AERSP 301 Shear of beams (Open Cross-section) Jose Palacios

AERSP 301Shear of beams

(Open Cross-section)

Jose Palacios

Shear of Open and Closed Section Beams

• Megson – Ch. 17• Open Section Beams

– Consider only shear loads applied through shear center (no twisting)– Torsion loads must be considered separately

• Assumptions– Axial constraints are negligible

– Shear stresses normal to beam surface are negligible• Near surface shear stress = 0• Walls are thin

– Direct and shear stresses on planes normal to the beam surface are const through the thickness

– Beam is of uniform section• Thickness may vary around c/s but not along the beam

– Thin-Walled• Neglect higher order terms of t (t2, t3, …)

• Closed Section Beams– Consider both shear and torsion loading

Force equilibrium: General stress, Strain, and Displacement Relationships

• S – the distance measured around the c/s from some convenient origin

• σz – Direct stress (due to bending moments or bending action of shear loads)

– Shear stresses due to shear loads or torsion loads (for closed section)

• σs – Hoop stress, usually zero (non-zero due to internal pressure in closed section beams)

zs = sz =

shear flow; shear force per unit length

q = t

(positive in the direction of s)

Force equilibrium (cont’d)

• From force equilibrium considerations in z-direction:

• Force equilibrium in s-direction gives

Stress Strain Relationships

• Direct stress: z and s strains z and s

• Shear stress: strains (= zs = zs)

• Express strains in terms of displacements of a point on the c/s wall

• vt and vn: tangential and normal displacements in xy plane

Not used

(1/r: curvature of wall in x-y plane)

Stress Strain Relationships

• To obtain the shear strain, consider the element below:

• Shear strain:

Center of Twist

• Equivalent to pure rotation about some pt. R

(center of twist [for loading such as pure torsion])

• For the point N

• Origin O of axes chosen arbitrarily, and axes undergo disp. u, v,

Center of Twist (cont’d)

• But

Equivalent to pure rotation about some pt. R (center of twist [for loading such as pure torsion])

Center of twist cont…

• Also from

• Comparing Coefficients with:

Position of Center of Twist

Shear of Open Section Beams

• The open section beam supports shear loads Sx and Sy such that there is no twisting of the c/s (i.e. no torsion loads)

• For this, shear loads must pass through a point in the c/s called the SHEAR CENTER– Not necessarily on a c/s member

• Use the equilibrium eqn.

• And obtaining z from basic bending theory

(no hoop stresses, s = 0)

Shear of Open Section Beams cont…

From:

Shear of Open Section Beams cont…

• Integrating with respect to s starting from an origin at an open edge (q = 0 at s = 0) gives:

• For a c/s having an axis of symmetry, Ixy = 0. Then eq. for qs simplifies to:

Shear sample problem

• Determine the shear flow distribution in the thin-walled z-section shown due to shear load Sy applied through its shear center (no torsion).

• Where is the shear center?

• And the centroid?

t

y

xh

h/21 2

43

Shear Flow Distribution (Sx = 0):

s

xyyyxx

yyys

xyyyxx

xyys ydst

III

ISxdst

III

ISq

0202


8

12

3

333 thI

thI

thI xyyyxx

dsyx

h

SydstIxdstI

III

Sq

sys

yy

s

xyxyyyxx

ys )48.632.10(

03002

Bottom Flange: 1-2, y = -h/2, x =-h/2 + S1

0 ≤ S1 ≤ h/2

dshShh

Sq

dsyxh

Sq

sy

sys

))2/(48.6)2/(32.10(

)48.632.10(

1

0 1312

03

Show this: EXAM TYPE PROBLEM


h

Sq)h(S

qS

hSSh

S

dshShh

Sq

y

y

sy

42.02/ 2@

0)0( 1@

)74.116.5(

))2/(48.6)2/(32.10(

21

11

12

13

0 1312

1

t

y

xh

h/21 2

43


222

23

2220323

42.342.342.0

)(48.6)(42.3(2

shshh

S

qdsshh

Sq

y

sy

In web 2-3:

y =-h/2 + S2

x = 0 for 0 ≤ S2 ≤ h

Symmetric distribution about Cx with max value at S2 = h/2 (y = 0) and positive shear flow along the web

Shear Flow S2 = 0h

Sq y42.0

2

Shear sample problemIn web 3-4:

y =h/2 x = S3 for 0 ≤ S3 ≤ h

330 3334 )(24.3)(32.10(3

qdshsh

Sq

sy

Shear Flow Distribution in z-section

Calculation of Shear Center

• If a shear load passes through the shear center, it will produce NO TWIST M = 0

• If c/s has an axis of symmetry, the shear center lies on this axis

• For cruciform or angle sections, the shear center is located at the intersection of the sides

Sample Problem• Calculate the shear

center of the thin-walled channel shown here:

Sample problem shear center

The shear center (point S) lies on the horizontal (Cx) axisat some distance ξs from the web. If a shear load Sy passes through the shear center it will produce no twist.

Let’s look at the shear flow distribution due to Sy:Since Ixy = 0 and Sx = 0

S

xx

ys dsty

I

Sq

0

)6

1(12

3

h

bthI xx Further:

Sy

s dsy

hb

h

Sq

03 61

12Then:


Along the bottom flange 1-2, y = -h/2

12

0 13

12

61

62/

61

12 1

s

hb

h

Sdsh

hb

h

Sq ySy

At point 2: S1 = b

b

hb

h

Sq y

61

6

22

Along the web 2-3 , y = -h/2 + S2


b

hb

h

Sssh

hb

h

S

qdssh

hb

h

Sq

yy

Sy

61

6

2261

12

2/61

12

2

22

23

20 223

23

2

At point 3: S1 = h

22

3

61

6qb

hb

h

Sq y

Along the top flange 3 - 4 , y = h/2


b

hb

h

Ssh

hb

h

S

qdsh

hb

h

Sq

yy

Sy

61

6

261

12

2/61

12

23

3

30 33

34

3

At point 4: S3 = b

04 q Good Check!


Shear Flow Distribution due to Sy

The moments due to this shear flow distribution should be equal to zero about the shear center

20 0 23112 22 dsqds

hq s

b h

S ξs

Sy

Solve for ξs to find the shear center location:

s

hyb h

syy dsb

hb

h

Sds

ssh

hb

h

Sdss

h

hb

h

S

0 22

0 0 2

22

23

112 61

6

2261

12

261

62

hb

h

bs

1

3 2

Documents

AERSP 301 Shear of beams (Open Cross-section) Jose Palacios