TOPIC 3 - SHEAR STRESS

SHEAR STRESS

Identify the shear stress, shear strain and modulus of rigidity

Shear stress has been defined as the stress component that acts in the plane of the
sectioned area. In order to show how this stress can develop, we will consider the effect
of applying a force P to the bar in fig 3.1.

P

AC
BD

Fig 3.1: Load acting on the bar

If the support is considered rigid, and P is large enough, it will cause the material of the
bar to deform and fail along the planes identified by AB and CD. A free-body diagram
of the unsupported center segment of the bar, fig 3.2, indicates that the shear force V =
P/2 must be applied at each section to hold the segment in equilibrium.

The average shear stress distributed over each sectioned area that develops this shear
force is defined by:

 aver V
A

where:
aver = average shear stress at the section, which is assumed to be the same at
each point located on the section
V = internal resultant shear force at the section determined from the equations
of equilibrium.
A = area at the section.

Shear Strain

- The measurement of the distortion  or change in shape of the

element/material.
- The unit of shear strain is the radian.

  G G = shear modulus of elasticity (also
called the modulus of rigidity).

Modulus Of Rigidity

• Shear stress ratio  compared to shear strain named modulus inside
elasticity shear. So that:

G=/

• G are also known as modulus of rigidity or shear modulus. Shear modulus
is equal Young modulus, E, for tension and direct compression. For mostly
material, E will be about 2.5 times G.

Explain The Factors Influencing Shear
Strength

• Type of loaded
Tension
Shear
Combined Shear and Tension

• Methods of fastening
Bolting
Rivetting
Pins
Keys
Welding/Soldering/Brazing
Bonding
Velcro
Magnetism

• Assembly
• Accuracy of positioning
• Ability to Hold components rigidly together

against all forces
• Requirement to separate components
• Retention of fastening over time

Calculate The Value Of Shear Stress, Shear Strain And
Modulus Of Rigidity

Shear Stress Calculation

Example 3.1

The wooden strut shown below is suspended from a 10mm diameter steel rod, which is
fastened to the wall. If the strut supports a vertical load of 5kN, calculate the average
shear stress in the rod at the wall and along the two shaded planes of the strut, one of
which is indicated as abcd.

c 20mm
b 40mm

d
a

5kN

Solution: aver = V

Average shear stress: A
i) for the rod
= 5000N
(5mm)2

= 63.7 N/mm2

ii) for the strut aver = V
A

= 2500N

(40mm)(20mm)
= 3.12 N/mm2

Example 3.2
A punch with a diameter of 19mm is used to punch hole in a 6mm steel plate. A force
of 116kN is required. What is the average shear stress in the plate and the average
compressive stress in the punch?

116kN

19mm 6mm plate

Solution:

Average shear stress for the plate: As = (19mm)(6mm) = 358mm2

aver = V
As

= 116 x 103N
(19mm)(6mm)

= 324 N/mm2

Average compressive stress in the punch:

c = P Ac = the cross-sectional area of the punch

Ac
= 116 x 103N

(19mm)2 / 4
= 409 N/mm2

Shear Strain Calculation

If the shear stress in a piece of wood is 100 Mpa
and if the modulus of rigidity, G = 85 GPa, find
the shear strain.

G = shear stress/shear strain
shear strain = Shear stress/G

= (100 x 10^6)/(85 x 10^9)
= 0.00117 radians.

Modulus Of Rigidity Calculation



Calculate the ultimate shear strength

The beam shown below is made of wood and is subjected to a resultant internal vertical
shear force at V = 3 N. (a) Determine the shear stress in the beam at point P and (b)
Calculate the maximum shear stress in the beam.

2mm P
5mm
1.5 mm
V = 3N
4 mm

(a) shear stress at point P
i) Section properties.

I = bh3 0.5mm A’
2mm
12 P
= (4mm)(5mm)3

12
= 41.7mm4

N A

Q= y’A’

12 4 mm

= [0.5mm + ½(2mm)](2mm)(4mm)
= 12mm3

ii) Shear stress

 = VQ

Ib
= 3N (12mm3)

4.17mm4(4mm)
= 0.216N/mm2

(a) maximum shear stress in the beam
i) Section properties

Q= y’A’ A’ A
12
2.5mm
= [2.5mm/2](4mm)(2.5mm) N
= 12.5mm3

ii) Shear stress

 = VQ 4 mm

Ib
= 3N (12.5mm3)

4.17mm4(4mm)
= 0.225N/mm2

Notice that this is equivalent to:
 = 1.5 V

A

= 1.5 3N

(4mm)(5mm)
= 0.225N/mm2

Translate The Single And Double
Shear

Single shear stress
Single shear stress formed when there was only a pair of

different force direction which subjected on some joint.

Doule Shear Stress
Double shear stress is more than a pair force subjected in
different direction acted at joint. This situation happen when more
than bars are connected.

List the examples Of Single And
Double Shear

PP

P/2 (a) V = P/2 V = P/2
P/2 Double shear steel joint (b)

P P

V = P/2 V = P/2
(b)
P/2 (a)
P/2

Double shear wood bonding

Illustrate Between The Single And
Double Shear

• Single shear carries all load on one face while
double shear carries it on two faces, so stress
is lower by a factor of 2 for a given load.

• For example shear stress on a cantilever pin is
V/A ( load/area, single shear) but on a pin
between two supports it is V/2A

Calculate The Single And Double
Shear Using Formula

Two bars of wood connected by using a bolt diametrical 13mm such
as under. Determine:

i. Shear stress in wood
ii. Shear stress in bolt

10kN 20mm 10kN
20mm

30mm 30mm

80mm

Solution xx
yy
30mm
20mm

plane area = 20x30mm2

Shear at xx and yy plane as figure above
i. Shear area which occurred in wood was area xx and yy plane

so shear area = 2 x 20x10-3 x 30x10-3m2
= 1.2x10-3m2

 wood  10  103 N
1.2  103 m2

 wood  8.33  106 N / m2

 wood 8.33MN / m2

i. Shear stress in bolt = F
A

10 103 N / m2
p 13103
  bolt 2

4

 bolt  75.34 106 N / m 2

 bolt  75.34MN / m 2

Two wood connected by using rivet diametrical 1.5 cm. If tension force
subjected was 30 kN, estimated shear stress in rivet.

30kN

30kN

Solution

Force 30 kN that are subjected cause rivet part on plane x1x2 try to
slide in lower part plane. Shear force 30 kN evacuate at surface x1x2
marking by dotted line further produce shear stress.

x1 x2 30kN

30kN

Shear Stress = Force /Area
.  = F/A

 30kN /  1.52

4
=16.98kN/cm2

(Change to kN/m2 unit)

Three wood press with rivet diametrical 1.5cm. If tension load that are
subjected was 50kN, estimated shear stress in rivet.

25kN 50kN
25kN

Solution

aa
bb

Caused of tension force that are subjected, there are tendency in rivet
to shear across plane as in figure with dotted line in section aa and bb.
Area that are bearing shear was double rivet cross sectional area
where area rivet was:

A   0.00152  0.177x103 m 2

4

So shear stress average in rivet was

F

A

 25x103 =141 MN/m2
0.1 7 7 x1 0 3

Three plate are connected with two rivet such as figure below. If shear
stress in rivet can not exceed more than 80MN / m2, determine
diameter of rivet.

10kN 10kN
10kN 10kN

Solution

5kN Shear Area A
5kN 10kN

Shear Area B

Shear above subjected with double shear and have two rivet , hence

area of shear was = 2  2  d 2 =A

4

F

A

So that 80 106  10 103

2 2  d 2

4

d 2  10 103 m

80   106

d=6.31mm