ASSIGNMENT 2 - STRUCTURE (GROUP 9)

DCC20053 : MECHANICS OF
CIVIL ENGINEERING
STRUCTURE

ASSIGNMENT 2 : TOPIC 5 (SHEAR STRESS)
LECTURER NAME : MADAM NORHAIZAH BINTI AMBIAH

GROUP 9

NAME REGISTRATION NUMBER

AZHAN FAKHRI BIN MOHAMAD 04DKA20F2067
AHMAD MIRZA AZWAR BIN MAT RIFFIN 04DKA20F2047

CONTENTS

 OBJECTIVE  SHEAR STRESS IN RECTANGULAR BEAM

 INTRODUCTION  SHEAR STRESS IN  BEAM

 TYPE OF STRESS  EXAMPLE (SHEAR STRESS IN  BEAM)

 SHEAR STRESS IN JOINT  CONCLUSION

 SHEAR STRAIN  REFERENCES

 MODULUS OF RIGIDITY

 EXAMPLE ( SHEAR STRESS IN JOINT )

 SHEAR STRESS IN BEAM

OBJECTIVE

1. Defined the relationship between stress and strain.
2. Analyse the stress and strain using related equations.
3. Determine and analyse the deformation of a rod of uniform or

variable cross section under one or several load.
4. Explain the effect of shear force on a loaded beam.
5. Using the shear stress formula for beams of section I.

INTRODUCTION

● Stress is associated with the strength of the material from which
the body is made.

● Stress is the internal force exerted by one part of an elastic body
upon the adjoining part.

● Stress has a dimension of Newton/m2.

● Any deformable material (ie. Rubber) tend to distort/deform when
subjected to a bending moment.

● All bending moment induce bending stress and all shear forces
induce shear stress.

TYPE OF STRESS

1. Direct stress or normal stress, σ -

 Stress that is evenly distributed over the entire cross-section.
 Normal stress includes tensile and compressive stress

TYPE OF STRESS

2. Shear stress, τ –

 Results when a force tends to make part of the body or one side of a plane
slide past the other.

SHEAR STRESS IN JOINT

 Joint is component that connect something in member of structure
 Load and moment will be transferred through this joint
 Joint will get action either or combination forces following

– Axial force
– Shear force
– Tension force
– Moment force
 Types of joint that common use in constructing is bolt, rivet and welding.

SHEAR STRESS IN JOINT

a) Single shear stress

 Established when there was only double opposite force direction
who act on something extension

  V = Shear force
A = Area

 = Shear stress

SHEAR STRESS IN JOINT

b) Double shear stress

 Occur when there was more than pair of opposite force direction
act on this extension. This situation arises when there were more
than two members continuing on.

= V  = Shear stress

n1An2 V = Shear force
n1 = number of extension
n2 = number of bolts / rivets
A = Area

SHEAR STRAIN

● Shear strain is a strain that acts parallel to the surface of a material that it
is acting on.

● Angle  is single measurement or elemental form change call strain shear.

Angle is measured in radiant, therefore had no dimension.
● Under the action of shear stresses, it does not tend to lengthen or shorten

in the x, y and z directions. Within other words the side length of the
element does not change otherwise the shear stress produces a change in
the shape of the element.

SHEAR STRAIN

 = L
L

 = Shear strain
L = Change of length of the body
L = Original length of the body

MODULUS OF RIGIDITY

 Shear Stress ratio  over with Shear Strain named Modulus
Rigidity, G.

 G is also known as the modulus of rigidity or the modulus of
shear. The shear modulus is equal to Young's modulus, E, for
tension and direct compression. For most materials, E is more less
2.5 times G.

 G = Modulus of rigidity

G=   = Shear stress
 = Shear strain

EXAMPLE ( SHEAR STRESS IN JOINT )

Three plates are connected by two rivets as figure under. If shear stress in rivet may
not exceed 80MN/m2, determine the diameter of rivet.

SOLUTION :

π × d2
A=

4

π × d2 = V

4 n1A n2

π × d2 10 × 103 F = 10 kN

=  = 80 MN/m2
4 (2) (80) (2)
= 80 N/mm2
nn12==
d2 = 39.789 2
d = 39.789
2
d = 6.31 mm

SHEAR STRESS IN BEAM

● When a beam get load with burden, both of bending moment, M and
shear strength, V will act on cross section.

● In this unit we will study about distribution of shear stress, that combined
with shear force, V.

● To see how shear stress action, consider a rectangular cross-sectional
beam with wide, b and height, h as Figure below.

SHEAR STRESS IN BEAM

 = FA ത

bI

 = Shear stress
F = Load / Shear force
ത = Distance from natural axis
b = Wide
I = Second moment area

SHEAR STRESS IN
RECTANGULAR BEAM

● Consider a beam of rectangular section subject to load. Take one cut at
any point as in figure below.

A = Area of shearing part
b = Width of the shearing part
h = Height / depth
ത = Distance from centroid of the shearing area
to the natural axis
y = Distance of the shearing area from the
natural axis
 = Shear stress
I = Second moment area

SHEAR STRESS IN  BEAM

● Consider I-shaped beam section as below .

A = Area of shearing part
b = Thickness of the web
h = height
B = Overall width of the section
D = Overall of the section
y = Distance from the shearing area from
the natural axis
 = Shear stress
I = Second moment area

EXAMPLE (SHEAR STRESS IN  BEAM)

● Determine the shear stress distribution over the cross section at
A, B, C and D as shown in the figure. Given F = 25 kN and I = 4008
cm4.

102.1mm

A

10mm

B
C

260.4mm D

6.4mm

SOLUTION :

STEP 1 : NEUTRAL AXIS, ഥ

ത =
2

ത = 260.4
2

ത = 130.2 mm

ത = 130.2 mm

STEP 2 : SECOND MOMENT AREA , IXX STEP 3 : SHEAR FORCE , kN
F = 25 kN
 IXX = 4008 cm4
= 40.08 × 106 mm

Ta 102.1mm Ta
Tb Tc Tc 10mm
260.4mm Tb

Td Td

6.4mm

ത = 130.2 mm Tf Te Te
Tf

Tg Tg

STEP 4 : SHEAR STRESS AT EACH POINT 102.1mm
Tc Tc
Ta = Tg = 0kN Ta Ta
1.Tb Tb 10mm
Tb

A = 10 × . ത = 125.2 mm Td Td

= 1021 mm2

ഥ = 130.2 - 260.4mm 6.4mm

= 125.2 mm
Te
b = 102.1 mm ത = 130.2 mm Tf Te Tf

ഥ Tg Tg
Tb =

Tb = × ( . )
. × ( . )

= 0.781 N/mm2

2. Tc

A = 102.1 × Ta 102.1mm Ta
125.2 mm Tc Tc 10mm
ഥ = 113002.12m- m 2 Tb
= Tb

= 125.2 mm ത = 260.4mm

b = 6.4 mm

ഥ Td Td
Tc =

6.4mm

Tc = × ( . ) ത = 130.2 mm Tf Te Te
( . × 6 )( . ) Tg Tf

Tc = 12.458 N/mm2

Tg

3. Td

A1 = 102.1 × 10 102.1mm
= 1021 mm2 Ta Ta

ഥ 1 = 130.2 – 10mm
Tb
Tc
= 125.2 mm ത 1 = 125.2 mm Tb
Tc
A2 = 120.2 × 6.4 ത 2 = 60.1 mm
= 769.28 mm2 260.4mm

ഥ 2 = . – Td Td


= 60.1 mm 6.4mm
Te
b = 6.4 mm
Tf
ത = 130.2 mm Tf Te

Td = × ( × . + . × . )

. × . Tg Tg

Td = 16.964N/mm2

STEP 5 : SKETCH SHEAR STRESS DIAGRAM

Ta = 0 N/mm2 102.1mm 0 N/mm2
Tb = 0.781 N/mm2 0.781 N/mm2
Tc = 12.458 N/mm2 12.458 N/mm2
Td = 16.964 N/mm2
Te = - 12.458 n/mm2 16.964 N/mm2
Tf = - 0.781 N/mm2
Tg = 0 - 12.458 N/mm2
- 0.781 N/mm2
A
0 N/mm2
10mm

BC

260.4mm D

6.4mm

CONCLUSION

Objective from assignment is to learn about shear stress in beam
and in joint. We able to use formula and equation related to shear
stress in beam to find shear stress value across I section beam. We
also learn about shear stress in joint for rivets. Joints are important
in civil engineering field because joints are components that
connect a member to form a structural framework. Connections
are needed to move various forms of loads and moments. It must
be designed so as not to fail to bear burdens and moments. Design
of beam also important to make sure beam able to support force
that is applied to it. In this assignment the objective is achieved.

REFERENCES

I. Yusof Ahmad, ‘Mekanik Bahan dan Struktur’ Penerbit UTM 2001
II. Hibbeler, R.C., Mechanics Of Materials, 8th Edition in SI units, Prentice

Hall, 2011.
III. Kekuatan Bahan – 1983 G.H. Ryder Universiti Pertanian Malaysia Mc Millan

Publishers (M) Sdn. Bhd.
IV. Kajidaya bahan – 1989 Mohamad Rashid b. Nabi Bax U.T.M.

THANK
YOU

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