Transverse shear stress

1. PRESENTATION ON
TRANSVERSE SHEAR STRESS
AMIT KUMAR SINGH
15ME63R30
MECHANICAL SYSTEMS DESIGN
BY-

2. Table of content
Concept
1. Transverse Shear Load
2. Transverse Shear stress
3. Difference between Bending and Shear stress
Assumptions
Derivation
Analysis in rectangular cross section
Other cross section example
Points to remember
References

3. TRANSVERSE SHEAR LOAD

4. Beamsareoftensubjectedtotransverseloadswhichgenerate
bothbendingmoments‘M’andshearforces‘V’alongthebeam.The
bendingmomentscausebendingnormalstressesσtoarisethrough
thedepthofthebeam,andtheshearforcescausetransverseshear-
stressdistributionthroughthebeamcrosssectionasshown
TRANVERSE SHEAR
STRESS

5. DifferencebetweenBending and Shearstress
Bending stress
It acts perpendicular to
plane of cross section.
Bending stress varies
linearly over the depth
of beam.
At extreme fibre bending
stress is maximum.
At neutral axis bending
stress is zero.
Shear stress
It acts parallel to the
plane of cross section of
beam.
It varies parabolically
over the depth of beam.
At extreme fibre it is
zero.
At neutral axis shear
stress have some value.

6. ASSUMPTIONS
 Shear stresses are uniformly distributed
across the width of the beam.
 The shear formula is applicable for prismatic
beams.
 Accuracy of shear formula for rectangular
beam is directly proportional to depth to
width (d/b) ratio .
 Beam should be of homogeneous material.

7. DERIVATION
To determine the shear stress distribution
equation, look at a loaded beam as Fig.

8. Consider
now a segmentof this
elementat distance y
above the N.A.up to
the top of the element,
Look at a FBD
of the element
dx with the
bending
moment stress
distribution
only,

9. stressesduetothebending
momentsonlyformacouple,thereforethe
forceresultantisequaltozerohorizontally.
𝐹𝑥=0
𝐴
𝜎𝑥1 𝑑𝐴- 𝐴
𝜎𝑥2dA+𝜏 𝑥𝑦 𝑏. 𝑑𝑥 = 0
Substituting the valuesof the stresses:
𝐴
𝑀
𝐼
𝑦𝑑𝐴 −
𝐴
𝑀 + 𝑑𝑀
𝐼
ydA+𝜏 𝑥𝑦 𝑏. 𝑑𝑥 = 0
(
𝑑𝑀
𝐼
) 𝐴
𝑦𝑑𝐴= 𝜏 𝑥𝑦 𝑏. 𝑑𝑥
Let the width of section at a distance y from the N.A. be a function of y
and call It ‘b’. Applying the horizontal equilibrium equation ,
gives:
In order for it to be in equilibrium, a shear stress τ must be
present.

10. Where,
P= The shear force carried by the section
I= Moment of inertia
b= The sectional width at the distance y from the N.A.
Q= The First moment of Area
M= Moment
Y= Distance of centroid of hatched area from N.A.
𝝉=
𝑷𝑸
𝒃𝑰
solving for τ,
𝜏 𝑥𝑦= (
𝑑𝑀
𝑑𝑥
)(
1
𝑏𝐼
) 𝐴
𝑦𝑑𝐴
(
𝑑𝑀
𝑑𝑥
) = Load ‘P’
𝐴
𝑦𝑑𝐴=Q

11. TRANVERSE SHEAR STRESS ANALYSIS
OF RECTANGULAR CROSS SECTION-

12. A= b[d/2-y]
𝑦= y+1/2[d/2 -y]
𝑦=1/2[d/2+y]
Q=A 𝑦=b/2[(
𝑑
2
)2
-𝑦2
]
𝐼 𝑁𝐴=
𝑏𝑑3
12
𝜏=
𝑃
𝑏𝑑3
12
𝑏
2
(( 𝑑
2
)2−𝑦2)
𝑏
𝜏 =
6𝑃
𝑏𝑑3
[(
𝑑
2
)2 − 𝑦2]
At external fibres y=d/2
Therefore 𝜏=0
𝜏 𝑚𝑎𝑥=
6𝑃
𝑏𝑑3 ×
𝑑2
4
𝛕 𝐦𝐚𝐱=
𝟑
𝟐
𝐏
𝐛𝐝
𝝉 𝒎𝒂𝒙=
𝟑
𝟐
[𝝉 𝒂𝒗𝒈]
WHERE 𝝉 𝒂𝒗𝒈 =
𝒍𝒐𝒂𝒅
𝑻𝒐𝒕𝒂𝒍 𝒄𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒂𝒓𝒆𝒂
𝑦
[d/2 - 𝑦]
d
b

13. Triangular cross section
𝜏 𝑚𝑎𝑥 = 3/2𝜏 𝑎𝑣𝑔
𝜏 𝑁𝐴=4/3 𝜏 𝑎𝑣𝑔

14. 𝜏 𝑚𝑎𝑥 =
9
8
𝜏 𝑎𝑣𝑔
𝜏 𝑚𝑎𝑥 =
9
8
𝜏 𝑎𝑣𝑔
𝜏 𝑁𝐴 = 𝜏 𝑎𝑣𝑔
Rectangular cross section with vertical and
horizontal diagonals

15. I section

16. POINTS TO REMEMBER
 In case of rectangular , square and circular cross section
shear stress is maximum at N.A.
 For square and rectangle
𝜏 ∝A 𝑦 (because
𝑃𝑏
𝐼 𝑁𝐴
=constant)
 For circular , triangular , and square of vertical and
horizontal diagonals
𝜏 ∝
A 𝑦
𝑏
(because
𝑃
𝐼 𝑁𝐴
=constant)
 For I-section
𝜏 ∝
1
𝑏
(because A 𝑦=constant) at junction of flange and
web.

17. REFERENCES
 Strength of Materials lecture by Prof: S .K. Bhattacharya
Department of Civil Engineering, IIT, Kharagpur.
 Mechanics of material by Timosenko.

18. Thank you!