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Axial Deformation

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Axial Deformation

1 Theoretical Background

1.1 Basic Theory of Axial Deformation

Definition 1



A plane linear member, when subjected to exterior loads and/or change of temperature, undergoes an axial deformation if after the deformation:

(a)   the axis of the member remains straight;

(b)   the cross-sections remain plane, perpendicular to the longitudinal axis of the beam and do not rotate about the same longitudinal axis after the deformation.

Equilibrium Equation

Figure 1 Equilibrium of an Elementary Beam Volume

The differential relation between the exterior load and the axial internal resultant is obtained from the equilibrium of the elementary volume:

(1)

where is the distributed loading parallel to the beam longitudinal axis and is the axial stress resultant .

Integrating equation (1) the stress resultant force in a particular cross-section is calculated as:

(2)

where is the value of the axial force at the origin of the integration interval.

Strain-Displacement Equation

Figure 1 Geometrical Aspects of the Axial Deformation

(a)    Undeformed Member and (b) Deformed Member

The extensional strain along the longitudinal axis of the beam is obtained using the notation shown in Figure 1:

(3)

The rest of the generalized strain tensor components, extensional and shear strains, are:

(4)

(5)

(6)

The expression of the generalized strain tensor is:

(7)

The displacement is obtained by integration the differential equation (3):

(8)

where is the displacement at the beginning at the integration interval.

The elongation is calculated using equation (8) as:

(9)

where is the total length of the bar.

Constitutive Equation

The constitutive equation reflects the relation between the stress and the strain. If the linear elastic material behavior is considered, applying the Hooks Law, the relation between the normal stress and extension strain is written:

(10)

where is the modulus of elasticity and is obtained performing tensile tests.

Considering the assumption that the cross-section of the bar is a small surface, the variation of the modulus of elasticity is negligible on this surface, and consequently, the constitutive equation (10) is expressed as:

(11)

Note: equation (11) implies that the normal stress varies only along the length of the member, but has a constant value on the entire cross-section.

The representation of the normal stress is shown in Figure 2.

Figure 2 Cross-Section Normal Stress Distribution

Note: equation (11) implies that the normal stress varies only along the length of the member, but has a constant value on the entire cross-section.

The rest of the stress tensor components are zero:

(12)

(13)

Consequently, the generalized stress tensor is:

(14)

Cross-Section Stress (Internal) Resultants

The following cross-sectional stress (internal) resultants are obtained using the stress distribution expressed by equation (11) through (13):

(15)

(16)

(17)

The relations between the normal stress and the cross-section resultants, and are derived using the notation shown in Figure 3.

Figure 3 Normal Stress and Stress Resultants

(a) Normal Stress and (b) Stress Resultants

If the axes and of the coordinate system intersect such that the axis passes through the cross-section centroid, the static moments and are zero and the axial force remains the only non-zero stress resultant:

(18)

(19)

(20)

Then, from equation (18) the normal stressis calculated as:

(21)

Note: It is concluded that a beam made from a linear elastic material undergoes an axial deformation if the axial force passes through the cross-section centroid.

Thermal Effects on Axial Deformation

The thermal strain was introduced as:

(22)

where is the thermal expansion coefficient and is the change in the member temperature.

The total elongation strain is the sum of the elongation strain induced by the exterior load action and thermal effects and is expressed as:

(23)

Then, the total elongation of the member is written as:

(24)

Uniform-Axial Deformation

Definition 2

The uniform axial-deformation element is a linear member characterized by:

(a)   a constant area along the entire length of the member;

(b)   is made of a homogeneous elastic material;

(c)    is subjected to a constant axial force

Uniform-axial deformed members (shown in Figure 4)

.

Figure 4 Member Exhibiting Uniform Axial-Deformation

Transcribing the requirements of the definition 2 the following expressions are obtained:

(25)

(26)

(27)

Note: Equation (27) implies the absence of the distributed load.

Rewriting equations (21), (23) and (24) previously obtained for the case of the member with uniform axial-deformation the following equations are obtained:

for axial stress (28)

for elongation strain (29)

for total elongation (30)

Axial flexibility and stiffness coefficients

the axial stiffness coefficient. (31)

axial flexibility coefficient (32)

Substituting the equation (32) into the total elongation expression (30), the total elongation of a bar under uniform-axial deformation is recast as:

(33)

Nonuniform-Axial Deformation

If any one of the assumptions contained in definition 2 is violated the axial deformation is called nonuniform-axial deformation. The most common cases of nonuniform-axial deformation are:

Member with non-homogeneous cross-section;

Member with variable Cross-Section;

Member loaded along its length.

The formulae described in the section 1.1 have to be adapted function of the situation.

Verification of the Members Subjected to Axial Deformation

The design formula used is the relationship between maximum normal stress and the allowable normal stress:

(34)

The formula (34) was used for a long period of time in a procedure known as the allowable-stress design. Due to the simplicity of application, this method is still commonly used in United States for the design of steel structures.

The allowable normal stress is defined by limiting the value of the normal stress in the axially deformed member. Dividing the yield stress pertinent to the material subjected to axial deformation by a safety factorthe allowable axial stress is calculated:

(35)

The safety factor is greater than one, usually taking values between 2 and 3. The yield stress for different materials is found in Appendix 1.2.

2 Solved Problems

Problem 2.1

A solid brass rod AB and a solid aluminum rod BC are connected together by a rigid coupler of negligible length at B as shown in Figure 2.1.a. The diameters and the modulus of elasticity of the two segments are d1 = 65 mm, d2 = 50 mm, E1 = 105 GPa and E1 = 69 GPa, respectively. The system is loaded by two concentrated loads, and , acting along the centroidal line of the system at point and , respectively . Calculate the axial stress existing in the two rods and the displacement at point B and C. Verify the rod segments.

Figure 2.1.a

A. General Observations

Two axial forces are acting on the rod:

and

The areas of the two rod segments are:

B. Calculations

B.1 Free-Body Diagram

The constraint located at point A is replaced by a horizontal reaction as shown in Figure 2.1.b.

Figure 2.1.b

B.2 Reaction Calculation

The equation of equilibrium used is the projection of all axial forces on the horizontal axis:

The equilibrium equation contains only one unknown reaction force,, and consequently, the system is statically determinate.

B.3 The Axial Force Diagram

The axial force diagram is drawn in Figure 2.1.c.

Figure 2.1.c

The axial force on the interval is a constant tension force, while on the interval is a constant compression force . It can be concluded that both segments of the rod are uniform-axial deformed members.

B.3 Stress and Strain Calculation

The stress and strain in the interval, where the rod is made of solid brass are obtained as:

tension stress

elongation strain



In a similar manner is calculated the stress and strain pertinent to the interval representing the aluminum made rod.

compression stress

elongation strain

B.4 Flexibility Coefficients

The flexibility coefficients are calculated as:

B.5 Calculation of the axial displacements

The displacement of the point, the right end of the brass segment, is calculated as:

where the displacement at the origin of the interval and, because the point is constraint against the horizontal movement.

The calculation of the displacement implies the knowledge of the displacement of the origin point of the interval.

The displacement in point is calculated as:

B.6 Verification of the Rod

The verification of the rod segments is conducted as:

for the brass segment

ok

where is the brass allowable stress, is the brass yield stress which can be found in Appendix 1.2, and is the safety factor . For this calculation and .

for the aluminum segment

ok

where is the brass allowable stress, is the aluminum yield stress which can be found in Appendix 1.2, and is the safety factor . For this calculation and.

Problem 2.2

Two uniform, linearly elastic members are held together at point B and the resulting two-segment rod is attached to rigid supports at ends A and C. A single external load P = 4000 kN is applied at joint B. Member (1) has a length m, diameter d1 = 120 mm and is made of steel with a modulus of elasticity = 200 GPa. Member (2) has a length = 1.8 m, diameter d2 = 150 mm and is made of an aluminum alloy with a modulus of elasticity =75 GPa. Conduct the following tasks: (a) verify the axial stress in both members and (b) calculate the axial displacement at point B.

Figure 2.2.a

A. General Observations

The areas of the two rod segments are:

B. Calculations

B.1 Free-Body Diagram

The constraint located at point A is replaced by a horizontal reaction as shown in Figure 2.2.b.

Figure 2.2.b

B.2 Reaction Calculation

The equation of equilibrium used is the projection of all axial forces on the horizontal axis:

The equation contains two (2) unknown reaction forces and. Consequently, the system is a statically indeterminate. An additional equation is necessary. This is equation is geometrical in nature and represents the fact that the total elongation of the beam is zero.

Considering that the calculation of the displacement starts at point A, and the axial displacement at the end B of the interval, representing the steel rod, is calculated as:

At the left end of the interval the axial displacement is expressed as:

This way the second equation is obtained as:

and by substituting the flexibility coefficients and into the expression of the axial displacement the following algebraic system is obtained:

The solutions, representing the two reactions, and , are found:

Substituting the numerical data the reaction forces are calculated as:

B.3 The Axial Force Diagram

The axial force diagram is drawn in Figure 2.2.c.

Figure 2.2.c

The axial force on the interval is a constant tension force, while on the interval is a constant compression force. It can be concluded that both segments of the rod are uniform-axial deformed members.

B.3 Stress and Strain Calculation

The stress and strain in the interval, where the rod is made of solid steel are obtained as:

tension stress

elongation strain

In a similar manner is calculated the stress and strain pertinent to the interval representing the aluminum made rod.

compression stress

elongation strain

B.4 Calculation of the axial displacement

The displacement of the point, the right end of the steel segment, is calculated:

where the displacement at the origin of the interval and, because the point is constraint against the horizontal movement.

B.5 Verification of the Rod

The verification of the rod segments is conducted as:

for the steel segment

ok

where is the brass allowable stress, is the brass yield stress which can be found in Appendix 1.2, and is the safety factor . For this calculation and.

for the aluminum segment

ok

where is the brass allowable stress, is the aluminum yield stress which can be found in Appendix 1.2, and is the safety factor . For this calculation and .

Problem 2.3

A rigid beam AB, shown in Figure 2.3.a, is supported by two vertical rods made of steel with a modulus of elasticity E=200 GPa. The support rod located at the end A has a diameter d1=25 mm. The weight of the beam AB is negligible and is loaded at point C with a concentrated force P = 60 kN. Calculated: (a) the diameter, d2, of the hanger located at the end B, considering that the relation between the vertical displacement at the ends of the beam , (b) under same condition the vertical displacement in node C, (c) if the hanger located at end B of the rigid beam has a diameter d2=20 mm, what should be the position of the concentrated load P for relation to be true, and (d) the axial stresses in the hangers considering the conditions stipulated in the previous question.

A. General Observations

A.1 The steel rod (1) has a length and diameter, while for the steel rod (2) only the length is known. Both rods are made of steel with a modulus of elasticity of steel.

Figure 2.3.a

A.2 The vertical concentrated force is located at point at a distance of from the left end. The total distance in-between the rigid beam supports is, while the distance from the application point to point is .

B. Calculations

B.1 Free-Body Diagram

The rigid beam is supported at ends and by two steel rods which are playing the supporting role for the beam. Sectioning the rods and replacing them by two corresponding axial forces and, the free-body diagram of the system is obtained as shown in Figure 2.3.b.

B.2 Reactions Calculation

Two moment equations are written:

Figure 2.3.b

The two equilibrium equations are containing two unknowns, the forces and, and consequently, the system is statically determinate.

The verification of the reaction forces is done using the equilibrium equation of the projection of the forces on the vertical direction:

B.3 Calculation required by question (a)

The axial forces in the rods are tension type forces:

Because the axial forces in the rods and they have constant areas, both rods are uniform-axial deformed members. The flexibility coefficients are obtained as:

The geometrical condition required is:

Because the rod diameter is always a positive value, from the two above solutions only the positive value is a valid solution. Consequently, the diameter of the rod (2) is calculated as:

B.4 Calculations required by question (b)

The vertical displacement is composed from two components: one elastic and equal with the vertical displacement and other rigid induced by the rigid rotation of the beam in the vertical plane. The elastic vertical displacement is calculated:

where is the flexibility coefficient of the rod (1).

The vertical displacement is obtained:

B.5 Calculations required by question (c)



The geometrical condition imposed is written as:

The flexibility coefficient of the rod (2) is calculated considering that the rod diameter is:

The position of the concentrated force measured from point is obtained by substituting the previously calculated flexibility coefficients:

B.6 Calculations required by the question (d)

The axial forces in the hangers are:

for rod (1)

for rod (2)

The axial stresses are obtained:

tension stress in rod (1)

tension stress in rod (2)

B.7 Verification of the rods

Considering that the steel yielding stress and the safety coefficient employed are and , respectively, the allowable steel stress is calculated as:

Consequently,

the rod (1) is ok

the rod (2) is ok

Problem 2.4

The system shown in Figure 2.a is composed of a rigid beam AD, pinned into the wall at point A, and two unequal linear elastic rods, BE and CF, made of steel with a modulus of elasticity . The steel rods lengths and areas are and, and, and, respectively. The system is loaded with by vertical forceacting at point D. The steel rods and forcelocations are determined by the following distances measured from point A: , and, respectively. What is the allowable force supported by the system if the allowable normal stress for steel is.

Figure 2.a

A. General Observations

The system illustrated in Figure 2.a is geometrical defined by the data contained in the text.

B. Calculations

B.1 Free-Body Diagram

The rigid beam is supported at the ends and by two steel rods, which are playing the supporting role for the beam, together with the pinned support located at point. Sectioning the rods and replacing them by two corresponding axial forces and, and substituting the constraints introduced by the pinned support by its corresponding reaction forces, and , the free-body diagram of the system is obtained and illustrated in Figure 2.b.

Figure 2.b

B.2 Reactions Calculation

The following equilibrium equations are written:

no axial force

The last two equilibrium equations contain three unknown reaction forces, and. The system is statically indeterminate. In order to find these unknown quantities one additional equation is necessary. This equation is obtained from the deformation compatibility condition schematically described in Figure 2.c.

Because the beam AD is rigid, purely geometric relations between the rod elongations, and, and the rotation angle are written as:

Figure 2.c

Using the elongation expressions the forces in the rods are calculated as:

where the stiffness coefficients, and, are calculated from the geometrical and material properties characteristics of the rods as:

Substituting the expressions of the rod forces into the last two equilibrium equations the three original unknowns, , and , are replaced with two unknowns, and. The equilibrium equations are then written as:

Solving the algebraic system, the two unknowns are found as:

The axial forces in the rods are calculated:

To find the force producing an allowable stress in the rods the following equations are written:

for rod (1)

for rod (2)

The allowable forcefor the entire system is:

The vertical displacements at point and are then calculated:

Problem 2.5

An axial load P is applied to a tapered circular rod of length L as shown in Figure 2.5.a. The variation of the rod radius along its length is expressed as:

where is the radius at the left end cross-section.

Symbolically express: (a) the axial stress, (b) the elongation strainand (c) the total elongation. Apply the above obtain expressions for the case when where P = 9 kN, L = 2.50 m, r0 = 0.50 m and the rod is made of an aluminum alloy for which E = 69 GPa.

Figure 2.5.a

A. General Observations

The tapered aluminum rod is subjected to a non-uniform axial deformation, because the area is not constant along the length of the beam.

The variation of the area along its axis is expressed as:

where is the area representing the cross-section at the left end of the rod (x=0).

B. Calculations

B.1 Free-Body Diagram

The free-body diagram is illustrated in Figure 2.5.b.

Figure 2.5.b

B.2 Reaction Calculation

The constraint located at the left end of the beam is replaced by a horizontal concentrated reaction force. The reaction force is calculated using the equilibrium equation involving the horizontal projection of all forces.

The system is statically determinate.

B.3 Expression of the Axial Stress

The axial force in the rod is constant and represents a tension force:

The general expression of the axial stress in the rod is obtained as:

where is the axial stress in the cross-section located at the left end of the rod.

B.4 Expression of the Axial Strain

where is the axial stress in the cross-section located at the left end of the rod.

B.5 Expression of the Rod Total Elongation

The total elongation is obtained as:

B.6 Numerical Application

The following data is substituted in the above obtained expressions:

The reaction force is:

The area at the left end cross-section:

The axial stress and elongation strain at the left end cross-section are:

The variation of the axial stress and elongation strain are calculated in a number of twenty-one sections:

The maximum values are calculated in the right end cross-section (x = L):

The total elongation of the rod is:

The variation of the axial stress is illustrated in Figure 2.5.c.

Figure 2.5.c

Problem 2.6

A magnesium-alloy rod (Emag = 45 GPa) of diameter dm = 30 mm is encased by a brass tube (Ebra = 100 GPa) with outer diameter db. Both bars have an equal length L = 500 mm. An axial load P = 40 kN is applied to the resulting bimetallic rod. Assuming that that the magnesium rod and the brass tube are securely bonded to each other calculate: (a) the outer diameter db of the tube, if three fourths of the load P is carried by the magnesium rod and one fourth by the brass tube and (b) the total elongation of the bimetallic rod. The bimetallic rod is illustrated in Figure 2.6.

A. General Observations

A.1 It is assumed, due to the bondage between the materials, that the deformation at the right end cross-section of the bimetallic rod is equal for both materials.

A.2 The reaction force located at the left end of the rod is not necessary to be calculated, because the distribution of the external force in-between the magnesium and brass cross-section is established.

for magnesium

for brass

Note: It can be remarked that the bimetallic rod it is statically determined system, but because the modulus of elasticity is not constant for the entire length of the rod, the rod is subjected to a nonuniform-axial deformation.

Figure 2.6

B. Calculations

B.1 Calculation of the brass exterior diameter

The equality of the deformation at the right end of the bimetallic rod is:

Using the flexibility coefficients and, corresponding to magnesium and brass rods the displacements equality is written as:

where and .

Substituting the forces and the flexibility coefficients into the above equation, the expression of the brass area is obtained:

The area of the magnesium rod is calculated as:

and consequently, the area of the brass rod is obtained:



The exterior diameter of the brass tube is calculated from the following equation:

because only the positive value makes sense.

B.2 Calculation of total elongation

The total elongation is equal to the displacement of the either material calculated at the right end of the rod. The calculation is conducted using the data pertinent for magnesium and is also verified using the brass data.

The verification using brass:

Note: It can be remarked that the deformation of both materials is 1.572*10-4 m and, this way, the geometrical condition imposed is also verified numerically.

Problem 2.7

A uniform circular cylinder of diameter d and length L is made of a material with modulus of elasticity E. It is fixed to a rigid wall at end A and subjected to a distributed external axial loading of magnitude p(x) per unit length, as shown in Figure 2.7.a. The axial stress,, varies linearly with x as shown in Figure 2.7.b. Determine: (a) the expression for the distributed loading, p(x) and (b) the expression for the axial displacement, u(x), of the cross section.

Figure 2.7

A. General Observations

A.1 The member is subjected to a nonuniform-axial deformation and it is statically determinate system.

A.2 The axial stress has the following expression (see Figure 2.7):

B. Calculations

The horizontal reaction force is calculated as:

The axial force pertinent to a particular cross-section is obtained from the equilibrium as:

The axial stress is expressed:

Comparing the above expression with the expression given by the problem the following equation is written:

and after the algebraic manipulation the integral becomes:

Consequently,

It can be concluded that the linear variation of the axial stress is induced by a constant axially applied load.

Problem 2.8

Two rods are stress-free when welded together at point and welded to two rigid walls at points and. The geometrical and material characteristics of the two rods are illustrated in Figure 2.8.a. Subsequently, rod (1) and rod (2) are heated by an amount and, respectively from their initial installation temperature. Determine an expression for the axial forces induced in each rod by the change in temperatures.

Figure 2.8.a

A. General Observations

A.1 The rods have the same modulus of elasticity, but different expansion coefficients and.

A.2 It can be remarked the absence of any external load and the existence of a change in temperature

B. Calculations

B.1 Free-Body Diagram

The free-body diagram is illustrated in Figure 2.8.b.

Figure 2.8.b

B.2 Reaction Calculation

The constraints located at the left and right end points,and, of the member are replaced by two horizontal reaction forces, and, respectively. The following equilibrium equation is written:

The system is statically indeterminate and in order to calculate the reaction forces an additional equation is necessary. This equation is the fact that the total elongation is zero:

The expression of the displacements and are:

where

and are the flexibility coefficients

The axial forces in the rods are:

and

The elongation is calculated as:

From the condition imposing that the total elongation to be zero, the reaction force is obtained:

Consequently, the reaction force is:

and the forces ion the rods are:

3 Proposed Problems

Problem 3.2

The three-part axially loaded member, shown in Figure 3.2, consists of a tubular segment (1) with outer diameter and inner diameter, a solid circular rod segment (2) with diameter and a third solid circular rod segment (3) with diameter.

Figure 3.2

All three applied loads shown are acting along the centroidal axis of the members. Considering that the rigid couplers have a negligible length determine: (a) the axial stresses in each one of the three respective segments, (b) the displacements in points B, C and D if all segments have equal lengths L1 = L2 = L3 = 0.50 m and the modulus of elasticity of the material is .

Problem 3.3

The diameter of the central one-third of a 50 mm diameter steel rod is reduced to 20 mm, forming a three-segment rod, as shown in Figure 3.3. For the loading shown, determine the displacements of the rod points B, C and D, respectively. The rod is made of a material which has the modulus of elasticity E = 200 GPa.

Figure 3.3

Problem 3.4

A column in a two-story building is fabricated from square structural steel tubing having a modulus of elasticity E = 210 GPa. The cross-sectional dimensions of the two segments are shown in Figure 3. Two axial loads acting along the centroidal axis of the column are applied to the column at levels A and B. Calculate: (a) the axial stress both segments of the column and (b) the total shortening of the column length

Figure 3.4

Problem 3.5

A three-segment stepped aluminum-alloy column is subjected to the vertical axial loads shown in Figure 3.5 The cross-sectional areas of the segments are A1 =3870 mm2, A2 =5810 mm2 and A3 =9035 mm2, respectively. The material modulus of elasticity is E = 69 GPa. Calculate: (a) the axial stresses in all three segments and (b) the vertical displacement of the column at nodes A, B and C under the given loading system.

Figure 3.5

Problem 3.6

A uniform rod is subjected to three axial loads acting as shown in Figure 3.6 and is made of a material with a modulus of elasticity E = 70 GPa. What is the minimum allowable diameter of the cylindrical rod if the displacement at the right end D and the maximum axial stress in the rod can not excide 5 mm and 80 MPa, respectively?

Figure 3.6

Problem 3.7

A 1.2 in diameter aluminum-alloy hangar, illustrated in Figure 3.7, is supported by a steel pipe with an inside diameter of=75 mm. The moduli of elasticity for the steel pipe and hanger are Esteel =MPa and Ealuminum =MPa, respectively. Determine the thickness of the steel pipe if the maximum axial displacement at the node C is 2.5 mm.

Figure 3.7

Problem 3.8

A 3.60 m rigid beam AB that weighs 0.80 kN supports an air conditioner that weighs Wair=45 kN. The beam is supported by hanger rods (1) and (2) located at its ends as shown in Figure 3.8.

Figure 3.8

Conduct the following calculations: (a) if the diameter of rod (1) is 95 mm what is the stress in the rod? (b) if the stress in rod (2) is to be the same as the stress in rod (1), what should the diameter of rod (2)? (c) what are the downward displacements at the ends of the rigid beam if the rods length are L1 = L2 = 1.80 m and they are made of steel with a modulus of elasticity E =210103 MPa?

Problem 3.9

A hanger rod CD is attached to a rigid beam AB. The beam is supported at its ends by two hanger rods. Assuming that all tree hangers are identical and are made from the same material, calculate: (a) the axial stress in all tree hangers, (b) the vertical displacement of the points A, B, C and D.

Figure 3.9

Problem 3.10

A commercial sign of weight W is supported by a structural system comprised from a rigid beam AB and a wire CD, as shown in Figure 3.10. The rigid beam has a negligible weight, while the wire has length L, cross-sectional area A, and modulus of elasticity E. Assuming that the attachment pin D is directly located above pin A, and when there is no load acting on the beam, the beam is in a perfect horizontal position calculate the following: (a) the axial stress in the wire CD when the sign is attached at points B and C of the beam, (b) the vertical displacement in point C of the beam.

Figure 3.10

Problem 3.11

The inclined rod AB, shown in Figure 3.12, is pinned to a fixed support at A and is pinned at the end B to a block that is forced to move only horizontally when the load P is applied. Determine: (a) an expression for the axial stress in the inclined rod as a function of P, L, E, A, and the angle, (b) an expression for the horizontal displacement at the end B.

Figure 3.11

Problem 3.12 A bimetallic bar is made by bonding together two homogeneous rectangular bars, each having a width b, length L and moduli of elasticity of the bars are E1 and E2, respectively. An axial force P is applied to the ends of the bimetallic bar at location (y= yp, z = 0) such that the bar undergoes an axial deformation only. Assuming the following data: L = 1.5 m, b = 50 mm, t1 = 25 mm, t2 = 15 mm, E1 = 70 GPa, E2 = 210 GPa and P = 48 kN. Calculate: (a) the normal stress in each material, (b) the value of yp, (c) the elongation of the bar.

Figure 3.12

Problem 3.13

A bimetallic bar, shown in Figure 3.13, undergoes an axial deformation. The bar has the following geometrical and material characteristics: L = 2.55 m, b = 50 mm and h1 = h2 = 1.50 mm and E1=210 GPa. Calculate: (a) the modulus of elasticity E2 if the load P is applied at 10 mm, and (b) the total elongation of the bar for a load of P = 9 kN.

Figure 3.13

Problem 3.14

A steel pipe is filled with concrete, and the resulting column is subjected to a compressive load P = 360 kN. The pipe has an outer diameter of 325 mm and an inside diameter of 305 mm. The elastic moduli of the steel and concrete are: Estel = 210 GPa and Econc = 25 GPa. Determine: (a) the stress in the steel and the stress in the concrete due to this loading, (b) the shortening of the column if its initial length is L = 3.70 m, (Ignore- radial expansion of the concrete and steel due to Poisson's ratio effect.)

Figure 3.14

Problem 3.15

A homogenous rod of length L and elasticity modulus E is a conical frustum with diameter d(x) that varies linearly from d0 at one end to 2*d0 at the other end, with d0 << L. An axial load P is applied to the rod, as shown in Figure 3.15. Determine analytical expressions for: (a) the stress distribution,, on an arbitrary cross section and (b) the elongation of the rod, .

Figure 3.15

Problem 3.16 A uniform circular cylinder of diameter d and length L is made of a material with modulus of elasticity E and specific weight. It hangs from a rigid ceiling as shown in Figure 3.16. Determine: (a) the expression of the axial stress, (b) the expression of the strain, (c) the expression of the displacement and (d) the compression force necessary to be applied in order to return the bar at its initial length.

Figure 3.16

Problem 3.17

A steel pipe with outer diameter do = 50 mm, and inner diameter di = 38 mm and a solid aluminum-alloy rod of diameter d = 19 mm form a three-segment system that undergoes axial deformation due to a single external load PC = 55 Kn acting on a collar at point C, as shown in Figure 3.17. Considering the following data: L1 = L2 = 0.75m, L3 = 1.20 m, E1 = 210 GPa and E2 = E3 = 69 GPa, calculate: (a) the axial stresses induced in all three segments and (b) determine the axial displacement at points B and C.

Figure 3.17

Problem 3.18

A three-segment rod is attached to rigid supports at ends A and D and is subjected to equal and opposite external loads P at nodes B and C, as shown in Figure 3.18. The rod is homogeneous and linearly elastic, with modulus of elasticity E. Assuming A1 = A3 = A and A2 = 2A, L1 = 2L and L2 = L3 = L, calculate: (a) the axial stresses in all tree segments and (b) the horizontal displacements at nodes B and C, respectively.

Figure 3.18

Problem 3.19

A rigid beam AD, supported by a pin at its end point D and attached by the two vertical steel rods at points A and C, is loaded by a vertical load P at point B. Neglecting the weight of the beam and assuming that the support rods are stress-free when P = 0, calculate: (a) the forces F1 and F2 in the support rods after load P is applied, (b) the deformation of the supporting and (c) the expressions previously obtained if A1 = A2 = 500, L1 = 1m, L2 = 2m , E1 = E2 = 210 GPa, P = 50 kN, a = 0.50 m and b = 1.5 m.

Figure 3.19

Problem 3.20

A rigid beam AD, shown in Figure 3.20, is supported by a smooth pin at B and by two vertical rods attached to the beam at points and C. Neglecting the weight of the beam and assuming that the rods are stress-free when P = 0. Considering that A1 = 650 mm2, A2 = 325 mm2, L1 = L2 = 1.25 m, a = 0.60 m, b = 1.25m, E1 = E2 = 69 GPa and P = 22.70 kN, determine: (a) the axial forces in the support rods, (b) determine the axial stress in each support and (c) calculate the elongation of the support rods.

Figure 3.20

Problem 3.21

A rigid beam AD, shown in Figure 3.21, is supported by three identical vertical rods that are attached to the beam at points A, C, and D and loaded in node B with a vertical concentrated load P. Assuming that A = 650 mm2, L = 1.50 m, a = 0.50 m, b = 1.00 m, c = 1.50 m, E = 210 GPa and P = 45.5 kN, calculate: (a) the axial forces in the support rods and (b) the vertical displacements at nodes A, B, C and D.

Figure 3.21

Problem 3.22

The rod composed from two segments as shown in Figure 3.22, is attached to rigid walls at A and C. Determine expressions for the stresses pertinent to both segments resulting from a uniform temperature increase of the entire rod. Numerical application: A1 = 1000 mm2, A2 = 1500 mm2, L1 = 2 m, L2 = 1.5 m, E1 = 210 Gpa, E2 = 120 GPa, a1 = 12*10-6 and a2 = 8.0*10-6 and = 30.

Figure 3.22

Problem 3.23

A three-segment rod shown in Figure 3.23 is rigidly attached to walls at points A and D. Subsequently, the middle segment is heated by an amount, while the segments (1) and (3) are kept at their original temperature. Using the notation shown in Figure 3.23 and considering the system free at stress at the installation, calculate the stresses in the segments and the axial displacements at points B and C.

Figure 3.23

Problem 3.24

A steel pipe with outer diameter do, and inner diameter di and a solid aluminum-alloy rod of diameter d form a three-segment system as shown in Figure 3.2 The system is considered stress free when is welded to the rigid supports at points A and D. The installation temperature is recorded. Subsequently, the aluminum rod is cooled by 100F (), while the steel pipe is held at the initial temperature. Assuming do = 50 mm, di = 38 mm, d = 19 mm, L1 = 1.25 m, L2 = L3= 0.75 m, E1 = E3 = 210 GPa, E2 = 69 GPa, a1 = a3 = 6.5*10-6 and a2

Figure 3.24

Problem 3.25

The mechanical system shown in Figure 3.25 is composed from two identical steel rods (A = 40 mm2, E = 200 GPa, a = 12.0*10-6) and a 'rigid' beam AC. The beam is supported by a smooth pin at B. Assuming the two rods stress-free after installation, determine: (a) the axial stresses induced in rods if their temperature is decreased by 50C and (b) the small angle q through which the beam AC would rotate due to this temperature change.

Figure 3.25

Problem 3.26

The steel rod of diameter 20 mm is held without any initial stresses between two rigid walls as illustrated in Figure 3.26. Determine the temperature drop DT at which the stress in the rod reaches 200 MPa. Use for the steel E=200 GPa and a = 12.0*10-6.

Figure 3.26

Problem 3.27

The bar AB, shown in Figure 3.27, is held between rigid supports and heated nonuniformly in such a manner that the temperature increase DT at distance x from end A is given by the expression DT (x) = DT1*(x/L)2, where DT1 is the increase in temperature at end B of the bar. Obtain a formula for the compressive stress in the bar. (Assume that bar has a length L and is made of a material with modulus of elasticity E and coefficient of thermal expansion a.)

Figure 3.27

Problem 3.28

The copper bar AB of length 1.00 m is placed in position shown in figure 3.28 at room temperature. A gap of 0.2 mm exists between the end A of the bar and a rigid restraint. Calculate the axial compressive stress in the bar if the temperature is raised 90F. Use for copper the following material constants: E=110 GPa and a = 9.8*10-6.

Figure 3.28

Problem 3.29

Three identical springs, 10 in. apart, are attached to a horizontal rigid bar at points A, B, and C, as illustrated in Figure 3.29. Three vertical loads of magnitudes 138 N, 45 N and 31 N act at points A, B, and C, respectively. Calculate the angle of rotation q (degrees) of the rigid bar if the spring stiffness k is 14 N/m.

Figure 3.34

Problem 3.30

The rigid bar ABCD, shown in Figure 3.30, is pinned at point B and supported by springs at A and D. The springs at A and D have stiffnesses k1 = 11.20 kN/m and k2 = 36.75 kN/m, respectively. The dimensions a, b, and c are 0.45 m, 0.90 m, and 0.35 m, respectively. A load P acts at point C. Determine the maximum permissible load Pmax if the angle of rotation of the bar due to the action of the load P is limited to 2.

Figure 3.30






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