beams: pure bending1 beams: pure bending (4.1-4.5) mae 314 – solid mechanics yun jing

Beams: Pure Bending 1

Beams: Pure Bending (4.1-4.5)

MAE 314 – Solid Mechanics

Yun Jing


Beams in Pure Bending Prismatic beams subject to equal and opposite couples acting in the same plane are in pure bending.


Pure vs. Non-Uniform Bending Pure bending: Shear force (V) = 0 over the section Non-uniform bending: V ≠ 0 over the section

Pure bending• Moment → normal stresses

Non-uniform bending • Moment → normal stresses • Shear force → shear stresses (Ch. 6)


Pure Bending: Assumptions Beam is symmetric about the x – y plane All loads act in the x – y plane


Pure Bending: Curvature Sections originally perpendicular to the axis of the member remain plane and perpendicular: “Plane sections remain plane.”

Sign convention Positive bending moment:beam bends towards +y direction Negative bending moment: beam bends towards -y direction

Right angle


Pure Bending: Deformation Since angles do not change(remain plane), there is noshear stress. The top part of the beam contractsin the axial direction. The bottom part of the beam expandsin the axial direction. There exists a line in the beam thatremains the same length called theneutral line. Set y = 0 at the neutral line. ρ = radius of curvature εx < 0 for y > 0 and εx > 0 for y < 0


Pure Bending: Axial Strain The length of DE is LDE = ρθ The length of JK is LJK = (ρ-y)θ Axial strain at a distance y fromthe neutral axis (εx):

Maximum compressive strain occurs on the upper surface. Maximum tensile strain occurs on the lower surface.

)( y

L

LL

L DE

DEJKx

yx

LDE

LJK


Pure Bending: Axial Strain

c = maximum distance between the neutral axis and the upper or lower surface When c is the distance to the surface in compression When c is the distance to the surface in tension

cmax

cmax


Pure Bending: Transverse Strain Recall there are no transverse stresses sincethe beam is free to move in the y and z directions. However, transverse strains (in the y and zdirections) exist due to the Poisson’s ratioof the material.

ρ' = radius of anticlastic curvature = ρ/υ

xy yy

xz yz


Pure Bending: Normal Stress Let us now assume that the beam is made of a linear-elastic material.

The normal stress varies linearly with the distance from the neutral surface.

EyE xx


Pure Bending: Normal Stress Recollecting that the applied loading is a pure moment, we calculate resultant loads on the cross-section. The resultant axial force must be equal to zero.

0 AA

x

A

x ydAE

dAEdA

0A

ydA

…which is the definition of the centroid, so the neutral axis is just the centroid of the section.

01

A

ydAA


Pure Bending: Normal Stress The resultant moment about the z-axis must be equal to the applied moment M.

MdAyE

dAyEdAyAA

x

A

x 2

Definition of the second moment of inertia, I

EI

M

I

Myx

Centroids and Moments of Inertia 13

Example ProblemA steel bar of 0.82.5-in. rectangular cross section is subjected to two equal and opposite couples acting in the vertical plane of symmetry of the bar. Determine the value of the bending moment M that causes the bar to yield. Assume m=36 ksi.


Centroids and Moments of Inertia (A.1-A.5)


Yun Jing


Review From Last Time Beams in pure bending (no shear forces) Plane sections remain plane +M, upper surface in compression +M, lower surface in tension Transverse strains exist due to υ

yx

EI

M I

Myx

y

zy Neutral Line

Radius of Curvature


Centroids To match with our axis convention use centroid in the y-direction.

Centroids for common shapes are located in the back cover of your textbook. If a cross section shape has an axis of symmetry, then the centroid passes through that axis of symmetry.

A

ydAy A


Centroids for Composite Shapes Choose a reference x-axis (normally the base of the shape). Divide the composite shape into individual shapes. Calculate the area for each shape. Find the location of the centroid for each shape relative to the reference x-axis.

i i

i ii

A

yAy

x

y

y=0

1y

2y

3y

1b2b

3b

1h

2h

3h

1A

2A

3AA1

A2

A3

Areah1b1

h2b2

h3b3

0.5h1

h1+0.5h2

h1+h2+0.5h3

0.5h12b1

h2b2(h1+0.5h2)

h3b3(h1+h2+0.5h3)

ii

i yA


Moments of Inertia Moment of inertia with respect to the x-axis

Moments of inertia for common shapes are located in the back cover of your textbook.

x

x

x

x’

Ix’

Ix

Ix

Ix


Moments of Inertia for Composite Shapes Divide the composite shape into individual shapes. Calculate I of each shape about its own centroid. Calculate I of each shape about the centroid of the entire composite shape using the parallel axis theorem:

x

1b2b

3b

1h

2h

3h

2A

x’1A

x’’

x’’’3C

C

1d

3d

2')()( iiixix dAII

'xIA1

A2

A3

2AdI x b1h1

3/12

b2h23/12

b3h33/12

b1h1d12

b2h2d22

b3h3d32

b1h1(d12+h1

2/12)

b2h2(d22+h2

2/12)

b3h3(d32+h3

2/12)


Example ProblemFind the location of the centroid and the moment of inertia of the sectionbelow (the moment is applied about the horizontal axis).5 in

3 in

0.5 in

1.5 in4 in 3 in


Stress in the Elastic Range Revisiting the beam subject to pure bending

For positive bending moment Maximum tensile stress = Mc2 / I Maximum compressive stress = Mc1 / I

For negative bending moment Maximum tensile stress = Mc1 / I Maximum compressive stress = Mc2 / I

The ratio I/c1 (or I/c2) is known as S, elastic section modulus. Larger values for S indicate a beam that is more resistant to bending.

c1

c2


Deformations in the Elastic Range Which shapes are most resistant to bending?

S-beam rectangular

Why? Large portion of the cross section is located far from the neutral axis, resulting in large values of S.

The deformation of a member due to bending can be calculated byEI

M

1


Example ProblemA beam with cross-section like the previous example is subjected to apure bending moment of M = -4500 in·lb. Find the maximum tensile andcompressive normal stresses. 5 in

3 in

0.5 in

1.5 in4 in 3 in


Example ProblemKnowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B.


Example ProblemThe beam shown is made of nylon with an allowable stress of 24 MPa intension and 30 Mpa in compression. Determine the largest couple M thatcan be applied to the beam.


Example ProblemTwo vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam.


Bending in Composite Materials (4.6)


Yun Jing


Normal Stress in Composite Materials Strain is still the same as homogenous material because it is not dependent on material properties. The stress distribution is no longer linear. Use different material properties in each section.


Normal Stress in Composite Materials

Top section: Bottom section: dA

yEdAdF

111

)(1122 dAn

yEdAynE

dAyE

dF

1

2

E

En

Original Shape New Shape


Bending in Composite Materials The new “homogenous” beam produces the same response to bending as the composite beam. We can now calculate the centroid and moment of inertia for this beam. To calculate the axial stresses, use σx for material 1 and nσx for material 2. Curvature of the composite member is

IE

M

1

1


Reinforced concrete beam Beams are reinforced by steel rods Concrete is very weak in tension, the steel rods will carry the entire tensile load Concrete in the beam acts effectively only in compression


Example ProblemA bar having the cross section shown has been formed by securelybonding brass and aluminum stock. Determine the largest permissiblebending moment when the composite bar is bent about a horizontalaxis. Ea = 70 GPa, σall,a = 100 MPa, Eb = 105 GPa, σall,b = 160 MPa.


Example ProblemKnowing that the bending moment in the reinforced concrete beam is +100kip.ft and that the modulus of elasticity is 3.625106 psi for the concrete and 29 106 psi for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.


Example Problem A flitched beam consists of two 50mm×200mm wooden beams and a 12 mm×80mm steel plate. The plate is placed centrally between the wooden beams and recessed into each so that, when rigidly joined, the three units form a 100mm×200mm section as shown in the figure. Determine the bending moment the beam can withstand when the maximum bending stress in the timber is 12MN/m2. What will then be the maximum bending stress in the steel. (E for the timber is 10GPa, for steel is 120GPa)

beams: pure bending1 beams: pure bending (4.1-4.5) mae 314 – solid mechanics yun jing

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