This document provides information about shear stresses and shear force in structures. It includes: - Definitions of shear force and shear stress. Shear force is an unbalanced force parallel to a cross-section, and shear stress develops to resist the shear force. - Explanations of horizontal and vertical shear stresses that develop in beams due to bending moments. Shear stress is highest at the neutral axis and reduces towards the top and bottom of the beam cross-section. - Derivations of formulas for calculating shear stress across different beam cross-sections. Shear stress is directly proportional to the shear force and beam geometry. - Examples of calculating maximum and average shear stresses for various cross-sections

1. SHREE SA’D VIDYA MANDAL
INSTITUTE OF TECHNOLOGY
DEPARTMENT OF CIVIL
ENGINEERING

2. Subject:-Mechanics Of Solids
Topic:-Shear stresses on

3. Presented by:-
Name
 Arvindsai
 Dhaval Chavda
 Fahim Patel
 Navazhushen Patel
Enrollment no.
130454106002
130454106001
140453106005
140453106008

4. Topics To Be Covered
1. Shear Force
2. Shear Stresses In Beams
3. Horizontal Shear Stress
4. Derivation Of Formula
5. Shear Stress Distribution Diagram
6. Numericals

5. Shear force
Any force which tries to shear-off the
member, is termed as shear force.
Shear force is an unbalanced force,
parallel to the cross-section, mostly
vertical, but not always, either the right or
left of the section.

6. Shear Stresses
To resist the shear force, the element will
develop the resisting stresses, Which is
known as Shear Stresses().
= =
Shear force
Cross sectional
area
S
A

7. Example:-
 For the given figure if we want to
calculate the ..
 Then it will be
Let shear force be S
=S/(bxd)
d
b
S

8. Shear Stresses In Beams
 Shear stresses are usually maximum at the
neutral axis of a beam (always if the
thickness is constant or if thickness at
neutral axis is minimum for the cross
section, such as for I-beam or T-beam ), but
zero at the top and bottom of the cross
section as normal stresses are max/min.
NA
NA
NA

9.  When a beam is subjected to a loading,
both bending moments, M, and shear
forces, V, act on the cross section. Let us
consider a beam of rectangular cross
section. We can reasonably assume that the
shear stresses τ act parallel to the shear
force V.
v
n
V
z
m
O
b
h

10.  Shear stresses on one side of an element are
accompanied by shear stresses of equal
magnitude acting on perpendicular faces of
an element. Thus, there will be horizontal
shear stresses between horizontal layers of
the beam, as well as, Vertical shear
stresses on the vertical cross section.
m
n

11. Horizontal Shear Stress
 Horizontal shear stress occurs due to the
variation in bending moment along the
length of beam.
 Let us assume two sections PP' and QQ',
which are 'dx' distance apart, carrying
bending moment and shear forces 'M and
S' and 'M+ ∆M and S+ ∆S‘ respectively as
shown in Fig.

12.  Let us consider an elemental cylinder P"Q" of
area 'dA' between section PP' and QQ' . This
cylinder is at distance 'y' from neutral axis.

13.  
F)(FQ'
F
be,shallcylinderinforce
horizontalunbalancedHence,
dAδσσ
Q'
F
dAxσ
p'
F
be,shallQ'and'P'at
stressesthesetodueforcesThe
xy
I
dMM
δσσ
be,will'Q'at
stressbendingSimilarly,
xy
I
M
σ
bewillP"atstressbendingHence,







14.  This unbalanced horizontal force is resisted
by the cylinder along its length in form of
shear force. This shear force which acts
along the surface of cylinder, parallel to the
main axis of beam induces horizontal shear
stress in beam.
Aaxy
I
dM
dA)(σd(σpF
Q
FHF



15. DERIVATION OF FORMULA: SHEAR
STRESS DISTRIBUTION ACROSS BEAM
SECTION
 Let us consider section PP' and QQ' as
previous.
 Let us determine magnitude of horizontal
shear stress at level 'AB' which is at
distance YI form neutral axis.
 The section above AA' can be assumed to
be made up of numbers of elemental
cylinder of area 'dA'. Then total unbalance
horizontal force at level of' AS' shall be the
summation of unbalanced horizontal forces

16. a
_
yx
I
dM
FM
dA.y
1yy
1yy
.X
I
dM
dA.xy
1yy
1yy I
dM
HF









17.  Here, y = distance of centroid of area
above AB from neutral axis, And a= area
of section above AB.
 This horizontal shear shall be resisted by
shear area ABA'B‘ parallel to the Neutral
plane. The horizontal resisting area here
distance of centroid of area above AB
from neutral axis and a=area of section
above AB.
 Ah = AB x AA’=b x dx
where ‘b’is width of section at AB.

18.  We know that shear force is defined as
S=dM/dx
 Therefore, horizontal shear stress
acting at any level across the cross
sections.
Ib
ya
x
dx
dM
dx.b
ay
I
dM
x
A
F
forceshearhorizontal
resistingshear
stressshearHorizontal
_
H
_
H
M
H




19. SHEAR STRESS DISTRIBUTION DIAGRAM
1.Rectangular section
2.Circular section
maxNA
maxNA

20. 3.Triangular section
4.Hollow circular section
h/2 max
avg

NA
maxNA

21. 5.Hollow Rectangular section
6. “I” section
maxNA
maxNA

22. 7. “C” section
8. “+” section
maxNA
maxNA

23. 9. “H” section
10. “T” section
maxNA
maxNA

24. Numericals

25. Rectangular section sum
Example-1: Two wooden pieces of a section
100mm X100mm glued to gather to for m a
beam cross section 100mm wide and 200mm
deep. If the allowable shear stress at glued
joint is 0.3 N/mm2 what is the shear force the
section can carry ?

26. KN4FeShear forc
KN4N4000F
10061067.66
50810000F
3.0
I.b
yFA
τ
Νοω.
mm50
2
100
Υ
2mm000,10100100A
joint.gluedabovebeamofpartconsider
2N/mm3.0τ
4mm61067.66
12
3200100
I
:Solution












100mm
100mm
100mm
wooden piece

27. Circular section sum
Example-2: A circular a beam of 100mm
Diameter is subjected to a Shear force of 12kN,
calculate The value of maximum shear Stress
and draw the variation of shear stress along
the Depth of the beam.

28. 22.03N/mm
1.531.33
1.33
Now.
21.53N/mm
7853.98
31012
A
F
27853.98mm2100
4
π
A
N3101212kNF
τmax
τavgτmax
τavg
:Solution









29. 22.03N/mmτmax
D =100mm
NA

30. I section sum
Example-3: A rolled steel joist of I section
overall 300 mm deep X 100mm wide has
flange and web of 10 mm thickness. If
permissible shear stress is limited to
100N/mm2, find the value of uniformly
distributed load the section can carry over a
simply supported span of 6m.
 Sketch the shear stress distribution across
the section giving value at the point of
maximum shear force.

31. 100mm
300mm
10mm
150mm
__
Y 
10
NA
25.63N/mminm 
261.63N/mmavg 
2100N/mmmax 

32. mm10b
(givan)3N/mm100τ
3mm257500
)7010140()14510110(yA
N.A.atmaximumbewillstressshear
4mm61055.64Ixx
I 3xxI 2xxI 1xxIxx
4mm61029.1802800
12
328010
I 2xx
4mm61013.2321451100
12
310110
2ahIgI 1xx
Solution:














33. kN/m56.83w
kN/mm64.83
6
36.501
l
W
wu.d.l.
)loadtotalkN(36.501W
2
W
68.250
2
W
Now, F
kN68.250F
N6.250679F
1061055.66
257500F
100
I.b
yFa
τ












34. Triangular section sum
Example-4: A beam of triangular section
having base width 150mm and height 200mm
is subjected to a shear force of 20kN the value
of maximum shear stress and draw shear
stress distribution diagram.

35. 2N/mm2
33.15.1
τave5.1τmax
2N/mm33.1
15000
31020
A
F
τave
2mm15000200150
2
1
bh
2
1
A
20kNFforceshear
Solution:










36. max=2N/mm2
avg=1.33N/mm2
200mm
150mm
h/2
2/3.h
NA

37. Cross section sum
Example-5: FIG Shows a beam cross section
subjected to shearing force of 200kN.
Determine the shearing stress at neutral axis
and at a-a level. Sketch the shear stress
distribution across the section.
50mm
100mm100mm 100mm
100mm
100mm
50mm
50mm X

38. 215.48N/mm
50610129.16
1005000310200
I.b
yFA
τ
50mmb
200kNF
100mm5050y
25000mm10050A
:lineb_batstressshear
4mm610129.6
]
12
3100100
[2]
12
330050
[xxI
:Solution














39. 2N/mm03.5
250610129.16
812500310200
I.b
yFA
τ
3mm812500
25)50(250100)100(50yA
:axisx_xatstressShear
2
mm/N09.3
250
50
15.48X









40. 5.03N/mm2
15.48N/mm2
3.09N/mm2
100mm100mm 100m
m
100mm
100mm
50mm
50mm X
50mm
NA

41. Inverted T section sum
Example-6: Shows the cross section of a beam
which is subjected to a vertical shearing force of
12kN.find the ratio the maximum shear stress to
the mean shear stress.
60mm
20mm
60mm
20mm

42. 46
2
3
2
3
xx2xx1
21
2211
2
2
2
1
2
1
mean
max
3
mm101.36
30)1200(5012
12
6020
10)(301200
12
2060
III
inertiaofMoment
30mm
12001200
501200101200
AA
yAyA
y
50mm20
2
60
y1200mm6020A
10mmy1200mm2060A
AxisNeutralofPosition
τ
τ
:findTo
N101212kNs:Given
:Solution


















43. 2.21
τ
τ
5
11.029
τ
τ
stressshearmeantostressshearmaximumofRatio
5MPa
2400
1012
areac/s
forceShear
τ
stressshearMean
11.029MPa
101.3620
2510001050
τ
axisneutralaboveareaConsider
bI
SAy
τ
axis,neutralatproducedstressshearMaximum
mean
max
mean
max
3
mean
6
3
max












44. max=11.29MPA
avg=5 MPA
min =2.21 MPA
20mm
60mm
20mm
NA

45. L section sum
Example-7: An L section 10mm X 2mm show in
the fig. is subjected to a shear force F. Find the
value Of shear force F if max. shear stress
developed is 5N/mm2.

46. 4208.09mm
24)(6.7716
12
382
2ahgI
xx2
I
4106.12mm26.77)(920
12
3210
2ahgIxxI
6.77mm
1620
416920
2
a
1
a
2
y
2
a
1
y
1
a
y
4mm
2
y
216mm28
2
a
9mm
1
y
220mm210
1
a
:Solution




















47. 2mm
2mm
10mm
10mm
6.77mm
__
Y 
3.23mm

48. 68.14NF
2314.21
46.11F
5
I.b
yFA
τ
46.11mm
2
1.23
2)(1.232.232)(10yA
N.A.atoccurwillstressshearMaximum
314.21mm
208.09106.12
III
3
4
xx2xx1xx











49. Tee section sum
Example-8: A beam is having and subjected to
load as shown in fig. Draw shear stress
distribution diagram across the section at point of
maximum shear force, indication value at all
important points.
100kN
A B
3m 3m

50. 100200 200
300
100
25
275mm
__
Y

51. K.N502
100F
4mm3750000275)-(350(500x100)xyA
:web&flangeofjunctiontheatstressShear
4mm91.02x10
6693.57x106322.91x10
12
2150)-100)x(275x(5003100x(300)
12
275)-(350x(500x100)3500x(100)I
mm275
80000
612x10
100)x(300100)x(500
100)x15x(300100)x350x(500
Y
XX











2

52. 2
9max
3
2
avg
2
9min
N/mm85.1
500x10x02.1
3781250x3)10x(50
I.b
yFA
τ
mm3781250
)5.12x25x100()75x100x500(yA
axis,neutralatstressShear
mmN/84.1
100
500x367.0
τ
to,increasesuddenlywillstressShear
mmN/367.0
500x10x02.1
3750000x3)10x(50
I.b
yFA
τ
mm500b







53. max=1.85N/mm2
avg=1.84N/mm2
min=0.367N/mm2
100
200 200
300
100
25
275mm
__
Y 
NA

54. I section sum
Example-9: Find the shear stress at the junction
of the flange and web of an I section shown in fig.
If it is subjected to a shear force of 20 kN.
100mm
200mm
20mm
100mm
20

55. 2
2
6
3
3
_
46
3
xx
mm/N57.4915.0x
2
100
=
to,increasedsuddenlywillstressShear
mm915.0
100x10x36.39
180000x10x20
b.I
_
yAF
mm18000090x)20x100(yA
:webandflangofjunetionatstreesShear
mm10x36.39
12
200x100
I
N.k20F
:Solution






56. 100mm
200mm
20mm
100mm
20NA
2N/mm0.915τmin 
2N/mm915.0avg 
2
N/mm75.4max


57. Rectangular section sum
Example-10: A 50mm x l00mm in depth
rectangular section of a beam is s/s at the ends
with 2m span the beam is loaded with 20 kN
point load at o.5m from R.H.S. Calculate the
maximum shearing stress in the beam.
20kN
R
B
B
0.5m
2.0m
RA
A

58. 2N/mm5.4
00.35.1
τave5.1τmax
2N/mm00.3
100x50
31015
A
F
τave
kN51520B
B
B
R
kN15R
5.1x202xR
at AmomentTaking
Solution:










59. max=4.5N/mm2
50mm
100mm NA