z between C and F, the bending moment is M = 20 000 (Z + 2) - 46 7002 = 40 000 - 26 7002z

    In the bending moment diagram positive bending moments are those which make the beam concave to the left, and are plotted to the left in the figure.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page16 Professeur: Vong Seng tudiant: Im David

    2. A constant load of 0 per unit length is

    applied on a simply supported beam as shown below. Draw the shear force and bending moment diagram by Intergration Method . Formulas used: dv/ dx = q dm/ dx = V .

    dv/ dx = q Here, q= - IntegratingV= - X + C1 V at x 0 is L/ 2 . Putting x 0 in above equation, We get C1= /2 ; V = / 2 (L - 2X)

    dm/dx =V = /2(L - 2X) M = /2( LX X2 + C2

    for a simply supported beam, bending moment is zero at the two ends.

    M = 0 at X = 0 C2 = 0

    M = ( LX X2) = X/ 2 (L X) We see that the expression for shear force and bending moment is the same using the two methods. It only remains to plot them.

    3.

    Plot shear and bending-moment diagrams for a simply supported beam with a uniformly distributed load; see Figure.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page17 Professeur: Vong Seng tudiant: Im David

    9.Traction et Compresion Simple 1. A tensile test is carried out a bar of mild steel of diameter 2 cm. the bar yields under a load of 80kN. It reaches a maximum load of 150kN, and breaks finally at a load of 70kN. Estimate : the tensile stress at the yield point; The ultimate tensile stress; The average stress at the breaking point , if the diameter is 1m.

    Solution:

    The original cross-section of the bar is

    The average tensile stress at yielding is then where Py = load at yield point

    The ultimate stress is the no minal stress at the maximum load

    Where Pmax = maximum load

    The cross-sectional area in the fractured neck is

    The average stress at the breaking point is

    Where Pf =final breaking load

    2. A steel bar of rectangular cross-section, 3 cm by 2 cm, carries an axial load of 30kN. Estimate the average tensile stress over a normal cross- section of the bar.

    Resolution

    The area of a normal cross-section of the bar is A = 0.03 x 0.02 = 0.6 x 10-3 m2

    The average tensile stress over this cross-section is then = P/A = 30 x 10-3 / 0.6 x 10-3 = 50 MN/m2

  • Insitute de Technologie du Cambodge Rapport RDM

    Page18 Professeur: Vong Seng tudiant: Im David

    3. A circular bar of length L _ 32 in. and diameter d _ 0.75 in. is subjected to tension by forces P (see figure). The wire is made of a copper alloy having the following hyperbolic stress-strain relationship:

    (a) Draw a stress-strain diagram for the material. (b) If the elongation of the wire is limited to 0.25 in. and the maximum stress is limited to 40 ksi, what is the allowable load P? Solution:

  • Insitute de Technologie du Cambodge Rapport RDM

    Page19 Professeur: Vong Seng tudiant: Im David

    10.Flexion Pure

    1. The rectangular tube shown is extruded from an aluminum alloy for which y=40 MPa, U=60 Mpa and E=165 GPa. Neglecting the effect of fillets, determine (a) the bending moment M for which the factor of safety will be 3, (b) the corresponding radius of curvature of the tube. 25mm

    M

    500mm

    500mm

    325mm

    The moment Inertia I= 1/12* (325)(500)3 - 1/12*(275)(450)3 = 12.97*10-4 m4 Allowable stress for a factor of safety of 3 and an ultimate stress of 60MPa,

    We have all = U/3 = 20 MPa Since all < y , tube remains in the elastic rang. a. Bending moment all = M*y / I

    so M = I/y * all = 103.8 kN/m

    b. Radius of curvature: We have 1/ = M / EI = 103.8*103 / 165*109 * 12.97*10-4 = 4.85*10-4 m-1

    Thus = 2061 m

    2. The I-section of Problem 9.2 is bent by couples of 2500 Nm about Cx and 500 Nm about Cy. Estimate the maximum bending stress in the cross-section, and find the equation of the neutral axis of the beam.

    Solution Ix = 1.641 x10-6 m4, Iy = 0.130 x10-6 m4 For bending about Cx the bending stresses in the extreme fibres of the flanges are

  • Insitute de Technologie du Cambodge Rapport RDM

    Page20 Professeur: Vong Seng tudiant: Im David

    For bending about Cy the bending stresses at the extreme ends of the flanges are

    On superposing the stresses due to the separate moments, the stress at the comer a is tensile, andof magnitude The total stress at the comer a 'is also 172.2 MN/m2, but compressive. The total stress at the comer b is compressive, and of magnitude

    The total stress at the comer b 'is also 20.0 MN/m2, but tensile. The equation of the neutral axis is given by

    The greatest bending stresses occur at points most remote from the neutral axis; these are the points a and a, the greatest bending stresses are therefore + 172.2 MN/m2

    3. A T-section of uniform thickness 1 cm has a flange breadth of 10 cm and an overall depth of 10 cm. Estimate the allowable bending moments about the principal axes if the bending stresses are limited to 150 MN/m2.

    Solution Suppose 7 is the distance of the principal axis Cx from the remote edge of the flange. The total area of the section is A = (0.10) (0.01) + (0.09) (0.01) = 1.90 x lO-3 m2

    On taking first moments of areas about the upper edge of the flange, = (0.10)(0.01)(0.005) + (0.09)(0.01)(0.055) = 0.0545 x 10-3m3

    The second moment of area of the flange about Cx is

    The second moment of area of the flange

    The second moment of area of the flange about

    The second moment of area of the flange

  • Insitute de Technologie du Cambodge Rapport RDM

    Page21 Professeur: Vong Seng tudiant: Im David

    The second moment of area of the web about Cx is

    For bending about Cx, the greatest bending stress occurs at the toe of the web, as shown in the figure. The maximum allowable moment is

    The bending stress in the extreme fibres of the flange is only 60. 4 MN/m2 at this bending moment. The second moment of area about Cy is

    The T-section is symmetrical about Cy, and for bending about this axis equal tensile and compressive stresses are induced in the extreme fibres of the flange; the greatest allowable moment is

  • Insitute de Technologie du Cambodge Rapport RDM

    Page22 Professeur: Vong Seng tudiant: Im David

    12.Flexion Gauche

    1.

    The I-section is bent by couples of 2500 Nm about Cx and 500 Nm about Cy. Estimate the maximum bending stress in the cross-section, and find the equation of the neutral axis of the beam.

    For bending about Cx the bending stresses in the extreme fibres of the flanges are

    For bending about Cy the bending stresses at the extreme ends of the flanges are

    The total stress at the comer a 'is also 172.2 MN/m2, but compressive. The total stress at the comer b is compressive, and of magnitude

    The total stress at the comer b 'is also 20.0 MN/m2, but tensile. The equation of the neutral axis is given by

    The greatest bending stresses occur at points most remote from the neutral axis; these are the points a and a < the greatest bending stresses are therefore + 172.2 MN/m'.

    2.

    The I-section of Problem 9.2 is bent by couples of 2500 Nm about Cx and 500 Nm about Cy. Estimate the maximum bending stress in the cross-

    For bending about Cx the bending stresses in the extreme fibres of the

    For bending about Cx the bending stresses in the extreme fibres of the

  • Insitute de Technologie du Cambodge Rapport RDM

    Page23 Professeur: Vong Seng tudiant: Im David

    13.Flexion Cissaillement

    1. Estimate the elastic section modulus and the maximurn longitudinal stress in a built-up I-girder, with equal flanges carrying a load of 50 kN per metre run, with a clear span of 20 m. The web is of thickness 1.25 cm and the depth between flanges 2 m. Each flange consists of four 1 cm plates 65 cm wide, and is attached to the web by angle iron sections 10 cm by 10 cm by 1.25 cm thick.

    The second moment of area of each flange about Cx is

    The second moment of area of the web about Cx is

    The horizontal part of each angle section has an area 0.00125 m2, and its centroid is 0.944 m from the neutral axis. Therefore the corresponding second moment of area is approximately

    The area of the vertical part of each angle section is 0.001093 m2, and its centroid is 0.944 m from the neutral axis. Therefore the corresponding second moment of area is approximately

    The second moment of area of the whole section of the angle section about Cx is then

    The second moment of area of the whole cross-section of the beam is then

    The elastic section modulus is therefore

    The bending moment at the mid-span is

    The greatest longitudinal stress is then

  • Insitute de Technologie du Cambodge Rapport RDM

    Page24 Professeur: Vong Seng tudiant: Im David

    2. A welded steel girder having the cross section shown in the figure is fabricated of two 280 mm *25 mm flange plates and a 600 mm * 15 mm web plate. The plates are joined by four fillet welds that run continuously for the length of the girder. Each weld has an allowable load in shear of 900 kN/m. Calculate the maximum allowable shear force Vmax for the girder.

    3. A box beam of wood is constructed of two 260 mm *50 mm boards and two 260 mm *25 mm boards (see figure). The boards are nailed at a longitudinal spacing s _ 100 mm. If each nail has an allowable shear force F _ 1200 N, what is the maximum allowable shear force Vmax?

  • Insitute de Technologie du Cambodge Rapport RDM

    Page25 Professeur: Vong Seng tudiant: Im David

    14.Flexion compose

    1. A masonry pier has a cross-section 3 m by 2 m, and is subjected to a load of 1000 kN, the line of the resultant being 1.80 m from one of the shorter sides, and 0.85 m from one of the longer sides. Find the maximum tensile and compressive stresses produced.

    P represents the line of action of the thrust. The bending moments are

    The cross-sectional area is

    For a point whose co-ordinates are (x, y) the compressive stress is

    The compressive stress is a maximum at B, where x = 1.5 m and y = 1 m. Then

    ThestressatD,wherex = -1Smandy = - lm, is

    which is the maximum tensile stress.

    2. A cylindrical brick chimney of height H weighs w = 825 lb/ft of height (see figure). The inner and outer diameters are d1 = 3 ft and d2 = 4 ft, respectively. The wind pressure against the side of the chimney is p =10 lb/ft2 of projected area.

    Determine the maximum height H if there is to be no tension in the brickwork.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page26 Professeur: Vong Seng tudiant: Im David

    3. A circular post and a rectangular post are each compressed by loads that produce a resultant force P acting at the edge of the cross section (see figure). The diameter of the circular post and the depth of the rectangular post are the same. (a) For what width b of the rectangular post will the maximum tensile stresses be the same in both posts? (b) Under the conditions described in part (a), which post has the larger compressive stress?

  • Insitute de Technologie du Cambodge Rapport RDM

    Page27 Professeur: Vong Seng tudiant: Im David

    15.Comportement au-del domaine Elastique

    1.

    2.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page28 Professeur: Vong Seng tudiant: Im David

    3.The cross-sectional dimensions of an I-section are shown in Figure 15.6; in the fully plastic condition, the centroidal axis Cx is a neutral axis of bending. The tensile fibres of the beam all carry the same stress or; the total longitudinal force in the upper flange is The compressed longitudinal fibres contribute moments of the same magnitudes. The total moment carried by the beam is therefore In the case of elastic bending we defined the elastic section modulus, Z, as a geometrical property, which, when multiplied by the allowable bending stress, gives the allowable bending moment on the beam.

    Then Z, is the plastic section modulus of the I-beam, and

    As a particular case consider an I-section having dimensions

    The elastic section modulus is approximately

    If M, is the bending moment at which the yield stress o, is first reached in the extreme fibres of the beam, then

    Thus, in this case, the fully plastic moment is only 10% greater than the moment at initial yielding. The ratio (Z&J) is sometimes called the shapefactor.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page29 Professeur: Vong Seng tudiant: Im David

    17.Flambement des Plces Longes

    1.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page30 Professeur: Vong Seng tudiant: Im David

    3.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page31 Professeur: Vong Seng tudiant: Im David

    16.Calcul des Dplacement de la pouter

    1.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page32 Professeur: Vong Seng tudiant: Im David

    2.

    3.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page33 Professeur: Vong Seng tudiant: Im David

  • Insitute de Technologie du Cambodge Rapport RDM

    Page34 Professeur: Vong Seng tudiant: Im David

    18.Torsion

    1.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page35 Professeur: Vong Seng tudiant: Im David

    2.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page36 Professeur: Vong Seng tudiant: Im David

    3.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page37 Professeur: Vong Seng tudiant: Im David

    19.Concentration des Contraintes

    2.

  • Insitute de Technologie du Cambodge Rapport RDM

    Page38 Professeur: Vong Seng tudiant: Im David

    v

  • Insitute de Technologie du Cambodge Rapport RDM

    Page39 Professeur: Vong Seng tudiant: Im David

    3.