Combined Loading

mi@seu.edu.cn

• Unsymmetric Bending（不对称弯曲）

• Tension & Bending（拉弯组合）

• Eccentric Compression（偏心压缩）

• Core of Cross-sections（截面核心区域）

• Core of Rectangular Cross-sections（矩形截面核心区域）

• Core of Circular Cross-sections（圆形截面核心区域）

• Tension & Torsion（拉扭组合）

• Bending & Torsion（弯扭组合）

• Tension, Bending & Torsion（拉弯扭组合）

Contents

2

• A circular bar subjected to a single type of load

3

Introduction

x z zM y I

*

Sxy z zF S I b

x pT I

• The principle of superposition is used to determine the

resultant stress & strain

• Prismatic bars are frequently subjected to several loads

simultaneously

• Conditions for the principle of superposition

- Linear elasticity & small deformation

4

Introduction

- No interaction between variously loads

Unsymmetric Bending

• Analysis of pure bending has been limited

to members subjected to bending moments

acting in a plane of symmetry.

• Will now consider situations in which the

bending couples do not act in a plane of

symmetry.

• In general, the neutral axis of the section will

not coincide with the axis of the couple.

• Cannot assume that the member will bend

in the plane of the couples.

• The neutral axis of the cross section

coincides with the axis of the couple

• Members remain symmetric and bend in

the plane of symmetry.

5

Wish to determine the conditions under

which the neutral axis of a cross section

of arbitrary shape coincides with the

axis of the moment as shown.

• Moment vector must be directed

along a principal centroidal axis

0

or 0 product of inertia

zy

z

yz

M yM z dA

I

yz dA I

• The resultant force and moment

from the distribution of

elementary forces in the section

must satisfy

0 applied couplex y zF M M M

• If neutral axis passes through centroid

0

or 0

zx x

z

M yF dA dA

I

y dA

• Stress distribution

zz

z

M yM M y dA

I

• Superposition is applied to determine

stresses in the most general case of

unsymmetric bending.

Unsymmetric Bending

6

, .x z z y z zM y I w M EI

• Construct a coordinate system.

• In x-y plane: positive Mz results in compression for y < 0.

• Bending stress and deflection:

• The same sign conventions can be used for bending in x-z plane.

• Positive My results in compression for z < 0.

• Bending stress and deflection: , .x y y z y yM z I w M EI

• Sign Convention

7

z

x

y

b

h

x

zF

,y z

yFa

Unsymmetric Bending

, ,

, , 0

zz y x

yz zx x x

yz y

y z x

y

M yM F x a a x L

M zI M yM z I I

M F x x LI

8

z

x

y

b

h

x

zF

,y z

yFa

Unsymmetric Bending

• Equation for Neutral axis:

00

0

0

0

tan

tan

ta

0

n

yzx

z y

y y

z z

z

y

y z

y

z

y y

z z

M zM y

I I

I Ix az

y I x I

I Ix a

I x

M

M

I

F

F

I a

I

• The neutral axis passes through the centroid of the cross-section.

• The maximum stresses occur at the two farthest points from the neutral axis.

• With the exception of Iy = Iz, i.e. for circular/square cross-section, the bending

stress and deflection cannot be calculated from the resultant moment.

• Bending in a single plane occurs if and only if the orientation of neural axis stays

the same for every cross-section, i.e. Mz/My=constant for a prismatic beam.

• In general, superposition should be resorted to determine both the bending stress

and the deflection. 9

0 0,y z

yM

zM

yF

zF

z

y

Unsymmetric Bending

A 1600 lb-in couple is applied to a

rectangular wooden beam in a plane

forming an angle of 30 deg. with the

vertical. Determine (a) the maximum

stress in the beam, (b) the angle that the

neutral axis forms with the horizontal

plane.

SOLUTION:

• Resolve the couple vector into

components along the principle

centroidal axes and calculate the

corresponding maximum stresses.

• Combine the stresses from the

component stress distributions.

• Determine the angle of the neutral axis.

Sample Problem

10

• Resolve the couple vector into components and

calculate the corresponding maximum stresses.

3 4112

3 4112

1 4

1600lb in cos30 1386lb in

1600lb in sin 30 800lb in

1.5in 3.5in 5.359in

3.5in 1.5in 0.9844in

The largest tensile stress due to occurs along

1386lb in 1.75in452.6ps

5.359in

z

y

z

y

z

z

z

M

M

I

I

M AB

M y

I

2 4

i

The largest tensile stress due to occurs along

800lb in 0.75in609.5psi

0.9844in

z

y

y

M AD

M z

I

• The largest tensile stress due to the combined

loading occurs at A.

max 1 2 452.6 609.5 1062psi

z

y

11

• Determine the angle of the neutral axis.

0 0 000 04

0

0

1386lb in 800lb in0 258.63 812.68

5.359in 0.9844

812.68tan 3.142

258.63

72.4

yzx

z y

o

M z y zM yy z

I I

y

z

12

z

y

0 0,y z

• Solution

2. Critical cross-section

• For the I-32a beam shown, L = 4 m, [σ] = 160 MPa, F = 80 kN, α =

5°. Analyze the strength condition.

; ;2 4 4

y zz y

F L F LLx M M

3. Neutral axis & maximum stresses

4 4

y yz zx

y z y z

M z F LM y F Lz y

I I I I

1. Decompose the load

Sample Problem

13

cosyF F

sinzF F z

y

z

y

F

2L 2L

xF

4. Strength check:

max 217.8MPa 160MPay z

y z

M M

W W

6. If α = 0， max 115.6z

z

MMPa

W

22

zy www 5. Deflections:

cosyF F

sinzF F z

y

Remark: The fact that the neutral axis is not necessarily perpendicular

to the resultant moment can greatly affect the stresses in a beam,

especially if the ratio of the principal moments of inertia is very large.

Under these conditions the stresses in the beam are very sensitive to

slight changes in the direction of the load and to irregularities in the

alignment of the beam itself.

14

• Stress due to eccentric loading found by

superposing the uniform stress due to a centric

load and linear stress distribution due a pure

bending moment

centric bendingx x x

z

z

M yP

A I

• Validity requires stresses below proportional

limit, deformations have negligible effect on

geometry, and stresses not evaluated near points

of load application.

Tension & Bending (Eccentric Tension)

15

, F P M Pd

• Strength condition:

max max, .

An open-link chain is obtained by

bending low-carbon steel rods into the

shape shown. For 160 lb load, determine

(a) maximum tensile and compressive

stresses, (b) distance between section

centroid and neutral axis

SOLUTION:

• Find the equivalent centric load and

bending moment

• Superpose the uniform stress due to

the centric load and the linear stress

due to the bending moment.

• Evaluate the maximum tensile and

compressive stresses at the inner

and outer edges, respectively, of the

superposed stress distribution.

• Find the neutral axis by determining

the location where the normal stress

is zero.

Sample Problem

16

• Equivalent centric load

and bending moment

160lb

160lb 0.65in

104lb in

z

P

M Pd

22

2

2

0.25in

0.1963in

160lb

0.1963in

815psi

x

A c

P

A

• Normal stress due to a

centric load

441 14 4

3 4

3 4

0.25

3.068 10 in

104lb in 0.25in

3.068 10 in

8475psi

z

zx

z

I c

M c

I

• Normal stress due to

bending moment

17

• Maximum tensile and compressive

stresses

max

max

815 8475 9260psi

815 8475 7660psi

• Neutral axis

0

3 4

0 2

0

160lb 3.068 10 in

0.1963in 104lb in

0.0240in

z

z

z

z

M yP

A I

IPy

A M

18

x

z

y

e A(yF,zF)

Fc

x

z

y

e A(yF,zF)

F

c m mz

myx

z

y

e A(yF,zF)

F

• Bars subjected to axial forces deviated a distance from axis.

1. Find the equivalent system of forces at the centroid.

Eccentric Compression

19

mz

my

x

z

y

e A(yF,zF)

F

B(y,z)

y

zO D1

D2.

2. Normal stress on cross-sections.

Nx

F F

A A

z Fx

z z

M y F y y

I I

y Fx

y y

M z F z z

I I

F F

z y

F F y y F z z

A I I

FN = -F，

My = -F · zF,

Mz = -F · yF,

2 2

2 2 1 F F

y y z z

z y

y y z zFI Ai I Ai

A i i

，

• Note the negative sign in bending stresses.

20

y

z

ay

az

A(yF,zF)

3. Equation of neutral axis.

0

2

0 0 z

y zF

ia y

y ；

F

y

yzz

iza

2

000

4. Strength condition:

• The intercepts of neutral axis:

2 2

0 0

2 2

0 1

1 0

F F

z y

F F

z y

y y z zF

A i i

y y z z

i i

21

View from top

max max, .

The largest allowable stresses for the cast

iron link are 30 MPa in tension and 120

MPa in compression. Determine the largest

force P which can be applied to the link and

the neutral axis.

SOLUTION:

• Determine an equivalent centric load and

bending moment.

• Evaluate the critical loads for the allowable

tensile and compressive stresses.

• The largest allowable load is the smallest

of the two critical loads.

• Superpose the stress due to a centric

load and the stress due to bending.

Sample Problem

22

• Determine an equivalent centric and bending loads.

0.038 0.010 0.028m

0.028 bending momentz

d

M Pd P

• Evaluate critical loads for allowable stresses.

377 30MPa 79.6kN

1559 120MPa 77.0kN

A

B

P P

P P

kN 0.77P• The largest allowable load

• Superpose stresses due to centric and bending loads

3 9

3 9

0.028 0.022377

3 10 868 10

0.028 0.0381559

3 10 868 10

z AA

z

z BB

z

PM cP PP

A I

PM cP PP

A I

From geometry,

23

0

0

2 9 3

0 0

2

0 0

868 10 3 10 = 10.33 mm

0.028

zy z

F

y

z yF

ia y

y

ia z

z

• Neutral axis

y

B

A

Cz

y5

10

• Find the maximum normal stress for the bar shown.• Solution:

1. Equivalent System of forces

FN = F =10 kN

Mz = 10×103×5×10-2 = 500 Nm

My = 10×103×2.5×10-2 = 250 Nm

2. Maximum normal stress

max 14 MPay AN z A

A

z y

M zF M y

A I I

10max

MPay Bz BN

B

z y

M zM yF

A I I

； 2

F

zy

y

ia

F

y

zz

ia

2

Intercepts:

5 cm

10 cm

Sample Problem

24

View from left

z

y

O

1

2

3

①

②

③

ay1

az1

• Intercepts of neutral axis:

• Portion of a cross-section within which eccentric

loading results in only compressive stresses.

22

22

;

yzy z

F F

yzF F

y z

iia a

y z

iiy z

a a

；

Core of Cross-sections

25

z

y

h

b

①

②

③

④A

B C

D

4

3 1

6b

6h

2

,12

22 h

A

Ii zz

12

22 b

A

Ii

y

y

• Take side AB as neutral axis

211

ba,a zy

1 1

1 1

22

06

yzF F

y z

ii by z

a a ，

• Take side BC as neutral axis

06 22

FF zh

y ，

• Through corner B, there exists

infinite number of neutral axes

we can take.

2

,zF

y

iy

a

z

y

Fa

iz

2

2 2

2 2

1 0

1 02 2

12 126 6

1 0

F B F B

z y

F F

F F

y y z z

i i

y zh bh b

y z

h b

Core of Rectangular Cross-sections

26

z

y

O

d

①

1

8

d

• Due to axial symmetry, only one neutral axis is

necessary to determine the radius of the core

area:

①:

2

22

1 1

,2

4

40

82

y z

y z

yzF F

y z

da a

di i

dii dy z

da a

，

• Note: no matter whatever shape of a cross-section area, the

corresponding core area is always solid and non-concave.

Core of Circular Cross-sections

27

• Simultaneous tensile and torsional loading of a circular shaft.

2

3

4

16

Nx

x

P

F P

A dT T

W d

28

T

P

T

P

• Solution

Tension & Torsion

yl

aA B

Cz

F

A1

A2

τ

Pz

z

W

T

W

M ,

τ

σ

A1

• Critical section: AA1

σ

A2

A

][4 22

][3 22

3

4

r

r

• Critical points on A: A1 & A2

2 2 2 2

3

2 2 2 2

4

1( ) 4( ) [ ]

1( ) 3( ) 0.75 [ ]

zr z

z P z

zr z

z P z

M TM T

W W W

M TM T

W W W

Bending & Torsion

29

View from left View from right1A

2A

• For the circular shaft shown, L = 50 cm, F = 8 kN, a = 37.5 cm, [σ] =

100 MPa. Find the minimum diameter of shaft AB based on

maximum shearing stress criteria.

yl

aA B

Cz

F

F

M

2 2

2 2 2 2

3

2 2

5 3

3

14 4 [ ]

5 10

3279.8

zr z

z P z

z

z

z

M TM T

W W W

M TW m

Wd mm

• Solution:

2 2

2 2 2 2

4

2 2

13 3 0.75 [ ]

0.75

zr z

z P z

z

z

M TM T

W W W

M TW

What about the maximum

distortion energy criteria?

max3 kNm, 4 kNmT M FL

Sample Problem

30

• Two forces P = 18 kN and F = 15 kN are applied to the shaft with a

radius of R = 20 mm as shown. Determine the maximum normal and

shearing stresses developed in the shaft.

Tension, Bending & Torsion – an Exercise

31

• Unsymmetric Bending（不对称弯曲）

• Tension & Bending（拉弯组合）

• Eccentric Compression（偏心压缩）

• Core of Cross-sections（截面核心区域）

• Core of Rectangular Cross-sections（矩形截面核心区域）

• Core of Circular Cross-sections（圆形截面核心区域）

• Tension & Torsion（拉扭组合）

• Bending & Torsion（弯扭组合）

• Tension, Bending & Torsion（拉弯扭组合）

Contents

32