chapter 11 - stress, strain and deformation in...

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 1

Axial Deformation and Stress of Bars (Chapter 11)

The 4 governing equations that must be satisfied are:

Static Equilibrium (from COLM):

Stress-Strain: Kinematics:

Boundary Conditions: Depends on the problem


Example 1. Consider an elastic axial bar as follows:Axial force F applied to cross-section at right end (x=L)Left end fixed so displacement B.C. of Beam is prismatic (constant cross-sectional area of A)

Determine: ux(x), and ux(x=L).

Equilibrium at x: ;Displace B.C.:

Kinematics Relation (strain-displacement):


Constitutive Relation:

Combine last three equations to obtain:

Thus or .

Integrate from 0 to x' to obtain the axial displacement:

Apply the boundary condition: => C = 0


Solution for axial bar in tension:

Strain is given by:

Stress is given by:

and displacement at end,

Note that is the elongation of a bar of length L with cross-sectional area A and Young’s modulus E and subjected to a tensile force of F as shown below.


Consider a truss structure as done in ENGR 211. By the method of joints, the force in bar EF is . If the bar has a cross-sectional area of 2 sq.

in., then the stress is given by . Suppose

the truss is made of steel. The elongation of the truss memeber is then

. In this

case, the bar is in compression and the negative indicates that the member has shortened.

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 6Example 2: Elastic bar constructed of two different bars, fixed between two walls and loaded with a force P applied at point B:

Determine: axial force and stress in each bar, axial displacement at point B.

For this problem, we have 4 relationships to satisfy:1. Equilibrium of horizontal forces at any point2. Kinematics (strain/displacements in horizontal direction)3. Stress-Strain (material properties)4. Boundary Conditions


Use a “free-body diagram” to determine equilibrium of the forces acting on a segment of the bar. Assume forces in the bars are P1 and P2 (positive). Take a free-body of the beam cross-section at point B (where the axial load P is applied):

Each stress can be written in terms of an equivalent force over an area

so that: and

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 8Equilibrium of the free body at B in terms of forces requires that

or (1)

NOTE:Any problem which can not be solved for the internal forces by force equilibrium alone requires additional equations (defining displacements) in order to complete the solution. Such a problem is called statically indeterminate.

The above problem is thus statically indeterminate!

We know from boundary conditions that the bar’s total elongation between the two fixed walls is zero. First, calculate the deformation (elongation) of bars 1 and 2:


#1 and 3 are not really free-body diagrams!!!

elongation of bar 1 = ; elongation of bar 2 =

Displacement B.C.

total elongation = 0 = + (2)

We now have two equations to solve for and :


Solution of the equations for and gives:

and

Stresses are:

Note: for P to right, is tensile (+) and is compressive (-).

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 11Displacement at point B is

Note that for P to the right (positive P) , is positive (to the right) as expected.

Special Cases: 1. , : (tension), (comp)2. (bar 1 has larger area), :

, (bar 1 carries more load) , (same!)

3. , (bar 1 has higher E, i.e., is "stiffer"): , (bar 1 carries more load)


General Solution Procedure for Bars1. Determine number of free-body diagrams required.

Generally need a cut in each bar. However, if applied load or bar properties (E, A, etc.) change along the bar, then you need a cut on either side of where the change occurs.

2. Draw free-body diagrams labeling all unknown forces, . Must be consistent by assuming + directions; must be consistent with equal and opposite forces on adjacent free-body diagrams where a cut was made).

3. Write appropriate COLM and COAM equations.4. Determine if problem is statically determinate or

indeterminate.5. If indeterminate, determine appropriate displacement

kinematic boundary conditions for the problem (elongations sum to zero, elongations opposite, one elongation multiple of another, etc.).


6. Determine elongations in terms of unknown (assumed) internal forces. Remember positive direction of elongation is in same direction as assumed member force .

7. Evaluate kinematic B.C. in terms of displacements (now in terms of unknown internal forces).

8. Solve equations obtained in steps 3 and 7 for the unknown internal forces.

9. If appropriate, plot member force for all bars, i.e., the structure (particularly when bars in series; will help you understand what is going on).

Evaluate elongations, strains and stresses in each bar (by substituting the internal forces). If you followed the "+"=tension sign convention rigorously, then + elongation means stretching and - means compression, + stress means tension, etc.


Example 3: Two elastic bars in parallel.

Determine: force, stress and deflection in each vertical bar.

As before, we have four governing equations: Equilibrium, Stress/strain, Kinematics and Boundary Conditions.


Use a “free-body diagram” to determine equilibrium of the forces acting in each bar. Assume force in the bars are P1 and P2 (positive). Take a free-body by cutting each bar below its fixed point:

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 16Equilibrium of the free body in terms of forces requires that

or (1)

Elongation of each bar is:

elongation of bar 1 =

elongation of bar 2 =

Displacement B.C.: = or (2)


We now have two equations and two unknowns (P1 and P2). Writing the two equations in matrix notation:

Note: for simple problems, you can use Cramer’s rule of determinants to solve the system of equations. Otherwise, you Maple, EES, calculator, etc..

Solve for P1 and P2 gives:


Stress in each bar is

and the deflection is:

Special Cases:

for all cases below!

1. , : (tension), (tension)


2. (bar 1 has larger area), :

, (bar 1 carries more load)

, (stresses same!)

is twice , but is 1/2 of , so stress is same

3. , (bar 1 has higher E, i.e., "stiffer"):

, (bar 1 carries more load)

, (stress not same)

Since (kinematic BC) and , strain is same in each bar; but is twice , so is twice .

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 21Class Exercise: The horizontal bar is rigid and pinned at it's left end. The horizontal bar rests on two vertical bars as shown and has a 1 Kip load at its right end. For each vertical bar, determine: 1) force, 2) stress, 3) displacement at it's top and 4) axial strain.

You can assume that motion is "small" so that the vertical bars remain vertical when loaded; also ends are rounded so they carry only axial forces.

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 22Example 4: Uniaxial elastic bar subjected to a uniform temperature increase of T:

The bar is fixed between two walls and has a constant cross-section A.

Determine: axial strain and stress in the bar and the force on the wall.


For this problem, we have 4 governing equations to satisfy:

1. Equilibrium :

2. Stress-Strain :

3. Kinematics: , ,

= axial strain measured/observed

(e.g., by a strain gage)

4. Boundary Conditions : &

or


Solution

1) =0 (zero because of B.C.)

2) Combine stress-strain, kinematics, and boundary condition to obtain

3) (bar is in compression)

P is the force in the bar; hence force on wall is also P.


Suppose we have an aluminum bar of area A=0.1 in2 with

E=10x106 psi and =6x10-6 (in/in)/F

and thermal loading of T=250F (250F temperature rise above the reference temperature)

Then

and

(compression)


Bar with Distributed Axial Load

Consider a bar with a distributed axial load [units of force/length].

Determine the differential equation relating the internal force, P, to the applied distributed load, . Note that the internal force P will be a function of x. First, consider a free-body of a differential section at any point x:


For equilibrium in x direction:

.

Divide by dx and take limit to obtain

Recall . Substitute P into

equilibrium to obtain the governing ODE .


Note that in obtained the last solution, we have combined the required governing equations:

1. Static Equilibrium (from COLM):

2. Constitutive (Stress-Strain):

3. Kinematics (Strain-Displacement):

and obtained a single second-order ODE

In solving this ODE, we will need two boundary conditions; one for the axial displacement ( ), and one for internal force (P, or stress ).


Two similar, but different, examples of axial loading:Case 1:

or,

Note: stress is constant (not a function of x)!


Case 2:

Note: total load on beam is

(same as case 1, but now distributed over length).

Governing ODE:

integrate

Need to apply a boundary condition to obtain . Recall that

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 31B.C. for P (internal axial force):

Do a free body of the bar at any x:

For x=100", P(100")=0

Applying B.C. for P (note: ):


Substituting into the reduced ODE:

Integrate above to obtain:

B.C. for : bar is fixed at left end, so

Hence,

Substituting EA values gives final displacement solution:


Axial displacement at x=100" (end):

Stress is given by:

Axial stress varies with x!!

Note on these similar but very different cases:


1. Concentrated load of 20,000 lb at end2. Distributed load of 200 lb/in (but total load of 20,000 lb)

Displacements completely different:Case 1: displacement varies linearly with x (0.1" at end)Case 2: displacement varies quadraticly with x (0.05" at

end).Stresses are completely different:

Case 1: stress is constant (10,000 psi)Case 2: stress varies linearly with x (10,000 psi at left

end, to 0 at right end)

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 35An Alternate Approach for Bars with Axial Distributed Load

The 4 governing equations that must be satisfied for 1-D are:




4. Boundary Conditions: Depends on the problem

In COLM, recall that is the body force per unit volume applied to the system. Consider a prismatic bar with cross-

section area, A, and length, L. Then .


If we multiply by A, then we can define the applied force per unit length of the bar as . Note that

.

Now take COLM and multiply by A/A:

COLM can now be written as:

1a. Static Equilibrium (from COLM):


Consider the problem where and A are both constant (not function of x). Let

1. Apply COLM: .

Integrate wrt to x to obtain: . Need B.C. for . At x=L, . Get this from free body:

Thus:


Apply B.C. to stress solution:

Solution for stress becomes:

or

2. Apply Constitutive Equation:

or

3. Apply Kinematics:

. Integrate wrt to x to obtain:


Need B.C. for . Bar is fixed at x=0, thus . Applying the B.C. to displacement solution:

Hence, solution for axial displacement is

Recall, the above solution for stress, strain and displacement is only valid for =constant= . If or A is a function of


x, then in Step 1/1a (solution of COLM equation), use and .

Note that this alternate approach to solving problems with a distributed axial load is really no different then the first approach. In the alternate approach, we solve the governing equations in succession:




In the first approach, we combined all equations to get a

second order ODE to solve: .

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 41Class Exercise: Distributed load of 200 lb/in over length of bar and 10,000 lb point force at end:

Determine:a)

b)

2001, W. E. Haisler Chapter 11: Axial Deformation and Stress of Bars 42Other Cases1. Assume a long bar which is slightly tapered with an end load F, and whose cross-sectional area is a function of x: A=A(x).

Stress-Strain: Kinematics:

Equilibrium at x:

Displacement B.C.:


Combine all constitutive and kinematics equations to obtain

or

Integrate from 0 to x':

Displace B.C.: so

In order to complete the solution, must have a specific case for A=A(x).


2. Non-constant, axial force at any cross-section, F=F(x).

Same as above except leave F inside of the integral.

F can be a function of x if gravity acts in the x direction.

chapter 11 - stress, strain and deformation in...

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