MECHANICS OF CHAPTER 6MATERIALS - Civil Department · PDF fileMECHANICS OF MATERIALS 6- 5...

MECHANICS OF MATERIALS

CHAPTER

6 Shearing Stresses in Beams and Thin-Walled Members


4 - 26- 2

Introduction

MyMdAFdAzMVdAF

dAzyMdAF

xzxzz

xyxyy

xyxzxxx

00

00

Distribution of normal and shearing stresses satisfies ( from equilibrium)

Transverse loading applied to beam results in normal and shearing stresses in transverse sections.

Longitudinal shearing stresses must exist in any member subjected to transverse loading.

When shearing stresses are exerted on vertical faces of an element, equal stresses exerted on horizontal faces

MECHANICS OF MATERIALSVertical and Horizontal Shear Stresses

4 - 3

MECHANICS OF MATERIALSShear Stress in Beams

Two beams glued together along horizontal surface

When loaded, horizontal shear stress must develop along glued surface in order to prevent sliding between the beams.


6- 5

Shear on Horizontal Face of Beam Element

Consider prismatic beam

Equilibrium of element CDC’D’

A

CD

ACDx

dAyI

MMH

dAHF

0

xVxdx

dMMM

dAyQ

CD

A

Let,

flowshear

IVQ

xHq

xI

VQH


6- 6

Shear on Horizontal Face of Beam Element

where

section cross full ofmoment second

above area ofmoment first

'

21

AA

A

dAyI

y

dAyQ

Same result found for lower area

HH

qIQV

IQV

xHq

dAyQQAA

)(zero) isNA wrt area ofmoment first (

021

flowshear

I

VQxHq

Shear flow,


6- 7

Example 6.1

Beam made of three planks, nailed together. Spacing between nails is 25 mm. Vertical shear in beam is V = 500 N. Find shear force in each nail.


6- 8

Example 6.1

46

2

3121

3121

36

m1020.16

]m060.0m100.0m020.0

m020.0m100.0[2

m100.0m020.0

m10120

m060.0m100.0m020.0

I

yAQ

SOLUTION:Find horizontal force per unit length or

shear flow q on lower surface of upper plank.

mN3704

m1016.20)m10120)(N500(

46-

36

IVQq

Calculate corresponding shear force in each nail for nail spacing of 25 mm.

mNqF 3704)(m025.0()m025.0(

N6.92F


6- 9

Determination of Shearing StressAverage shearing stress on horizontal face of

element is shearing force on horizontal face divided by area of horzontal face.

xtx

IVQ

Axq

AH

ave

If width of beam is comparable or large relative to depth, the shearing stresses at D’1 and D’2 are significantly higher than at D, i.e., the above averaging is not good.

Note averaging is across dimension t (width) which is assumed much less than the depth, so this averaging is allowed.

On upper and lower surfaces of beam, tyx= 0. It follows that txy= 0 on upper and lower edges of transverse sections.

qtIt

VQave ;


6- 10

Shearing Stresses txy in Common Types of Beams

For a narrow rectangular beam,

AV

cy

AV

IbVQ

xy

23

123

max

2

2

For I beams

web

ave

AV

ItVQ

max


6- 11

Example 6.2

Timber beam supports three concentrated loads.

MPa8.0MPa12 allall

Find minimum required depth d of beam.


6- 12

Example 6.2

kNm95.10kN5.14

max

max

MV


6- 13

Example 6.2

2

261

261

3121

m015.0

m09.0

d

d

dbcIS

dbI

Determine depth based on allowable normal stress.

mm246m246.0

m015.0Nm1095.10Pa 1012 2

36

max

dd

SM

all

Determine depth based on allowable shear stress.

mm322m322.0

m0.0914500

23Pa108.0

23

6

max

dd

AV

all

Required depth mm322d


6- 14

Longitudinal Shear Element of Arbitrary Shape

Have examined distribution of vertical components txy on transverse section. Now consider horizontal components txz .

So only the integration area is different, hence result same as before, i.e.,

Will use this for thin walled members also

IVQ

xHqx

IVQH

Consider element defined by curved surface CDD’C’.

AdAHF CDx 0


6- 15

Example 6.3

Square box beam constructed from four planks. Spacing between nails is 44 mm. Vertical shear force V = 2.5 kN. Find shearing force in each nail.


6- 16

Example 6.3SOLUTION:

Determine the shear force per unit length along each edge of the upper plank.

lengthunit per force edge mmN8.7

2

mmN6.15

mm10332mm64296N2500

4

3

qf

IVQq

Based on the spacing between nails, determine the shear force in each nail.

mm44mmN8.7

fF

N2.343F

For the upper plank,

3mm64296

mm47mm768mm1

yAQ

For the overall beam cross-section,

4

41214

121

mm10332

mm76mm112

I


6- 17

Shearing Stresses in Thin-Walled Members

Consider I-beam with vertical shear V.

Longitudinal shear force on element is

xI

VQH

ItVQ

xtH

xzzx

Corresponding shear stress is

NOTE: 0xy0xz

in the flangesin the web

Previously had similar expression for shearing stress web

ItVQ

xy

Shear stress assumed constant through thickness t, i.e., due to thinnness our averaging is now accurate/exact.


6- 18


The variation of shear flow across the section depends only on the variation of the first moment.

IVQtq

For a box beam, q grows smoothly from zero at A to a maximum at C and C’ and then decreases back to zero at E.

The sense of q in the horizontal portions of the section may be deduced from the sense in the vertical portions or the sense of the shear V.


6- 19


For wide-flange beam, shear flow qincreases symmetrically from zero at Aand A’, reaches a maximum at C and then decreases to zero at E and E’.

The continuity of the variation in q and the merging of q from section branches suggests an analogy to fluid flow.


6- 20

Example 6.4

Vertical shear is 200 kN in a W250x101 rolled-steel beam. Find horizontal shearing stress at a.

3mm258700

mm2.122mm6.19mm108

Q

Shear stress at a,

m0196.0m10164

m107.258N1020046

363

It

VQ

MPa1.16


Work out this example of a wide flange beam (Doubly symmetric)


6- 22

Unsymmetric Loading of Thin-Walled Members

Beam loaded in vertical plane of symmetry, deforms in symmetry plane without twisting.

ItVQ

IMy

avex

Beam without vertical plane of symmetry bends and twists under loading.

ItVQ

IMy

avex


Bending+Torsioneffect

Bending+Torsioneffect

Pure Bending


6- 24

Unsymmetric Loading of Thin-Walled Members

Point O is shear center of the beam section.

If shear load applied such that beam does not twist, then shear stress distribution satisfies

FdsqdsqFdsqVIt

VQ E

D

B

A

D

Bave

F and F’ form a couple Fh. Thus we have a torque as well as shear load. Static equivalence yields,

VehF

Thus if force P applied at distance e to left of web centerline, the member bends in vertical plane without twisting. Net torsional moment is Fh-Ve = 0, so shear stresses due to bending shear only, and not due to torsional shear.

If load not applied thru shear center then net torsional moment exists, so total shear stress due to bending shear & torsional shear (ref. open thin walled torsion)

MECHANICS OF MATERIALSFacts about Shear Center

When force applied at shear center, it causes pure bending & no torsion.

Its location depends on cross-sectional geometry only.

If cross-section has axis of symmetry, then shear center lies on the axis of symmetry (but it may not be at centroid itself).

If cross section has two axes of symmetry, then shear center is located at their intersection. This is the only case where shear center and centroid coincide.

MECHANICS OF MATERIALSWant to find shear flow and shear center of thin-walled open cross-sections.

For I and Z -sections s.c. at centroid.

For L and T -sections s.c. at intersection of the two straight limbs, i.e., where bending shear stresses cause zero torsional moment.

Thin-walled cross sections are very weak in torsion, therefore load must be applied through shear center to avoid excessive twisting


6- 27

Example 6.5

2121

21212

1212

23

233

hbtht

hbtbthtIII

fw

ffwflangeweb

10061501003

63 22

btht

bte

fw

f mm40e

Determine location of shear center of channel section with b = 100 mm, h = 150 mm, and t = 4 mm

VhFe

IhbVt

dshstIVds

IVQdsqdstF

f

b b

f

bb

fxz

4

22

0 000

s


6- 28

Example 6.6Determine shear stress distribution for

V = 10 kN

ItVQ

tq

Shearing stress in the web,

MPa3.18

6222

12

2/max

2

11

1

hsxyxy

wfw

wf

wxy thbtht

shtshbt

V

ItVQ

Shearing stresses in the flanges,

MPa3.13

15.01.0615.0004.01.0N100006

612

)(

22

hbtht

VbbI

Vh

sI

VhhstItV

ItVQ

fw

Bxz

fff

xz

q

ss1

fw

BxyBxz

tt

becauseonly

)()(

MECHANICS OF MATERIALSShear center of a thin walled semicircular cross-section

(a) Find shear stress ( xq ) at an angle , i.e., at section bb

Find the first moment of the cross-sectional area between point a and section bb

sincos 2

0

trrdtrydAQ

trV

ttrtrV

ItVQ

x

sin22/

sin3

2

s VVy

MECHANICS OF MATERIALS(b) Find the shear center (S)

Moment about geometric center of circle O, due to the shear force is Ve

Shear stress acting on element dA

Corresponding force is xq dA and moment due to this force is

)( dArdM xo

rVtrdtr

VrdMM oo4sin2

0

trddA

reVeM o

4

trV

x

sin2

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