UNSYMMETRICAL BENDING

Moment applied along principal axis

dAyMM

dAzMM

dAFF

AZZR

AyyR

AxR

0 ;

0 ;

0 ;

If y and z are the principal axes. ∫ yz dA = 0(The integral is called the product of inertia)

UNSYMMETRICAL BENDING (cont.)

• Moment arbitrarily applied

• Alternatively, identify the orientation of the principal axes (of which one is the neutral axis) • Orientation of neutral axis:

y

y

z

z

I

zM

I

yM= +

= +

tantany

z

I

I

EXAMPLE

The rectangular cross section shown in Fig. 6–33a is subjected to a bending moment of 12 kN.m. Determine the normal stress developed at each corner of the section, and specify the orientation of the neutral axis.

EXAMPLE (cont.)

• The moment is resolved into its y and z components, where

• The moments of inertia about the y and z axes are

Solutions

mkN 20.7125

3

mkN 60.9125

4

z

y

M

M

433

433

m 10067.14.02.012

1

m 102667.02.04.012

1

z

y

I

I

EXAMPLE (cont.)

• For bending stress,

• The resultant normal-stress distribution has been sketched using these values, Fig. 6–33b.

Solutions

(Ans) MPa 95.4102667.0

1.0106.9

10067.1

2.0102.7

(Ans) MPa 25.2102667.0

1.0106.9

10067.1

2.0102.7

(Ans) MPa 95.4102667.0

1.0106.9

10067.1

2.0102.7

(Ans) MPa 25.2102667.0

1.0106.9

10067.1

2.0102.7

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

E

D

C

B

y

z

z

z

I

zM

I

yM

EXAMPLE (cont.)

Solutions

m 0625.02.0

95.425.2

z

zz

(Ans) 4.791.53tan102667.0

10067.1tan

tantan

3

3

y

z

I

I

• The location z of the neutral axis (NA), Fig. 6–33b, can be established by proportion.

• We can establish the orientation of the NA using Eq. 6–19, which is used to specify the angle that the axis makes with the z or maximum principal axis.

GENERAL CASE

GENERAL CASE

GENERAL CASE

z

SHEAR FORMULA

'' where

'

AyydAQ

It

VQ

A

SHEAR IN BEAMS

Rectangular cross section• Shear –stress distribution is parabolic

A

V

yh

bh

V

5.1

4

6

max

22

3

SHEAR IN BEAMS (cont)

Wide-flange beam• Shear-stress distribution is parabolic

but has a jump at the flange-to-web junctions.

EXAMPLE 1

A steel wide-flange beam has the dimensions shown in Fig. 7–11a. If it is subjected to a shear of V = 80kN, plot the shear-stress distribution acting over the beam’s cross-sectional area.

EXAMPLE 1 (cont)

• The moment of inertia of the cross-sectional area about the neutral axis is

• For point B, tB’ = 0.3m, and A’ is the dark shaded area shown in Fig. 7–11c

Solutions

4623

3

m 106.15511.002.03.002.03.012

12

2.0015.012

1

I

MPa 13.13.0106.155

1066.01080

m 1066.002.03.011.0''

6

33

'

''

33'

B

BB

B

It

VQ

AyQ

EXAMPLE 1 (cont)

• For point B, tB = 0.015m, and QB = QB’,

• For point C, tC = 0.015m, and A’ is the dark shaded area in Fig. 7–11d.

• Considering this area to be composed of two rectangles,

• Thus,

Solutions

MPa 6.22

015.0106.155

1066.010806

33

B

BB It

VQ

33 m 10735.01.0015.005.002.03.011.0'' AyQC

MPa 2.25

015.0106.155

10735.010806

33

max

C

cC It

VQ

SHEAR FLOW IN BUILT-UP BEAM

• Shear flow ≡ shear force per unit length along longitudinal axis of a beam.

I

VQq

q = shear flowV = internal resultant shearI = moment of inertia of the entire cross-sectional area

SHEAR FLOW IN BUILT-UP BEAM (cont)

EXAMPLE 2

Nails having a total shear strength of 40 N are used in a beam that can be constructed either as in Case I or as in Case II, Fig. 7–18. If the nails are spaced at 90 mm, determine the largest vertical shear that can be supported in each case so that the fasteners will not fail.

EXAMPLE 2 (cont)

• Since the cross section is the same in both cases, the moment of inertia about the neutral axis is

Case I • For this design a single row of nails holds the top or bottom flange onto

the web. • For one of these flanges,

Solutions

433 mm 205833401012

125030

12

1

I

(Ans) N 1.27205833

3375

90

40

mm 33755305.22'' 3

V

VI

VQq

AyQ

EXAMPLE 2 (cont)

Case II• Here a single row of nails holds one of the side boards onto the web.• Thus,

Solutions

(Ans) N 3.81205833

1125

90

40

mm 11255105.22'' 3

V

VI

VQq

AyQ

SHEAR FLOW IN THIN-WALLED BEAM

• Approximation: only the shear-flow component that acts parallel to the walls of the member will be counted.

SHEAR FLOW IN THIN-WALLED BEAM (cont)

• In horizontal flanges, flow varies linearly,

• In vertical web(s), flow varies parabolically,

x

b

I

Vtd

I

txbdV

I

VQq

22

2/2/

2

2

42

1

2y

ddb

I

Vt

I

VQq

EXAMPLE 3

The thin-walled box beam in Fig. 7–22a is subjected to a shear of 10 kN. Determine the variation of the shear flow throughout the cross section.

EXAMPLE 3 (cont)

• The moment of inertia is

• For point B, the area thus q’B = 0. • Also,

• For point C, since there are 2 points of attachment:

• The shear flow at D, because there are 2 points of attachment :

Solutions

0'A

mkNmmkN

mm

mmkN

I

VQq cc /7.48/0487.0

10797.1

1750010

2

1

2

146

3

mkNmmkN

mm

mmkN

I

VQq DD /8.82/0828.0

10797.1

2975010

2

1

2

146

3

462 10797.135105027010212

1mmmmmmmmmmmmI

3

3

2975010503535102

352''

17500105035''

mmmmmmmmmmmmmm

AyQ

mmmmmmmmAyQ

D

C

SHEAR CENTRE

• Shear center is the point through which a force can be applied which will cause a beam to bend and yet not twist.

• The location of the shear center is only a function of geometry of the cross section and does not depend upon the applied load.

P

dFe f

PedFM fA

SHEAR CENTRE (cont)

P

dFe f

EXAMPLE 4

Determine the location of the shear center for the thin-walled channel section having the dimensions shown in Fig. 7–25a.

EXAMPLE 4 (cont)

• The cross-sectional area can be divided into three component rectangles—a web and two flanges.

• q at the arbitrary position x is

• Hence, the force is

Solutions

b

hthhbtthI

6222

12

1 223

bhh

xbV

bhth

txbhV

I

VQq

6/6/2/

2/2

bhh

Vbdxxb

bhh

VdxqF

bb

f

6/26/

2

00

EXAMPLE 4 (cont)

• Summing moments about point A, Fig. 7–25c, we require

• As stated previously, e depends only on the geometry of the cross section.

Solutions

(Ans) 23/

6/22

2

bh

be

bhh

hVbhFVe f