Chapter 11 - Stress, Strain and Deformation in...

2000, W. E. Haisler 1

Beam Bending (Chapter 13)

Classical bending of beams is characterized by the following assumptions/restrictions:

Geometry - long and slender prismatic beam along x axis


Applied Loading - Transverse forces or tractions normal to the x axis. Moments about the y or z axis.

Kinematics - Predominate deflection is normal to the x axis. Predominate strain is in the axial (x) direction. Assume small strain and rotations of beam so that “axial strains vary linearly over cross-section” or “plane sections remain plane”. We will explain these assumptions shortly.

In terms of the general elasticity problem, the above can be stated as follows:

2000, W. E. Haisler 3 Equilibrium . If the cross-sectional dimensions are small

compared to the beam length, then applied transverse tractions (in y and z directions) will be small compared to the resultant internal stress in the x direction. --> small transverse loads produce large axial stresses. Thus we assume that the only major stress is xx (all other stresses are zero or negligible).

Equilibrium reduces to . Implies that xx = xx (y,z).

Stress-Strain . We assume a linear isotropic material so that and

Kinematics/Displacements . Includes strain-displacement

relation and the kinematic (displacement)

boundary conditions for ux.


Equilibrium of BeamsAxial force, shear and moment diagrams

External transverse loading and/or bending moments produce internal axial and shear stresses. These can be put in terms of equivalent shear and moment resultants acting on the cross section:


Assume the bar has a cross-sectional area of A as below and the x-axis is along the centroid of the beam cross-section:

Axial force in x direction =

Moment about z axixs =


Shear force in y direction =


The internal forces and moments must be in equilibrium on a “free-body diagram.” Consider a segment x (free-body) from a beam such as:


Sum forces in x direction: . Divide by

x to obtain (as x 0): .


Sum forces in y direction:

Divide by x to obtain

Take limit for x. The integral term becomes (since is a constant for ). Hence:

Sum moments about the center of the differential element:


Divide by x and take limit. Note that is a constant as and hence the integral terms cancel. Thus

Note: Vy(x) Mz (x)zero constantconstant linear in xlinear in x quadratic in x


You can combine last two equations to get:

The differential equations for shear and moment can be integrated. Consider the moment equation and integrate from some point x0 to x1. We assume that has already been determined.

or

Note: Last equation says that Bending Moment is related to the “area under the shear diagram.”


Shear and Moment Diagrams are important because:

they provide information on internal shear forces and bending moments due to the applied loads,

the internal shear and bending moment can be related to internal axial stress and shear stress ,

they identify the location of maximum internal shear and moment (and stress), and

they will be used to determine the transverse deflection,


Shear and Moment Diagrams can be obtained in two ways:

1. Using the differential equations of equilibrium

, ,

2. Drawing a free-body diagram and useing equations of equilibrium ( Forces = 0, Moments = 0).


1. Cantilever beam with end moment, m

Sum forces in x and y direction to obtain P(x)=0 and . Sum moments at any point x to obtain

.


2. Cantilever beam with end load, F.

Sum forces in y direction to find . Sum moments at any point x to obtain or

.


The shear and moment diagrams are:


3. Cantilever beam with uniform distributed load, po

Sum forces in y direction to obtain .


Sum moments at any point x to obtain

or

Shear and bending moment diagrams are given by:


3a. Cantilever beam with uniform distributed load, po. Use the integration method to obtain V&M diagrams.

We know the shear at x=L is 0, so lets integrate from x=L to

any point x:

or

or


Next, integrate the shear equation to obtain the moment equation.

We know the moment is zero at the free end, x=L, so integrate from x=L to any point x:

or

or


4. Cantilever beam with 2 shear forces

Make a free-body by cutting at some point x to right of the 75 lbf load. Sum forces and moments about left end of free-body to obtain:


Make a free-body by cutting at some point x to the left of the 75 lbf load. Sum forces and moments about left end of free-body to obtain:


Now draw the shear and moment diagrams.


5. Simply Supported Beam with Point Load

From equilibrium: R1 + R2 = P and R1(8) - R2(12) = 0. Thus R1 = P (12/20) and R2 = P (8/20).


6. Simply Supported Beam with Distributed Normal Load

From equilibrium: R1 = R2 = poL/2.


7. Simply Supported Beam of length L=10 ft, with distributed normal load po =50 lb/ft (up) and force P = 200 lb (down) at x=6 ft.

For equilibrium: R1 = 170 lb, R2 = 130 lb

Obtain the shear diagram by drawing free-body diagrams for the segment to left and right of the force P:


. Free-body of beam to left of 200 lb force:


. Do a free-body to right of 200 lb force:


Alternately, we could determine the moment diagram using differential equation of equilibrium

We can determine the moment diagram by integrating the shear diagram. Consider the left segment of the beam and integrate from x=0 to any point x 6 (note: M(0) = 0):

The above equation is valid only from x=0 to 6. At x=6 ft, the above moment equation gives M(6 ft) = -120 ft lb.

2000, W. E. Haisler 31We can now integrate the shear diagram for the right portion of the beam (from x=6 to 10) knowing that M(6) = -120.


Determining location of maximum momentFor the left segment (x=0 to 6) we have

and

The point of the maximum moment can be obtained from

calculus, ie, where . Solving for x, we

obtain x=170/50=3.4 ft. At this point .

Note that from the differential equation relating moment and

shear, , so location of maximum moment is

where shear . From the shear equation above, is zero at x=170/50=3.4 ft. Same result!!


Some books may use a different sign convention for Shear:

From equilibrium, the change in direction of the assumed positive shear changes both the shear and moment equilibrium equations:


Chapter 13 -Analysis of Linear Elastic Solids - ContinuedBeam Bending - Stress and Displacement

Classical bending of beams is characterized by the following assumptions/restrictions:

Geometry - long and slender prismatic beam along x axis

A

L

hw

L is largecompared to h and w

cross section may be any shapex

y

z

The x axis is located at the centroid of the cross-section.


Applied Loading - External transverse loading and/or bending moments produces internal shear and moment resultants acting on the cross section

Txxx

y

xy

Txy

xy

xy

P

V

M

==>

Stress Distribution on cross-section ==> Force & Moment Resultants

==>

z


Equilibrium at any point x:

Resultant Axial force in x direction =

Resultant Moment about z axis =

Resultant Shear force in y direction =


Stresses - Major stress is Txx .

Kinematics - In order to obtain a “feel” for the kinematics (deformation) a beam subjected to bending loads, it is informative to conduct some experiments. The following photograph show a long beam with a square cross-section. Straight longitudinal lines have been scribed on the beam’s surface which are parallel to the top and bottom surfaces (an thus parallel to a centroidally-placed x-axis). Lines are also scribed around the circumference of the beam so that they are perpendicular to the longitudinals (these circumferential lines form flat planes as shown). The longitudinal and circumferential lines form a square grid on the surface. The beam is now bent by moments at each end as shown in the


lower photograph. After loading, we note that the top line has stretched and the bottom line has shortened. If measured carefully, we see that the longitudinal line at the center has not changed length. The longitudinal lines now appear to form concentric circular lines. We also note that the vertical lines originally perpendicular to the longitudinal lines remain straight and perpendicular to the longitudinal lines. If measured carefully, we will see that the vertical lines remain the same length. Each of the vertical lines (as well as the planes they form) has rotated and, if extended downward, they will pass through a common point which forms the center of the concentric longitudinal lines. The flat planes originally normal to the longitudinal axis remain flat planes and remain normal to the deformed longitudinal lines. The squares on the surface are now quadrilaterals and


each appears to have tension (or compression) stress in the longitudinal direction (since the horizontal lines of a square have changed length) and perhaps also some shear stress (since opposite vertical lines of a square have rotated different amounts).


Thus, to begin the theoretical development, we make some kinematic assumptions based on the experimental observation. We assume that the predominate deflection is normal to the x axis. Predominate strain is in the axial (x) direction. Assume small strain and rotations of beam so that “axial strains vary linearly over cross-section” or “plane sections remain plane”.


In terms of the general elasticity problem, the above can be stated as

1. If the cross-sectional dimensions are small compared to the beam length, then applied transverse tractions (in y and z directions) will be small compared to the resultant internal stress in the x direction. --> small transverse loads produce large axial stresses. Thus we assume that the only major stress is Txx (all other stresses are zero or

negligible). The stress tensor reduces to .


Equilibrium (Conservation of Linear Momentum reduces to

. This implies that Txx=Txx(y,z) for any point x.


2. Stress-Strain. We assume a linear isotropic material so that stress and strain are linearly related to each other:

Txx = E xx and yy = zz = - xx = - (/E) Txx. If Txx=Txx(y,z), then xx=xx(y,z) also.

3. Strain-Displacement. .

4. Kinematic assumptions: All deformation is described by the displacement, , and rotation of the centroidal axis. Vertical displacement is function of x only:


uoy

ux

(x)

y

(x)=rotation of mid-surface

x

y

=- y

=duoy/dxx

y

normal remains normalto deformed center line andremains a straight line

ux(x,y) = axial displacement of any point in the beam. uoy(x) = transverse displacement of centroidal axis.


Write all the pertinent equations (for bending about z axis only):

1. Txx= E xx Txx= Txx(x,y) and xx = xx(x,y)

2.

3. Equilibrium at any point x:

Substitute for stress and strain (1 & 2) into (3).


x

y y

z

A

We assume that E=E(x), ie, E is constant over cross-section. Remove E and uoy(x) from integral to obtain


.

Define = moment of inertia about z axis.

Thus

or

Note:

Solve for Txx from above:


Remember: M=M(x), uoy=uoy(x), and Txx = Txx(x,y).


1. Cantilever beam with end load, P

x

P

LAssume EI = constant

M(x)

xL

M(x) = P(L-x)PL

V(x)

xL

V(x) = PP

. Integrate ODE twice to obtain


C1 and C2 =

constantsApply B.C. (x=0 is fixed from displacement and rotation):

=> C2 = 0

=> C1 = 0

Thus transverse displacement is

Note:


Solve for stress:

For a given cross-section with I about z axis, max stress is at x=0 and y = +/- max value. At y = +max value (top surface), Txx is negative (compressive).


2. Cantilever beam with uniform distributed load, po

x

po

LAssume EI = constant

M(x)

xL

M(x) = p

V(x)

xL

V(x)=pPL

o(L-x)PL2/2

o(L-x)2 /2

. Integrate ODE twice to obtain


Apply B.C. (x=0 is fixed from displacement and rotation): => C2 = 0

=> C1 = 0


Note:


Solve for stress:

For a given cross-section with I about z axis, max stress is at x=0 and y = +/- max value. At y = +max value (top surface), Txx is negative (compressive).


3. Cantilever beam with 2 shear forces

x

PQ

a LAssume EI = constant

M(x)

xL

V(x)

xL

P

P-Q

a

P(L-a)

a

PL-Qa

1. Integrate ODE with moment from x=0 to x=a. B.C. are displacement and slope = zero at x=0. Note: the two constants of integration equal zero.


for

2. Integrate ODE with moment from x=a to x=L.

for

B.C. are that displacement and slope at x=a in above must match the displacement and slope solution for at the point x=a:

u(x=a) = (a2/6EI) [P(3L-a) - 2Qa] and du/dx (x=a) = (1/2EI) [P(2La-a2) - Qa2].


Solve for the two constants: and

Substitute constants into uoy and rearrange to obtain

for


Solve for the stress: . Substitute

approriate M(x) consistent with x (ie,

).


4. Simply Supported Beam with Point Load

x

P

a NO MOMENTS AT ENDSWITH SIMPLE SUPPORTSL

Assume EI = constant

From equilibrium: R1 + R2 = P and R1a - R2(L-a) = 0. Thus R1 = P (L-a)/L and R2 = P (a/L).

M(x)

xL

V(x)

xL

P(a/L)

-P(L-a)/L

P(a/L)(L-a)


5. Simply Supported Beam with Distributed Normal Load

x

NO MOMENTS AT ENDSWITH SIMPLE SUPPORTS

po

L

Assume EI = constant

From equilibrium: R1 = R2 = poL/2.M(x)

xL

V(x)

xL

L/2poM(x)=(po/2) x (x-L)

. Integrate ODE twice to

obtain


Apply B.C. (uoy = 0 at x=0 and x=L): => C2 = 0

=> C1 =

poL3/(24EI)


Note:


Solve for stress:

For a given cross-section with I about z axis, max stress is at x=L/2 and y = +/- max value. At y = +max value (top surface), Txx is negative (compression).


Centroids and Moments of Inertia

A1

A2

x

y

d1d2

(centroid of A1)

(centroid of A)

A = A1+A2y2

y1y

(centroid of A2)

reference axis

or or

the transfer theorem


Determination of shear stress due to bending stress:

xy

z

x x+dxdx

y=h

TxxTxx(x+dx)

(x)

ct(y) Tyx(y)

y

c

y

z

h

Txy

y

dyt(y)

dA = t dy

end view

Sum forces in x direction for an x-y plane located at y. From beam bending, we know:

and and


Sum forces due to stresses in x direction for the cross section located from any point y=h to outer most point (y=c).

Note the (y) indicates values evaluated at a specific y point. Divide by dx in above. Note that the Txx terms become

Dividing equilibrium equation in x direction by t(y) gives


The integral term is a geometrical term and we define

Q(y=h) = = first moment of the cross-sectional

area from the point y=h to the outer boundary (y=c). Note that in general the thickness may vary so that, t=t(y).

Thus the shear stress at any point x,y becomes:


For a cross-section where t = constant, the integral for Q results in a qudratic function in y (shear stress is zero at top and bottom surface and a parabola through the cross-section).Consider a rectangular cross-section of width b and height h. Calculate the moment of inertia about the z and y axes:

y

zydy

b

h


Similarly,

For a circular cross-section of diameter D, we find Iyy = Izz = D4/64 and J= Iyy + Izz = D4/32.


Consider a rectangular cross-section of width t and height 2c Now calculate Q(h):

y

zh

c

c

t


Note: at top or bottom, at center, Q is 0 at top and bottom, max at center, and varies quadraticly from top to bottom.

For the rectangular cross section, Txy is max at center and

given by: .


Consider an “I” section:

1 in

1 in

1 in

12 in

y

z

8 in

y,z axesat centroid

Determine: 1. Centroid (location of y,z axes)2. Moments of inertia Iyy and Izz3. Q(y=0)

Chapter 11 - Stress, Strain and Deformation in...

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