MECHANICS OF MATERIALS CHAPTER 6 SHEARING STRESSES IN BEAMS AND THIN-WALLED MEMBERS

Kinh Tế - Quản Lý - Công Nghệ Thông Tin, it, phầm mềm, website, web, mobile app, trí tuệ nhân tạo, blockchain, AI, machine learning - Cơ khí - Vật liệu MECHANICS OF MATERIALS CHAPTER 6 Shearing Stresses in Beams and Thin- Walled Members MECHANICS OF MATERIALS 4 - 26- 2 Introduction     MyMdAF dAzMVdAF dAzyMdAF xzxzz xyxyy xyxzxxx              0 0 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 MATERIALS Vertical and Horizontal Shear Stresses 4 - 3 MECHANICS OF MATERIALS Shear 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. MECHANICS OF MATERIALS 6- 5 Shear on Horizontal Face of Beam Element Consider prismatic beam Equilibrium of element CDC’D’         A CD A CDx dAy I MM H dAHF    0 xVx dx dM MM dAyQ CD A   Let, flowshear     I VQ x H q x I VQ H MECHANICS OF MATERIALS 6- 6 Shear on Horizontal Face of Beam Element where sectioncrossfullofmomentsecond aboveareaofmomentfirst '''' 2 1      AA A dAyI y dAyQ Same result found for lower area HH q I QV I QV x H q dAyQQ AA            )( zero)isNAwrtareaofmomentfirst( 0 21  flowshear    I VQ x H q Shear flow, MECHANICS OF MATERIALS 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. MECHANICS OF MATERIALS 6- 8 Example 6.1             46 2 3 12 1 3 12 1 36 m1020.16 m060.0m100.0m020.0 m020.0m100.02 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. m N3704 m1016.20 )m10120)(N500( 46- 36      I VQ q Calculate corresponding shear force in each nail for nail spacing of 25 mm. mNqF 3704)(m025.0()m025.0(  N6.92F MECHANICS OF MATERIALS 6- 9 Determination of Shearing Stress Average shearing stress on horizontal face of element is shearing force on horizontal face divided by area of horzontal face. xt x I VQ A xq A H 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. qt It VQ ave     ; MECHANICS OF MATERIALS 6- 10 Shearing Stresses txy in Common Types of Beams For a narrow rectangular beam, A V c y A V Ib VQ xy 2 3 1 2 3 max 2 2             For I beams web ave A V It VQ   max   MECHANICS OF MATERIALS 6- 11 Example 6.2 Timber beam supports three concentrated loads. MPa8.0MPa12  allall   Find minimum required depth d of beam. MECHANICS OF MATERIALS 6- 12 Example 6.2 kNm95.10 kN5.14 max max   M V MECHANICS OF MATERIALS 6- 13 Example 6.2     2 2 6 1 2 6 1 3 12 1 m015.0 m09.0 d d db c I S dbI     Determine depth based on allowable normal stress.   mm246m246.0 m015.0 Nm1095.10 Pa1012 2 3 6 max     d d S M all  Determine depth based on allowable shear stress.   mm322m322.0 m0.09 14500 2 3 Pa108.0 2 3 6 max    d d A V all  Required depth mm322d MECHANICS OF MATERIALS 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 I VQ x H qx I VQ H     Consider element defined by curved surface CDD’C’.     A dAHF CDx  0 MECHANICS OF MATERIALS 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. MECHANICS OF MATERIALS 6- 16 Example 6.3 SOLUTION: Determine the shear force per unit length along each edge of the upper plank.    lengthunitperforceedge mm N 8.7 2 mm N 6.15 mm10332 mm64296N2500 4 3    q f I VQ q Based on the spacing between nails, determine the shear force in each nail.  mm44 mm N 8.7        fF N2.343F For the upper plank,     3 mm64296 mm47mm768mm1   yAQ For the overall beam cross-section,     4 4 12 14 12 1 mm10332 mm76mm112  I MECHANICS OF MATERIALS 6- 17 Shearing Stresses in Thin-Walled Members Consider I-beam with vertical shear V. Longitudinal shear force on element is x I VQ H  It VQ xt H xzzx       Correspondi...

MECHANICS OF MATERIALS

CHAPTER

Beams and Walled Members

Thin-MECHANICS OF MATERIALS

Introduction

 y  M M

dA F

dA z

M V

dA F

dA z

y M

dA F

x z

xz z

x y

xy y

xy xz

x x

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

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

MECHANICS OF MATERIALS

Vertical and Horizontal Shear Stresses

MECHANICS OF MATERIALS

Shear 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.

MECHANICS OF MATERIALS

Shear on Horizontal Face of Beam Element

Consider prismatic beam

A

C D

x

dA

y I

M M

H

dA H

x dx

dM M

M

dA y Q

C D

H q

x I

VQ H

MECHANICS OF MATERIALS

Shear on Horizontal Face of Beam Element

where

section cross

full of moment second

above area

of moment first

' 2

y

dA y Q

Same result found for lower area

q I

Q V I

Q V x

H q

dA y Q

Q

A A

zero) is

NA wrt area of

moment first

(

0

2 1



flow shear

H q

Shear flow,

MECHANICS OF MATERIALS

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

3 12

1

3 12

1

3 6

m 10 20 16

] m 060 0 m 100 0 m 020 0

m 020 0 m 100 0 [ 2

m 100 0 m 020 0

m 10 120

m 060 0 m 100 0 m 020 0

SOLUTION:

Find horizontal force per unit length or

shear flow q on lower surface of

upper plank

m

N 3704

m 10 16.20

) m 10 120 )(

N 500 (

4 6 -

3 6

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

m N q

F  ( 0 025 m )  ( 0 025 m )( 3704

N 6 92



F

MECHANICS OF MATERIALS

Determination of Shearing Stress

Average shearing stress on horizontal face of

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

x t

x I

VQ A

x q A

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

q

t It

VQ

MECHANICS OF MATERIALS

Shearing Stresses txy in Common Types of Beams

For a narrow rectangular beam,

A V

c

y A

V Ib

VQ

xy

2 3

1 2

3

max

2 2

MECHANICS OF MATERIALS

Example 6.2

kNm 95

10

kN 5 14 max

max





M V

2 6 1

3 12 1

m 015 0

m 09 0

d d

d b c

I S

d b I

246 0

m 015 0

Nm 10 95 10 Pa

10

3 6

322 0

m 0.09

14500 2

3 Pa 10 8 0

2 3

6 max

MECHANICS OF MATERIALS

Longitudinal Shear Element of Arbitrary Shape

Have examined distribution of vertical

components t xy on transverse section Now consider horizontal components

H q

x I

MECHANICS OF MATERIALS

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

length unit

per force edge

mm

N 8 7 2

mm

N 6 15 mm

10332

mm 64296 N

2500

4 3

I

VQ q

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

44 mmmm

N 8

mm 47 mm 76 8mm 1







 A y Q

For the overall beam cross-section,

4

4 12

1 4 12

1

mm 10332

mm 76 mm

MECHANICS OF MATERIALS

Shearing Stresses in Thin-Walled Members

Consider I-beam with vertical shear V.

Longitudinal shear force on element is

x I

Shear stress assumed constant through

thickness t, i.e., due to thinnness our

averaging is now accurate/exact

MECHANICS OF MATERIALS

Shearing Stresses in Thin-Walled Members

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

I

VQ t

q  

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.

MECHANICS OF MATERIALS

Shearing Stresses in Thin-Walled Members

For wide-flange beam, shear flow q increases symmetrically from zero at A and 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

MECHANICS OF MATERIALS

Example 6.4

Vertical shear is 200 kN in a

W250x101 rolled-steel beam Find

horizontal shearing stress at a.

3 mm 258700

mm 2 122 mm

6 19 mm 108

10 200

4 6

3 6 3





MECHANICS OF MATERIALS

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

MECHANICS OF MATERIALS

Unsymmetric Loading of Thin-Walled Members

Beam loaded in vertical plane of symmetry, deforms in

symmetry plane without twisting

It

VQ I

It

VQ I

My

ave



MECHANICS OF MATERIALS

Bending+Torsion

effect

Bending+Torsion effect

Pure Bending

MECHANICS OF MATERIALS

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

F ds

q ds

q F

ds q

V It

D

B A

D B



F and F’ form a couple Fh Thus we have a torque as

well as shear load Static equivalence yields,

Ve h

F 

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

MECHANICS OF MATERIALS

Facts 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 MATERIALS

Want 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

MECHANICS OF MATERIALS

Example 6.5

2 12

1

2 12

1 2 12

1 2

2 3

2 3

3

h bt h

t

h bt bt

h t I

I I

f w

f f

w flange

150

100 3

b t e

f w

f

mm 40

e 

I

hb Vt

ds

h st I

V ds I

VQ ds

q ds

t F

f

f

b b

f xz

2

1 1

w

w f

w

xy

t h bt h

t

s h t s h bt V It

6

1 2

) (

2 2

Vb b

I Vh

s I

Vh h

st It

V It

VQ

f w

B xz

f f f

B xy B

)

( 

MECHANICS OF MATERIALS

Shear 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

t r

t r

t r V It

VQ x

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

) ( dA r

rd t

r

V r

Ve

t r

V x





   2 sin