Download pdf - Rapport Rdm

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

Insitute de Technologie du Cambodge Rapport RDM

Page2 Professeur: Vong Seng Étudiant: Im David

RAPPORT

3. Contrainte

1. Au point d’un material de deux dimension de la contrainte est difini par 𝜎𝜎x = 60.0 MN/m2, traction 𝜎𝜎y = 45.0 MN/m2, compression Txy = 37.5 MN/m2, contrainte cissaillement:

Claculer 𝜎𝜎x’, 𝜎𝜎y’, 𝜏𝜏x’y’. Evaluer la valeur et la direction de la contrainte principal et la valeur maximum de la contrainte cissaillement.

Résolution:

On a :

1/2(𝜎𝜎x + 𝜎𝜎y ) = 1/2 (60.0 - 45.0) = 7.5 MN/m2

1/2(𝜎𝜎x + 𝜎𝜎y ) = 1/2 (60.0 +45.0) = 52.5 MN/m2

Alors,

Et

Donce :





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 orientée

par le vecteur

Résolution:

1. Le centre du cercle est à un point d'abscisse . Pour une facette orientée 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 réalité :

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

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

Si on pose: , on a :

et



4.Déformation;

1.



2.



3.



5.propièté mécanique des Materiaux

1,

2.



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

670 10 ,xε−= × 6420 10xyγ −= × .Determine the following quantities: (a) the strains for

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



6.Caractéristiques Géométriques d’un Section

1.



2.



3.



7.Type d’Appuis.

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

Σ FX

= 0 ⇒ RAX

= 0

In Y Direction

Σ FY = 0 ; R

AY+R

BY – 100 –160 = 0

⇒ RAY

+RBY

= 260

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

Σ MZ

= 0 ; We get,

Σ 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 Σ 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.



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+ 4×2= Rcy× 2 Rcy = 22.5kN RDy = 10+5×4 - 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) Σ MA = 0 12 X 3 – RBy * 9 = 0 RBy = 4 kN = RBx Σ MB = 0 12 * 6 – RAy * 9 = 0 RAy= 8kN Σ Fx = 0 RAx – 9 – 4 = 0 RAx = 13kN So,

RA= (82 + 132 )1/2= √233 kN

RB = (42 + 42 )1/2= 4 √2 kN



8.Eléments de Réduction

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.



2. A constant load of ω0 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= - 𝜔𝜔𝜔𝜔 IntegratingV= - 𝜔𝜔𝜔𝜔 X + C1 V at x 0 is – 𝜔𝜔𝜔𝜔L/ 2 . Putting x 0 in above equation, We get C1= 𝜔𝜔𝜔𝜔/2 ; V = 𝜔𝜔𝜔𝜔/ 2 ×(L - 2X)

dm/dx =V = 𝜔𝜔𝜔𝜔/2×(L - 2X)

M = 𝜔𝜔𝜔𝜔/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 = 𝜔𝜔𝜔𝜔( LX –X2) = 𝜔𝜔𝜔𝜔 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.



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 𝜎𝜎 = P/A = 30 x 10-3 / 0.6 x 10-3 = 50 MN/m2



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:



10.Flexion Pure

1. The rectangular tube shown is extruded from an aluminum alloy for which 𝜎𝜎y=40 MPa, 𝜎𝜎U=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 𝜎𝜎all = 𝜎𝜎U/3 = 20 MPa Since 𝜎𝜎all < 𝜎𝜎y , tube remains in the elastic rang. a. Bending moment 𝜎𝜎all = M*y / I

so M = I/y * 𝜎𝜎all = 103.8 kN/m

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

Thus 𝜌𝜌 = 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



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



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



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



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



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?



14.Flexion composée

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.



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?



15.Comportement au-delà domaine Elastique

1.

2.



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.



17.Flambement des Plèces Longes

1.



3.



16.Calcul des Déplacement de la pouter

1.



2.

3.





18.Torsion

1.



2.



3.



19.Concentration des Contraintes

2.



v



3.