EMA 3702 Mechanics & Materials Science (Mechanics of ... · Mechanics & Materials Science (Mechanics of Materials) Chapter 5 Beams for Bending . Design of beams for mechanical or

EMA 3702

Mechanics & Materials Science

(Mechanics of Materials)

Chapter 5 Beams for Bending

Design of beams for mechanical or civil/structural

applications

Transverse loading in most

cases for beams

Types of loading:

• Concentrated load Pi

• Distributed load w (N/m)

• Uniformly distributed

• Non-uniformly distributed

Introduction

Common Beam Configurations

Simply supported Overhanging beam Cantilever beam

Continuous beam Fixed beam

Statically Determinate Beams

Statically Indeterminate Beams

Beam fixed at one end

& simply supported at

the other

Example: a simply supported beam with

two transverse loading forces of P1

& P2 and uniformly distributed load w

At an arbitrary cross-section C, draw

body diagram (FBD) for each side:

• Internal shear force V (or V’) that

creates shear stress in cross-section C

• Bending moment M (or M’) that

creates normal stress in cross-section C

I moment of inertia for cross-section C

y Distance from the neutral surface

c Maximum distance from the neutral surface

Equilibrium Analysis for Beam under

Transverse Loads

and 𝜎 = −𝑀𝑦

𝐼 𝜎𝑚𝑎𝑥 =

𝑀𝑐

𝐼

The bending moment at any point

is positive (+) when the external

moments acting on the beam tend to

bend the beam at the point as

indicated (“ends up”)

The shear at any point of a beam is

positive (+) when external forces

(loads and reactions) acting on the

beam tend to shear off the beam at

that point as indicated (“right down”)

Sign Conventions for Internal Shear &

Bending Moment (1)

Positive shear if

trying to twist the

element clockwise

Positive bending

moment if trying to

bend the element

concave up (“ends

up”)

Sign Conventions for Internal Shear &

Bending Moment (2)

Shear and Bending-Moment Diagrams (1)

Example 1

For the simply supported

beam DE under load P at the

midpoint F

Distribution of shear (force) & bending moment along the

axis will help easily determine the maximum shear & normal

stresses that will occur, which is critical to structure design

P 0.5L 0.5L

D E F

Draw FBD for entire DE.

Equilibrium for DE gives:

∑Fy = 0

∑ME = 0

RD = RE = 0.5P

D E F

RD RE

P

RD + RE = P

0.5PL = RDL

Draw FBD for the two sections

when left of F :

Equilibrium gives:

∑Fy = 0

∑M = 0

As to sign of V and M :

V – positive

M - positive

E

P

RE = 0.5P

F

M’

V’

x

y Shear and Bending-Moment Diagrams (2)

D

RD = 0.5P

x M

V = 0.5P

= 0.5Px

D E F

RD = 0.5P RE = 0.5P

P

V = RD = 0.5P

M = 0.5Px

Shear and Bending-Moment Diagrams (3) Draw FBD for the two sections

when right of F :

Equilibrium gives:

∑Fy = 0

∑M = 0

Therefore

M = 0.5PL – 0.5Px = 0.5P(L-x)

As to sign of V and M :

V - negative, M - positive

= 0.5P D

P

RD = 0.5P

M

V

x

= 0.5P(L-x)

x

y

D E F

RD = 0.5P RE = 0.5P

P

V = P - RD = 0.5P

M + Vx = P 0.5L

E

RE = 0.5P V’

M’

0.25PL

0.5L L

0.5P

-0.5P 0.5L L

V

x 0

Shear Diagram

M

x 0

Moment Diagram

P 0.5L 0.5L

D E F

V = 0.5P

M = 0.5Px

V = -0.5P

To the left of F

To the right of F

To the left of F

To the right of F M = 0.5P(L-x)


M

x

0.25PL

0.5L L 0

Moment Diagram

Based on the bending

moment diagram drawn,

for a prismatic beam

with uniform cross-

section and moment of

inertia I and largest

dimension from neutral

axis c,

maximum axial normal

stress for this beam can

be determined:

At mid-point F of beam

x = 0.5L, Mmax = 0.25PL I

cM maxmax


P 0.5L 0.5L

D E F

I

PLc

I

cPL

4

)25.0(

For a simply supported beam

DE with localized bending

moment M0 applied at center

point F as illustrated, please

draw the shear & bending

moment diagram.

Class Exercise

L

D E

FBD for the entire beam DE is

drawn:


∑Fy = 0

∑ME = 0

Therefore,

M0 0.5L

F

D E

RD RE

E M0

F RD + RE = 0

RDL + M0 = 0

RD = -M0/L RE = M0/L

Class Exercise

FBD for section to the left of

the center point is drawn.

Equilibrium gives:

∑Fy = 0

∑MD = 0

V = -M0/L (negative due to

V trying to rotate counter

clockwise)

M = -M0x/L (negative due to

bending concave down or

“ends down”)

D E

RD = M0 /L M0

V

x 0

Shear Diagram

-M0/L

0.5L L

M

x 0

Moment Diagram

-0.5M0

0.5L L

D

RD = M0/L

V

M

x

V = RD

RDx = M

RE = M0 /L

Class Exercise

FBD for section to the right

of the center point is drawn:

Equilibrium gives:

∑Fy = 0

∑ME = 0

V = -M0/L (negative due to

V trying to rotate counter

clock wise)

M = M0(L-x)/L (positive due

to bending concave up or

“ends up”)

V

x 0

Shear Diagram

-M0/L

0.5L L

D E

M0 RD = M0 /L RE = M0 /L

E

RE = M0/L V

M x

V = RE

V(L-x) = M

M

x 0

Moment Diagram

-0.5M0

0.5L L

0.5M0

Load – Shear Relationship for Beam (1)

When x 0

For a simply supported beam AB

subjected to distributed load w,

choose an arbitrary element

section CC’ with width of Δx,

FBD for section CC’ is shown.

Equilibrium along vertical y axis

wdx

dV

The slope of shear diagram (V vs. x)

equals negative local distributed

load per unit length

𝐹𝑦 = 0 V – (V+ΔV) – wΔx = 0

ΔV = – wΔx

Integrating between C and D

Load – Shear Relationship for Beam (2)

D

C

x

xCD wdxVV

Or

V

x

VC

C 0

Shear Diagram

D

VD

curve -under area xwVV CD

Note:

Not applicable when there

is concentrated load as shear

curve will NOT be

continuous

wdx

dV wdxdV

RA

wx

V RB

wΔx

V

Shear - Bending Moment Relationship (1)

or

Equilibrium for moment about C’

0' CM

Therefore,

Vdx

dMWhen x 0

The slope of moment diagram (M

vs. x) equals local shear force

𝑀 + ∆𝑀 +𝑤∆𝑥∆𝑥

2= 𝑀 + 𝑉∆𝑥

∆𝑀 = 𝑉∆𝑥 − 𝑤(∆𝑥)2

2

∆𝑀

∆𝑥= 𝑉 −

1

2𝑤∆𝑥

Integrate between C and D

Vdx

dM

D

C

x

xCD VdxMM

Or

curveshear under area CD MM

Note:

Not applicable when

localized bending moment is

applied between C and D as

M–x will NOT be continuous

V

x

VC

C 0

Shear Diagram

D

VD

M

x

MC

C 0

Moment Diagram

D

MD

VdxdM

Shear - Bending Moment

Relationship (2)

RA

wx

V M

Please draw the shear &

bending moment diagram

for a simply supported

beam with uniformly

distributed load w.

Class Example (1)

w

L

D E

Free body diagram for the

entire beam DE is drawn.


∑Fy = 0

∑MD = 0

RD = RE = 0.5wL

D E

RD RE

w

RD + RE = wL

wL • 0.5L= REL

Class Example (2) Define coordinate system

For shear at D,

VD = RD = 0.5wL

It is positive.

At any point x, shear force is

V

x

0.5wL

0

Shear Diagram

-0.5wL

wxwLwdxVxVx

D 5.0)(0

M

x 0

0.125wL2

Moment Diagram For bending moment,

at D, M = 0

At any point x, the bending

moment is

2

005.05.05.0)( wxwLxdxwxwLVdxMxM

xx

D

0.5L

0.5L

D E

RD = 0.5wL RE= 0.5wL

w

E

Design of Prismatic Beams for Bending (1)

Design of a long (or large-span) beam is often controlled

by consideration of normal stress, which largely depends

on |Mmax|

As a result, the largest normal stress:

If allowable stress is known, the minimum allowable value

for section modulus S = I/c is:

𝜎𝑚𝑎𝑥 =𝑀𝑚𝑎𝑥𝑐

𝐼 𝜎𝑚𝑎𝑥 =

𝑀𝑚𝑎𝑥𝑆

𝑆𝑚𝑖𝑛 =|𝑀𝑚𝑎𝑥|

𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

Design of Prismatic Beams for Bending (2)

Design goal: for the same materials and same support/brace

mechanism, the beam design with the smallest cross-

sectional area (weight per unit length) that meets safety

standard should be selected.

General steps

1. Determine the allowable stress (either from design

specification or from ultimate strength/safety factor).

2. Draw the shear and bending moment diagram and

determine |Mmax|.

3. Calculate the minimum allowable section modulus Smin.

4. Specify the dimension based on the cross-section shape

and relation with Smin.

Class Example For a wide-flange (W-shapes)

prismatic beam to support the

P = 15000 lb of load as shown,

the allowable normal stress is

24000 psi. Please select an

appropriate wide-flange.

Maximum bending moment

P

8 ft

allowablemS

M max

inkipinlbPLM 1440)128(15000max

Maximum axial normal stress

3

2

3

max 6024000

101440in

inlb

inlbMS

allowable

Minimum section modulus

Class Example

For the various wide-flange

beams that has S in the

range:

The best is W1640 due to

smallest cross-section area

(or weight per length)

Shape S (in3) A (in2)

W2144 81.6 13.0

W1850 88.9 14.7

W1640 64.7 11.8

W1443 62.6 12.6

W1250 64.2 14.6

W1054 60.0 15.8

360inS

P

8 ft

For a wide-flange (W-shapes)

prismatic beam to support the

P = 15000 lb of load as shown,

the allowable normal stress is

24000 psi. Please select an

appropriate wide-flange.

Nonprismatic Beams

Prismatic beam: Beam with constant (or uniform) cross-

section along its length

• Normal stress at critical sections (i.e., with largest

bending moment |Mmax|) should, at most, be equal to

allowable stress overdesigned at other parts

w

w

Prismatic beam

Nonprismatic beam

Nonprismatic beam: Beam with

varying cross-section

• Local dimension (or section

modulus) could be varied based

on local bending moment so

that material/cost could be

saved

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 5 Beams for Bending

Homework 5.0

Read textbook chapter 5.1, 5.2, and 5.3 and give an

honor statement confirm reading


Homework 5.1

For the beam, please draw the shear and bending moment

diagrams and determine the equations of the shear and

bending moment curves

w

L

D E


Homework 5.2

For the beam, please draw the shear and bending moment

diagrams and determine the equations of the shear and

bending moment curves

P a L-a

D E F


Homework 5.3

For the beam and loading, please (a) draw the shear and

bending moment diagram, and (b) calculate the maximum

normal stress due to loading on a transverse section at F.

Knowing the cross-section for the beam is square with

b=h=10 cm

1 m

D E F

2 kN/m 10 kN

1 m 1 m


F

Homework 5.4

For the beam and loading shown, design the width b of the beam knowing the material used has allowable stress of 10 MPa and the beam has height of h = 10 cm.

1 m

D E

2m

2 kN/m

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EMA 3702 Mechanics & Materials Science (Mechanics of ... · Mechanics & Materials Science (Mechanics of Materials) Chapter 5 Beams for Bending . Design of beams for mechanical or