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MECHANICS OF MATERIALS
CHAPTER
6 Shearing Stresses in Beams and Thin-Walled Members
MECHANICS OF MATERIALS
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.
MECHANICS OF MATERIALS
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
MECHANICS OF MATERIALS
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,
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
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
MECHANICS OF MATERIALS
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 ;
MECHANICS OF MATERIALS
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
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
kNm95.10kN5.14
max
max
MV
MECHANICS OF MATERIALS
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
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
IVQ
xHqx
IVQH
Consider element defined by curved surface CDD’C’.
AdAHF 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 6.3SOLUTION:
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
MECHANICS OF MATERIALS
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.
MECHANICS OF MATERIALS
6- 18
Shearing Stresses in Thin-Walled Members
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.
MECHANICS OF MATERIALS
6- 19
Shearing Stresses in Thin-Walled Members
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.
MECHANICS OF MATERIALS
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
MECHANICS OF MATERIALS
Work out this example of a wide flange beam (Doubly symmetric)
MECHANICS OF MATERIALS
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
MECHANICS OF MATERIALS
Bending+Torsioneffect
Bending+Torsioneffect
Pure Bending
MECHANICS OF MATERIALS
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
MECHANICS OF MATERIALS
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
MECHANICS OF MATERIALS
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