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Bending Stress

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• Pure Bending vs. Nonuniform Bending（纯弯曲与横力弯曲）

• Assumptions for Pure Bending（纯弯曲基本假设）

• Neutral Surface & Neutral Axes（中性层与中性轴）

• Kinematics（几何关系）

• Hooke’s Law（物理关系）

• Static Equivalency（静力等效关系）

• Pure Bending Normal Stress Formula（纯弯曲正应力公式）

• Normal Stress Strength Condition（正应力强度条件）

• Stress Concentrations（应力集中）

• Bending of a Composite Beam（复合梁弯曲）

• Bending of a Curved Beam（曲梁弯曲）

Contents

2

• Shearing Stresses in a Rectangular Beam（矩形梁切应力）

• Effect of Shearing Stress/Strain（切应力和切应变效应）

• Shearing Stresses in a Wide-flange Beam（宽翼缘梁切应力）

• Shear Flow in a Thin-walled Beam（薄壁梁剪力流）

• Shearing Stresses in a Circular Beam（圆截面梁切应力）

• Shearing Stresses in an Equilateral Triangular Beam（等边三角梁切应力）

• Shearing Stress Strength Condition（切应力强度条件）

• Rational Design of Beams（梁的合理设计）

• Nonprismatic and Constant-strength Beams（非等直梁和等强度梁）

• Unsymmetric Loading of Thin-Walled Members and Shear Center

（薄壁梁的非对称弯曲与剪力中心）

Contents

3

S 0 constF M ，

• Pure bending (CD)

S 0 0F M≠ ， ≠

• Nonuniform bending (AC & DB)

Pure Bending vs. Nonuniform Bending

4

P

BPP

DCA

a aPl

P

P

SF

PaM

MM

• Before:

• After:

Deformation Characteristics

• Straight longitudinal lines turns into curves.

• Longitudinal lines get shortened under compression and lengthened

under tension.

• Cross-section lines remain straight and perpendicular to longitudinal

curves.5

• Assumption of uniaxial stress state: individual

longitudinal layers are under uniaxial

tension/compression along beam axis, without

stresses acting in between.

• Plane assumption: under pure bending, cross-

sections of beams remain planar and

perpendicular to beam axis and only rotate a small

angle.

Assumptions for Pure Bending

6

• Neutral Axes: intersecting lines of the neutral surface & cross

sections.

Neutral Surface & Neutral Axes

• Neutral Surface the longitudinal layer under neither tension nor

compression.

Neutral Axis

Neutral Surface

z

Neutral Surface

Neutral Axis

z

• Before:

• After:

7

ba

1 2

1 2

o1 o2

y

( )y d d

yd

y

Neutral

Surface

1

2

2

ρ

dθ

o1 o2

a b

Kinematics

8

• The y-coordinate is

measured from the proposed

neutral axis.

y

z

• Normal stress acting on a longitudinal layer is linearly proportional

to its distance from the neutral surface, positive for layers under

tension / negative for layers under compression.

• Remark: the above equation can only be used for qualitative analysis

of stresses in bending beams since it is difficult to measure the

curvature of radius (ρ) of individual longitudinal layers.

Hooke’s Law

9

y

y E y E

y

z

0 NA A

E EF dA ydA Ay

AA

y yzdAE

dAzM

0

2 1

zZ z

A Az

z

z

ME EM y dA y dA I

EI

M yEy

I

• Neutral axis passes through the centroid:

EIz : flexural rigidity2

zA

I y dA : second moment of cross-section w.r.t. z.

i i

i

y A y A• for an arbitrarily defined y-coordinate:

Static Equivalency

10

0.y

• Normal stress on cross-sections:

: bending section modulus

• Maximum normal stress on cross-sections:

.z zM y I

max 2z z zW I y I h

max maxz z z zM y I M W

Pure Bending Normal Stress Formula

11

• The neutral axis passes through the centroid of the cross-sectional

area when the material follows Hooke’s law and there is no axial

force acting on the cross section.

• Our discussion is limited to beams for which the y axis is an axis of

symmetry. Consequently, the origin of coordinates is the centroid.

• Because the y-axis is an axis of symmetry, it follows that the y-axis

is a principal axis. So is the z-axis.

• Remarks:

• A strain gauge is placed under cross-section C of a simply supported

beam shown. Under the concentrated load P, the strain gauge reads ε

= 6× 10-4. Find the magnitude of P for E = 200 GPa.

Sample Problem

12

P

DCA B

1 m

0.4 m0.5 m 20 mm

40 mm

• Solution:

3 4200 10 6 10 120 MPa

640 N m

0.5 0.5 0.4 0.2 640 N m

3.2 kN

C

C C z

C A

E

M W

M R P P

P

13

• Solution:

• Find the support position (a) at the condition of minimum

“maximum normal stress” for the overhanging I-beam shown below,

under uniformly distributed load q.

2A BF F ql

2

2

10,

2

1 1,

2 2

M qx x a

M qx ql x a x a l a

al

a

qFA FB

x

Sample Problem

1. reaction force at the

supports. Due to symmetry:

2. Equation of bending

moments:

14

22

2 22 2 2 2

1 1

2 2 2 2 2

1 10

2 8 2 4 2

0.207

qa l lq ql a

qa ql qlaa la l a l l l

a l

⊕

2

8 2

ql qla

2

2

qa

2

2

qa

• Equating the absolute

value of the negative and

positive moment

extremities results in

minimum bending

moments and hence

minimum “maximum

normal stress.”

3. Diagram of bending moments:

15

• Find the maximum tensile and compressive stress in the T-beam

shown below.

2.2 m 1 m

q = 10 kN/m20 mm

20 mm

80 mm

120 mm

y1

z

z1

y

1. Centroid (neutral surface, neutral axis):

80 20 10 20 120 60+2052 mm

80 20 20 120

i i

i

A yy

A

By the Parallel Axis Theorem:

4644

23

23

z

m107.64mm10764

52801202012

1202010522080

12

2080I

)()(

2

' 'z z zzI I Ad

2. Moment of inertia:

Sample Problem

• Solution:

16

AD B C

3.8 kNm

5 kNm

-

+

xD = 0.87 m

4. Maximum normal stress (At cross-section B)3 -3

6

max -6

3 36

max 6

( 5 10 ) ( 52 10 )34 10 Pa 34 MPa

7.64 10

( 5 10 ) [(140 52) 10 ]57.6 10 Pa 57.6 MPa

7.64 10

0 23.27KNAy BM F

FBFA

x

3. Reaction forces and diagram of

bending moments

10 3.2 8.73KNA BF F

2

2

0,2.2 m2

2.2 2.2 m,3.2 m2

A

A B

qxM x F x x

qxM x F x F x x

17

AD B C

3.8 kNm

5 kNm

-

+

xD = 0.87 m

3 36

max 6

3 3

max 6

(3.8 10 ) [(140 52) 10 ]43.8 10 Pa 43.8 MPa

7.64 10

(3.8 10 ) [ 52 10 ]

7.64 10

• Maximum tensile stress: lower edge of cross-section D (43.8 MPa).

18

• Maximum compressive stress: lower edge of cross-section B (-57.6

MPa).

5. Maximum normal stress (At cross-section D)

6. Maximum normal stress

• For brittle materials

• For ductile materials

Normal Stress Strength Condition

max

maxz

M

W

max max

max max

, z z

M M

W W

19

• The maximum positive and negative bending moments in a beam

may occur at the following places: (1) a cross section where a

concentrated load is applied and the shear force changes sign, (2) a

cross section where the shear force equals zero, (3) a point of support

where a vertical reaction is present, and (4) a cross section where a

couple is applied.

• Three types problem that are

typically addressed by strength

analysis: Strength check

Cross-section design

Allowable load

• The maximum tensile stress and the maximum compressive stress

sometimes don’t occur on the same cross-section.

Remarks on Strength Condition

• Usually, the allowable bending stress is slightly higher than the

allowable uniaxial tensile/compressive stress. This is because the

bending stress only takes extremities at the upper/lower edges of

bending beams while the maximum axial stress is uniformly

distributed on bar cross-sections.

20

l

PP2

bz

h

(b)

P1

hz

b

(a)

• The dimension and material of the two cantilever beams shown are

identical. Find the allowable load ratio of these two beam based on

the normal stress strength condition: P1/P2 = ?

Sample Problem

21

• Solution：

max 1 1 1max 1 3 2

1

max 2 2max 2 2

2

1max 1 max 2

2

12 2 6

6

z

z

M Pl Pl

bh h bhW

M P l

hbW

P h

P b

22

• Two identical rectangular beams are placed together and subjected to

a concentrated load as shown. Find the allowable load [P] if the

allowable normal stress is given as []. What is [P] if the two beams

are pinned together?

b

2h

2h

l

P P

Sample Problem

23

• Solution

2 2

1

2

6 24

b h bhW

1. when the beams are not pinned together, each beam has its

own neutral surface and carries half of the bending moments.

24

max maxmax 2

1 1

2 12

2

M M Pl

W W bh

2[ ][ ]

12

bhP

l

2. After the beams are pinned, there exists only one neutral surface

maxmax 2

2

6

[ ][ ]

6

M Pl

W bh

bhP

l

• It can be seen that the load carrying ability are doubled after pinning.

25

• In ancient China, the typical aspect ratio of the cross-section of

rectangular beams is given as h:b = 3:2. If beams were made from

circular trees, employing the strength theory prove that the above

ratio is close to the optimal aspect ratio.

• Proof

h

b

222 dhb 2 2 2( )

6 6z

bh b d bW

2 2

0 26 2 3

zW d b d hb

b b

Sample Problem

Optimal aspect ratio means that Wz

achieves the maximum value.

26

• Given P = 20 kN, [] = 140 MPa. Compare the material

consumption for the following three types of cross-sections: (1)

rectangle with h/b = 2; (2) circle; (3) I-shaped.

Diagram of bending moment.

P = 20 kN

1 ml

maxmax

33 3max

6

20 10m 143 cm

140 10

z

z

M

W

MW

(1) For rectangular cross-section2

2

1 72 cm6

z

bhW A

Sample Problem

• Solution:

-20 kN.m

x

M

27

(2) For circular cross-section

32

211.3 cm 100 cm32

z

dW d A

(3) For I-shaped cross-section

Check the table for I-beam:

• I-beam consumes the least material while circular beam costs the

most.

• The maximum stress in the I-beam exceeds the maximum allowable

stress less than 5%. This is allowable in engineering practice.

3 2

3141cm 26.1cmzW A

28

• For the casting iron T-beam shown, the allowable tensile stress [+]

= 30 MPa, allowable compressive stress [ -] = 60 MPa, moment of

inertia Iz = 7.63× 10-6 m4. Analyze the strength condition.

• Solution:

M

A

4 kN.m

C B D

2.5 kN.m

Sample Problem

Diagram of bending moment. 29

2.5 kN 10.5 kN

z

1 m 1 m 1 m

4 kN9 kN

A B

C D

88 mm

52 mm

max a

max

2.5 8828.8MP [ ]

2.5 52[ ]

z

z

I

I

max a

max a

4 5227.3MP [ ]

4 8846.1MP [ ]

z

z

I

I

• For cross-section C:

• For cross-section B:

• The strength condition of the beam is satisfied.30

• For the T-beam shown below, the allowable tensile and compressive

stress are given as [σ+] and [σ -] respectively. Find the optimal ratio

for y1/y2. (C denotes the centroid of the beam cross-section.)

C

y1

y2

z

A

P

Sample Problem

31

• Solution:

max 1max

max 2max

1max

2max

[ ]

[ ]

[ ]

[ ]

z

z

M y

I

M y

I

y

y

• The maximum bending moment occurs at the fixed end A. Make the

upper and lower edge of cross-section A reach [σ+] and [σ -]

simultaneously:

32

Stress Concentrations

Stress concentrations may occur:

• in the vicinity of points where the loads are appliedmax

max

z

z

MK

W

• in the vicinity of abrupt changes in cross section

Stress-concentration factors for flat

bars with fillets under pure bending

Stress-concentration factors for flat

bars with grooves under pure bending33

Bending of a Composite Beam

Bimetallic beam Reinforced concrete Beam Sandwich beam

; y y

E E

• At the contact surface the

stresses in the two materials are

different.

• The y-coordinate is measured

from the proposed neutral axis.34

Wood-steel beam

Bending of a Composite Beam

1 2

1 2

1 2

1 2

0

0

N

A A

A A

i i i

i

F

dA dA

E EydA ydA

E A y

• Bending stress & Moment-curvature relationship2 1 1

i

i zZ i z i

A Ai i i z i

i

ii i i i z

i z i

i

E MM y dA y dA E I

E I

EyE E M y

E I

35

.i i i i i

i i

y E A y E A • For an arbitrarily defined y-coordinate:

• This equation determines the exact position of neutral axis.

Approximate Theory for a Sandwich Beam

3 3 3

1 2

33 3

1

3

2

1 1 1

2 2 2

,12 12

12 2 61 1

12

6

z c z c

cz

z c c c

z

z c

b bI h h I h

b h hI t t

I bh h h

E I E t

E I E h

• Provided that: 1 2E E

1 1

2 2

72 318

0.8 15

z

z

E I

E I

• If E1 = 72 Gpa (Al), E2 = 800 Mpa (Plastic), 2t/hc = 1/15:

1 1 2 2 1 1

1 2

1 1 2 2 1 1 1

1

, 0

z z

z z z

i i zi z z

z z z z

M M

E I E I E I

E E M yM y M y

E I E I E I I

• Provided that: 1 1 2 2z zE I E I

• A conservative theory.

36

Sample problem

• Determine the maximum normal stress in

the faces (Al, E1 = 72 Gpa) and the core

(E2 = 800 Mpa ) using: (a) the general

theory for composite beams, and (b) the

approximate theory for sandwich beams. M

= 3.0 kN-m.

• (a) the general theory

• (b) the approximate theory

• Solution:

3 3 12 4 12 4

1 2

3 2

1 1 2 2

1 2

1 2max max1 1 2 2 1 1 2 2

12.017 10 m , 56.25 10 m12

910.2 10 kN m

2 219.0 MPa, 0.198 MPa

z c z

z z

z z c

z z z z

bI h h I

E I E I

E M h E M h

E I E I E I E I

1 2max max1

220.0 MPa, 0

z

z

M h

I

37

Sample problem

• Calculate the largest tensile and compressive

stresses in the wood (E1 = 1500 ksi) and the

maximum and minimum tensile stresses in the

steel (material E2 = 30,000 ksi) M = 60 kip-in.

1 1 1 2 2 2 1 2

1 2

1 2

0 1500 4 6 3 30000 4 0.5 0.25

6.25

5.031 in, 1.469 in

E A y E A y h h

h h

h h

• Solution:

1 21 11 1

1 1 2 2 1 1 2 2

2 2 2 22 2

1 1 2 2 1 1 2 2

2 2

1

0.51.31 ksi, 0.251 ksi

0.55.030 ksi 7.62 ksi

zz

A Cz z z z

z z

C Bz z z z

C C

C

E M hE M h

E I E I E I E I

E M h E M h

E I E I E I E I

E

E

1

20C

• Neutral axis:

• Stresses along line A, C and B

38

Bending of a Curved Beam

• Change of arc length: JK r r y y y

• Longitudinal strain:JK y y r

JK r y r

• Hooke’s law:y E y E r E

Er y r

• Length of neutral surface remain unchanged:

• Geometry

39

y

y

y

y y y

0 1

1 1,

NA A A

A

A

r E EF dA dA dA

r r

A dA

dA A r

r

22 2 22

2 ,

zA A A A

z

zz

ry E E E r rM y dA dA dA dA

r r r

ME EAr A A r

A r

M rM y

A r y A r r

y E r E

y r

• Neutral axis:


1

Ar rdA

A • Distance between C and centroid:

• Static equilibrium and bending stress

40

• The change in curvature of the neutral surface :


1 1

, 1 1z zM

A r EA

M

rE

• Radius of neutral surface for various cross-sectional shapes..

41

• Determine the largest tensile and compressive stresses for a

curved rectangular bean shown below, knowing that

2

2

2ln

2

5.9686 in.2

ln2

0.0314 in.2

ln2

r h

A r h

A

dA dr r hb b

r r r h

A h

dA r h

r r h

hr r

r h

r h

• It is necessary to calculate ρ with enough significant figures in order

to obtain the usual degree of accuracy.

Sample Problem

2.5 in., 1.5 in., 6 in., 8 kip in., 1500 ksi.b h r M E

• Solution:

42

• Stresses approximated by the

theory for a straight bar:

max

min

6.75 7.86 ksi

5.25 9.30 ksi

zzM r

A r r

r

M y

y

r

A r

max,min

28.53 ksi

z

z

M h

I

• Largest tensile and compressive stresses

1 1

0.00758866zM

EA r

• The change in curvature of the neutral surface :

43

Shearing Stresses due to Transverse Loads

• If friction among beams is small, they will bend independently.

• The bottom surface of the upper beams will slide with respect to

the top surface of the lower beams.

• Horizontal shearing stresses must develop along the glued

surfaces in order to prevent the sliding.

• Because of the presence of these shearing stresses, the single solid

beam is much stiffer and stronger than the separate beams. 44

• Two assumptions:

Shearing Stresses in a Rectangular Beam

- Shearing stresses acting on the cross section are parallel to shear

force.

- Shearing stresses are uniformly distributed across the width of the

beam, although they may vary over the height.45

SF

dxn n1

z

a a1

F1 F2

dF

'

y

y

1

12 1

ave ave

ave

d

shear fl w

dd

d

o

z

Az

z z

A z

S z

z

Sz

dMdF F F y A

I

dFy y

bdx I b I b

F SdFf b

dx

FM

x

S

I

y A


• Horizontal shear forces/stresses

46

a 1a y

SF

S F

1 2

( ),

z

z z z

z

M y M dM y

I I

τhz

b

y

＋

－

22 2 2

2

2

22

S 2

Save 3 2

Save ave ave avemin max

'd 'd2 4

2 4 31

2 2

123

2 0; 02

b h h

zb y y

b hS y A b y A y

b hF y

F yy

bh A hb

Fh

A


• Vertical shearing stresses:

• If the width of the beam is comparable or large

relative to its depth, the shearing stresses at C1 and

C2 are significantly higher than their midpoint.

• Theory of elasticity shows that, for h 4b, the

maximum shearing stress does not exceed by more

than 0.8% than the average value.

47

• If the shear force is constant along beam axis, warping is the same at

every cross section.

Effect of Shearing Stress/Strain

48

2

S

2

min

Smax

31

2 2

2 0

30

2

F yG

GA h

h

F

GA

• In portions of the beam located under a distributed or concentrated

load, normal stresses will be exerted on the horizontal faces of a

cubic element of material, in addition to the stresses.

• Uniaxial stress state is violated due to the existence of

shearing stress..

• Plane hypothesis is violated due to the existence of shear

strain.

• The error involved, however, is small for the values of

the span-depth ratio encountered in practice.

• Warping does not substantially affect the longitudinal

strains even when the shear force varies continuously along

the length.

• Thus, under most conditions it is justifiable to use the

flexure formula for nonuniform bending, even though the

formula was derived for pure bending.

Effect of Shearing Stress/Strain

49

Shearing Stresses in a Wide-flange Beam

• The shearing stresses in the web of a wide-flange beam act only in

the vertical direction and are larger than the stresses in the flanges.

• The shearing stresses in the flanges of the beam act in both vertical

and horizontal directions.

• The shear formula cannot be used to determine the vertical shearing

stresses in the flanges.

• However, the shear formula does give good results for the shearing

stresses acting horizontally in the flanges.

50

51

y

z

b

h

ttf

h1y

Flange

Web

1

2 2221 1

*2 2 2 2S S

1 1

331 3 3 3

1 1

2 2Smin 1 1

2 2 2Smax 1 1

2 4 4 2 4

48

1

12 12 12

28

08

zA

z

z z

z

z

z

h hb h tS ydA y

F S Fb h h t h y

I t I t

b t hbhI bh bh th

Fy h bh bh

I t

Fy bh bh th

I t

• Shear force in the web

• For beams of typical proportions, shear force in the web is greater than 90% of the

total shear force; the remainder is carried by shear in the flanges.

1

1 1

32

2 2 2S 1S Web 1 1 1

2

2 2 2S1 11 1 max min

14

8 3 4

3 3 2 23 8 3

h

A hz

z

F hF dA t dy t bh bh th h t

I t

Fth thbh bh th

I t

min

min

max


52

y

z

b

h

t

h1

y

Flange

Web

1

22

*2 2S S

2 2Smax 1 1

min

2 4

48

28

2 0

zA

z

z z

z

b hS ydA y

F S Fbh by

I I

Fy h bh bh

b

I

b

h

b

y

• Discontinuities exist along the web/flange boundaries.

• The ratio between the minimum shearing stress in the web and the maximum stress

in the flange is b/t.

• In practice, one usually assumes that the entire shear load is uniformly carried by

the web ( = FS/Aweb).

• We should note, however, that while the vertical shearing stress in the flanges

can be neglected, its horizontal component has a significant value that will be

determined as follows.


tf

• Consider a segment of a wide-flange beam

subjected to the vertical shear FS.

z z S z

z z

dM S F dx SdF

I I

ave ave

S zzx xz

f z f

F SdF

t dx I t

• The corresponding average shearing stress

• NOTE: 0xy

0xz

in the flanges

in the web

• Previously found a similar expression for

the shearing stress in the web

ave

S zxy

z

F S

I t

dF

dx

dxSF

ft

• The longitudinal shear force on the vertical cut


53

1

1 1

d 2

2

24

4

z fs

S z S

z f z

S

z

f S

f

z

S y A st h

F S shF

I t I

bhFs b

I

bht Ff t

I

2

2 2 1

2

2

22

S f f SS z

z z z

f S

z

F bt h bht FF S

I t I t I t

bht Ff t f

I

max

max

2 2 4

2 4

S fS z

z z

fS

z

F bt h t h hF S

I t I t

bthtF hf

I t

54

t

SF

max 2

2 1

21 ,

4

f

f

tht

bt t


• For a wide-flange beam, the shear flow

increases symmetrically from zero at A

and A’, reaches a maximum at C and the

decreases to zero at E and E’.

• The continuity of the variation in f and

the merging of f from section branches

suggests an analogy to fluid flow.

Shear Flow in a Wide-flange Beam

SF

2 12f f

1f1f

1f 1f

2 12f f

• The variation of shear flow across the

section depends only on the variation of

the first moment.

aveS z

z

F Sf t

I

• The sense of f in the horizontal portions

of the section may be deduced from the

sense in the vertical portions or the

sense of the shear FS.

55

• For a box beam, f grows

smoothly from zero at A to a

maximum at C and C’ and then

decreases back to zero at E.

Shear Flow in a Box Beam

SF

ff

56

57

Shear Flow in a Thin-walled Beam

• The shearing stress formulae can be used to determine shearing

stresses in thin-walled beams, as long as the loads are applied in a

plane of symmetry of the member.

• In each case, the cut must be perpendicular to the surface of the

member, and the shearing stress formulae will yield the component

of the shearing stress in the direction of the tangent to that surface.

• The other component may be assumed equal to zero, in view of the

proximity of the two free surfaces.

d

d

d

dzs

sz

s

z

sz zS

z

dMtdx dN y A

I

dN

tdx I t

FM

I

A S

x t

y

Shear Flow in a Thin-walled Beam

• Horizontal shearing stress

• Shear flow

S z

z

F SdFf t

dx I

58

Knowing that the vertical shear

is 50 kips in a rolled-steel beam,

determine the horizontal

shearing stress in the top flange

at the point a.

SOLUTION:

• For the shaded area

34.31in 0.770in 4.815in 15.98inzS

• The shearing stress at a

3

4

50kips 15.98in2.63ksi

394in 0.770in

S z

z

F S

I t

Sample Problem – Shearing Stresses in Flanges

4394inzI

59

• A beam is made of three planks, nailed together.

Knowing that the spacing between nails is 25 mm and

that the vertical shear in the beam is FS = 500 N,

determine the shear force in each nail.

Sample Problem – Shear Force in a Web

6 3

6 4

6 3

-6 4

0.020m 0.100m 0.060m 120 10 m

16.20 10 m

(500N)(120 10 m ) N3704m16.20 10 m

z

z

S z

z

S Ay

I

F Sf

I

SOLUTION:

• Determine the horizontal force per unit length or shear

flow f on the lower surface of the upper plank.

• Calculate the corresponding shear force in

each nail for a nail spacing of 25 mm.

0.025 0.025 3704 92.6 NF f

60

A square box beam is constructed

from four planks as shown.

Knowing that the spacing between

nails is 1.5 in. and the beam is

subjected to a vertical shear of

magnitude FS = 600 lb, determine

the shearing force in each nail.

SOLUTION:

• Determine the shear force per

unit length along each edge of

the upper plank.

• Based on the spacing between

nails, determine the shear

force in each nail.

Sample Problem – Shear Force in Flanges

61

• For the upper plank

30.75in. 3in. 1.875in. 4.22inzS A y

• For the overall beam cross-section

4 4 41 1

12 124.5in 3in 27.42inzI

SOLUTION:

• Determine the shear force per

unit length along each edge of

the upper plank.

3

4

600lb 4.22in lb2 92.3

27.42in in

lb46.15

in

S z

z

F Sf

I

f

• Based on the spacing between

nails, determine the shear

force in each nail.

lb

46.15 1.5in 69.225lbin

F f

Sample Problem – Shear Force in Flanges

62

Shearing Stresses in a Circular Beam

• The shearing stresses can no longer be

assumed parallel to the y-axis.

• On the boundary of the cross section, the

shearing stress must act tangent to the

boundary.

• Only the neutral axis is an exception.

- Shearing stresses are concurrent at the intersection of boundary

tangent and y-axis.

- The projection of shearing stresses on y-axis are uniformly

distributed across the width of the beam.

z

τmax

y

dy

y’

A

• However, at a horizontal line we

may further assume:

63

z

τmax

y

dy

y’

A

2 2

2 21

2 4

4

2 22 2 2 2 2

23/2 3/2

2 2 2 2

' '

2 4 ' 4

2 24 4

3 3

y d z d y

zA y y z d y

y d y d

y y y y

y d

y y

S y dA y dy dz

y d y dy d y dy

d y d y

Shearing Stresses in a Circular Beam

3/22 2

2 2* SSS

44 2 2

3

S SS

min max 24

24 4

3

3π 64π 64 2 4

2 4

43 8 32 0, 0π 4 3π 64

z

z

F d y F d yF Sy

I b dd d y

dF F

Fy d y

d Ad d

64

Shearing Stresses in a Hollow Circular Beam

* 2 3

3*SS S

max 4

0 π 2 4 3 2 3

2 3 40

3π 4 2

z

z

z

S y Ay r r r

F rF S Fy

I b Ar r

• Solid circular cross-sections

• Hollow circular cross-sections

* 3 3

2 2 1 1 2 1

3 3

S 2 1

max 4 4

2 1 2 1

2 2

2 2 1 1S

2 2

2 1

0 2 3

2 30

π 4 2

4

3

zS y A y A y r r

F r ry

r r r r

r r r rF

A r r

65

• For a thin-walled circular beam:

2 2

2 2 1 1S Smax 2 2

2 1

4 2

3

r r r rF F

A Ar r

Shearing Stresses in an Equilateral Triangular Beam

1

23 3

2

1 1

2*S SS

33

S Smax

1 1

12 2 3 36

1 2

2 3 3

3 12

36

3 3

22

z

zA

z

z

bh bhI bh h

by h yyS ydA A y b y h y

h h

yt y b

h

F by h y h F y h yF Sy

I t bhbh by h

F F

bh Ay h

• Although the theory for maximum shearing stresses in beams is

approximate, it gives results differing by only a few percent from

those obtained using the exact theory of elasticity.

66

y

2 3h

2 3y

y

*

Smax

max

[ ]z

z

F S

I b

Shearing Stress Strength Condition

67

• Solution

• Diagram of shearing

forces & bending

moments

SF40 kN

40 kN

40 kN mM

• A circular beam is subjected to

a uniformly distributed load q

= 20 kN/m. The allowable

normal and shearing stresses

are [σ] = 160 Mpa, [τ] = 100

MPa. Find the minimum

required beam diameter.

A B

4 m

20 kN/m

Sample Problem

68

• For normal stress:

36max

max 3

40 10[ ] 160 10

π

32137 mm

z

M

dW

d

36max

max 2

4 4 40 10[ ] 100 10

3 3

4

26.1mm

SF

dA

d

• For shearing stress:

min 137mmd

69

• Normal stress plays the most important role in satisfying the strength

condition of beams under bending.

max

z max

[ ]zM

W

• Minimize the maximum bending moments by proper arrangements

of the form and position of loading and constraints.

Rational Design of Beams

• Proper design of cross-sections to maximize bending section

modulus.

70

+

Fl/4

F

l

F/2

l/4

F/2

l/4

+

Fl/8

q q

0.2l 0.2l

+

ql2/8

+

ql2/40

Rational Design of Loads & Constraints

71

3

32

0.125

z

dW

Ad

22 2 4

2

0.5

z

A hW

h

Ah

0.35zW Ah

• Compare a square with a circle:

2

0.1676

z

bhW Ah

2

0.14776 12

z

hhW Ad Ad

Rational Design of Cross-sections

• Among beam section choices which have an acceptable section

modulus, the one with the smallest weight per unit length or cross

sectional area will be the least expensive and the best choice. 72

• For materials with [+] = [ -], symmetric cross-sections may be

used such that the maximum tensile and compressive stress are

equal in magnitude at the upper/lower edges.

• For materials with [+] < [ -], i.e. casting irons, cross-sectional

neutral axis should deviate toward the tensile side.

y1z

y2

C

P

Symmetry vs. Asymmetry

1max

2max

[ ]

[ ]

y

y

73

• The maximum normal stress stays the same for every cross-section.

max max( ) ( ) [ ]M x W x

Nonprismatic and Constant-strength Beams

74

Unsymmetric Loading of Thin-Walled Members

• Beams loaded in a vertical

plane of symmetry result in

deforms in the symmetry

plane without twisting.

• Beams without a vertical

plane of symmetry bend

and twist under transverse

loading.

*

Save, zz

z z

F SM y

I I t

*

Save, zz

z z

F SM y

I I b

75

• When the force P is applied at a

distance e to the left of the web

centerline (shear center), the member

bends in a vertical plane without

twisting.

• If the shear load is applied such that the

beam does not twist, then the shearing

stress distribution satisfies

• F and F’ indicate a couple Fh and the

need for the application of a torque as

well as the shear load.

VehF

Unsymmetric Loading of Thin-Walled Members

*

Save S, ,

D B E

z

z B A D

F SF f ds F f ds f ds F

I t

SF

SF

SF

dF fds

76

• Determine the location for the shear center of

the channel section with b = 4 in., h = 6 in.,

and t = 0.15 in.

• Determine the shearing stress distribution for

FS = 2.5 kips applied at the shear center.

• Solution

Sample Problem

2

3 2

web flange

* 2 2

S S S S

0 0

3

0

2

S

1 12 2 6

12 2 12

3

2 4 6

3 4in.1.6in.

6in.62 2

3 3 4in.

1

12z

b b b

z

z z z

hI I I th bt th h b

F S F F thb F bhF fds ds st ds

I I I h h b

F h b be

hh b

b

bt

F

77

• The maximum shearing stress in the flanges

• The maximum shearing stress in the web

• Determine the shearing stress distribution for

FS = 2.5 kips applied at the shear center.

S 2.5kipsF

*

S S S

S Smax 21

12

2 2

6

62 6

6 2.5kips 4in2.22ksi

0.15in 6in 6 4in 6in

z

z z z

B

F S F F hhst s

I t I t I

F hb F b

th h bth h b

* 1 1 1 12 2 4 8max

* 1S S8S

max 2112

4

4 3 4

6 2 6

3 2.5kips 4 4in 6in3.06ksi

2 0.15in 6in 6 6in 6in

z

z

z

S bt h ht h ht h b

F ht h b F h bF S

I t th h b t th h b

78

• Pure Bending vs. Nonuniform Bending（纯弯曲与横力弯曲）

• Assumptions for Pure Bending（纯弯曲基本假设）

• Neutral Surface & Neutral Axes（中性层与中性轴）

• Kinematics（几何关系）

• Hooke’s Law（物理关系）

• Static Equivalency（静力等效关系）

• Pure Bending Normal Stress Formula（纯弯曲正应力公式）

• Normal Stress Strength Condition（正应力强度条件）

• Stress Concentrations（应力集中）

• Bending of a Composite Beam（复合梁弯曲）

• Bending of a Curved Beam（曲梁弯曲）

Contents

79

• Shearing Stresses in a Rectangular Beam（矩形梁切应力）

• Effect of Shearing Stress/Strain（切应力和切应变效应）

• Shearing Stresses in a Wide-flange Beam（宽翼缘梁切应力）

• Shear Flow in a Thin-walled Beam（薄壁梁剪力流）

• Shearing Stresses in a Circular Beam（圆截面梁切应力）

• Shearing Stresses in an Equilateral Triangular Beam（等边三角梁切应力）

• Shearing Stress Strength Condition（切应力强度条件）

• Rational Design of Beams（梁的合理设计）

• Nonprismatic and Constant-strength Beams（非等直梁和等强度梁）

• Unsymmetric Loading of Thin-Walled Members and Shear Center

（薄壁梁的非对称弯曲与剪力中心）

Contents

80

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