This presentation discusses transverse shear stress in beams. It begins with an introduction distinguishing bending stress from shear stress. The assumptions and derivation of the shear stress formula are then outlined. Analysis is shown for rectangular cross sections, where shear stress is highest at the neutral axis and zero at extreme fibers. Other cross section shapes are briefly discussed, including their maximum shear stress ratios. Key points are recapped about shear stress distribution across different cross section geometries. References are provided for further reading.

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!