STRENGTH OF MATERIALS :

STRESS
To be presented by : Group 1
INTRODUCTION
• Strength of materials, also called Mechanics
of materials, is a subject which deals with the
behavior of solid objects subject to stresses
and strains.

• The study of strength of materials often

refers to various methods of calculating the
stresses and strains in structural members,
such as beams, columns, and shafts.
 STRESS
Stress is defined as the strength of a material
per unit area, it is also called the unit strength. It
is the force on a member divided by area, which
carries the force, formerly express in psi, now in
N/mm2 or MPa.

𝐴𝑝𝑝𝑙𝑖𝑒𝑑 𝑓𝑜𝑟𝑐𝑒
𝑆𝑡𝑟𝑒𝑠𝑠 =
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝐴𝑟𝑒𝑎
 STRESS
INTERNAL FORCES AXIALLY LOADED BARS
01 02
• Compression • Centroidal Axial Load
• Tension • Saint Venant’s Principle
• Shear • Inclined Plane
• Torsion • Procedure in Stress Analysis
• Bending • Example of Axial (Normal Stress)
Force

SHEAR STRESS BEARING STRESS

03 Example and Explanation of
04 Example and Explanation
Shear Stress of Shear Stress
 Internal
Forces
01
 COMPRESSION FORCE
● Compression is a squeezing or pushing force applied to a material.
● This strain frequently shortens or compress the material.

● Example : Squeezing a sponge in your hand – you’re generating a

compression.
𝑚
● Units : Newtons (N) – 𝑘𝑔 2
𝑠
 TENSION FORCE
● Tension is a force that cause substance to stretch. This strain
frequently lengthens materials.
● This force is often given as “pulling force”.

● Example :Two men pulling a rope – they’re stretching the rope by

creating tension.
𝑚
● Units : Newtons (N) – 𝑘𝑔 𝑠2
 SHEAR FORCE
● Two nearby forces pushing or pulling each other, but not directly
opposing each other. By moving an object’s molecules apart
sideways, a shearing stress slice or rips it.

● Examples : Pruning Shear cutting through a branch, paper-cutter

cutting a paper. (The branch and the paper is “subjected to a shear
loading”)

Shear force is combination of tensile force

and compression force. It can be observed
when you apply both the forces to a body
on alternate faces.
 SHEAR FORCE
● Sign Convection : ● Similarly, a shear force that has
the tendency to move the left
● A shear force that tends to side of the section downward
or the right side upward will be
move the left of the section
considered a negative shear
upward or the right side of the force.
section downward will be
regarded as positive.
 TORSION FORCE (ROTATING)
● Torsional forces are forces that cause an object to twist or rotate
about its longitudinal axis.
● In torsional forces, the twisting moment is called torque.
● Example: Wringing out a damp washcloth
Units:
• Torsional force: Newton (N) or lb
• Torsional moment: N-m or lb-ft
• Torsional stress: N/m2 or lb/ft2

● Torsional forces result in shear stress within an object.

● Explanation: Torsional force causes the layers of an object to slide
or shear past each other, creating shear stress.
 BENDING MOMENT
● Bending moments are internal forces that occur in a structural
element when it is subjected to a load or a couple that causes it to
bend or curve.
● When a beam bends, the top part of the beam experiences
compression, while the bottom part experiences tension.

Units: N-m (SI) or

ft-lb (ENG)

● The magnitude of the bending moment varies along the length of

the beam.
● Bending moments are crucial in analyzing and designing structures
such as bridges, buildings, and beams.
 BENDING MOMENT
● Sign Convection :
● A bending moment is considered positive if it tends to cause
concavity upward ( sagging).

● While a bending moment is considered negative if it tends to cause

concavity downward ( hogging).
 02
Axially
Loaded Bars
 CENTRODIAL AXIAL LOAD
• Axial load is the term used to describe the load that pushes against the
body’s centroidal axis. Along the longitudinal axis of the member, the
centroidal axis of the body passes through the gravitational center. The
load moves along a route that is parallel to the body’s cross-sectional
plane.
 CENTRODIAL AXIAL LOAD
• The stress for the axial loading can now be estimated because the
free body diagram demonstrates that the forces add up to zero.
Since stress is computed by dividing force by area, in the example
above, the cross-sectional area that is perpendicular to the load P
would be used.
𝐹
𝜎 = 𝐴
where:
σ = normal stress
F = force
A = area
• When the cross-sectional area stays constant, the stress resulting
from an axial load would be constant.
 SAINT VENANT’S PRINCIPLE
• According to Saint Venant’s Principle, if a point on an axially
loaded cross-section is far enough from the place where the load
was applied or if the cross-section of the member has any
discontinuities, the stress measured at that point will be uniform.

• By calculating σ, we assume that we are reasonably far away from

the point of application or any discontinuity such that the normal
stress is uniform.

• During rapid changes in the material’s cross section, this increase

in stress, also known as a stress riser, also happens.
 SAINT VENANT’S PRINCIPLE

• Stress and strain distribution becomes uniform at a distance from

the load application points.
 INCLINED PLANE
• If a piece of material is sliced along a plane that is inclined at an
angle to the direction of loading, a system of forces must be
applied to the plane in order to maintain equilibrium.

• To obtain a clear picture of the stresses in a bar, we mist consider

the stresses acting on an inclined section of a bar.
 INCLINED PLANE

The Force P can be resolved into components:

Normal Force N perpendicular to the inclined

plane, 𝑁 = 𝑃𝑐𝑜𝑠𝜃
Shear Force V tangential to the inclined plane,
𝑉 = 𝑃𝑠𝑖𝑛𝜃
PROCEDURE IN STRESS ANALYSIS
1. Equilibrium Analysis

• FBD
• Find the reaction using:
Σ𝐹𝑥 = 0, Σ𝑦 = 0, Σ𝑀 = 0
• Compute for internal forces

2. Stress Computation

• Determine the Area/Cross sectional area of the member.

• Determine the type of forces: Normal or Tangent to the area.
1.0𝑃
𝜎 ≈
𝐴
EXAMPLE OF AXIAL OR NORMAL STRESS
• The bar ABCD in Fig. (a) Consists of three cylindrical steel
segments with different lengths and cross-sectional areas. Axial
loads are applied as shown. Calculate the normal stress in each
segment and make a free-body diagram.
EXAMPLE OF AXIAL OR NORMAL STRESS
• At section 1 : • At section 2 :

(+) →, ← (– )Σ𝐹𝑥 = 0 (+) →, ← (– ) Σ𝐹𝑥 = 0

𝐹𝐴𝐵 – 4000 𝑙𝑏. = 0 – 𝐹𝐵𝐶 + 9000 𝑙𝑏. – 4000 𝑙𝑏. = 0
𝐹𝐴𝐵 = 4000 𝑙𝑏. 𝐹𝐵𝐶 = 5000 𝑙𝑏.
EXAMPLE OF AXIAL OR NORMAL STRESS
• At section 3 :

(+) →, ← (–) Σ𝐹𝑥 = 0

𝐹𝐶𝐷 – 7000 𝑙𝑏. = 0
𝐹𝐶𝐷 = 7000 𝑙𝑏.
• Solving for Normal Stress :

• AB = 4000 lb.
P
𝜎𝐴𝐵 ≈ ( )𝐴𝐵
A
4000 lb. • CD = 7000 lb.
𝜎𝑨𝑩 = 2
= 3333.33 psi P
1.2 in
𝜎𝐶𝐷 ≈ ( )𝐶𝐷
A
• BC = 5000 lb 7000 lb.
σ𝐂𝐃 = 1.6 in2 = 4375 psi
P
𝜎𝐵𝐶 ≈ ( )𝐵𝐶
A
5000 lb.
𝜎𝑩𝑪 = 2 = 2777.78 psi
1.8 in
 03
Shear Stress
 SHEAR STRESS
● is defined as a force that induces material deformation due to
slippage along a plane or planes parallel to the imposed stress. It is
tangent to the area (tangent means that the force is parallel to the
cross-sectional area), and it arises whenever applied loads cause
one section of the body to slide past its adjacent section.
SHEAR STRESS
RIVET CONNECTION
SHEAR STRESS
BOLT CONNECTION
SHEAR STRESS
SHEAR STRESS
SHEAR STRESS
FORMULA FOR SINGLE SHEAR
𝑣
𝜏=
𝐴

FORMULA FOR DOUBLE SHEAR

𝑣 𝑃
𝜏= =
𝐴 2𝐴
SHEAR STRESS
 SHEAR STRESS
AREA IF THE SHEAR IS THE RIVET OR BOLT
𝜋𝑑 2
𝐴=
4

AREA IF THE SHEAR IS THE THICKNESS OF THE PLATE

𝐴 = 2𝜋𝑟(𝑡)
 SHEAR STRESS
Sample Problem

The lap joint shown in the figure above is fastened by four rivets with
a diameter of ¾ inches. Find the maximum load P that can be applied
if the working stresses are 14ksi for shear in the rivet. Assume that
the applied load is distributed evenly among the four rivets.
 SHEAR STRESS
Area if the shear is the Rivet or Bolt

𝜋𝑑²
𝐴=
4

Area if the shear is the thickness of the plate.

𝐴 = 2 𝜋r(t)
SHEAR STRESS
04
Bearing Stress
 BEARING STRESS
The bearing stress is the force that pushes a structure divided by
the area. We can define it as a contact pressure between two
separate bodies. If we apply a compressive stress on a body, then
there will be a reaction as internal stress in the material of the body.
This internal stress is known as the bearing stress.
BEARING STRESS
FORMULA FOR BEARING STRESS
 BEARING STRESS
Soil Pressure Beneath A Pie
 BEARING STRESS
Rivet Connection
BEARING STRESS

FORMULA
BEARING STRESS
 BEARING STRESS
The lap joint is connected by three 20 mm diameter rivets
assuming that the axial load P= 50 kN is distributed equally
among the three rivers. Find the bearing stress between a plate
and a rivet.

3𝑃𝑏 = 𝑃

𝐴𝑏 = 𝑑𝑡 = 20 25 = 500𝑚𝑚2 𝜎𝑏 = 33.33 𝑀𝑃𝑎

THANK YOU!
 MEMBERS :
LEADER:
● Calibo, Mary Ann

Members:
● Blase, Louie Dayne
● Castillote, Heidi
● Drio, Darlene Faye
● Gutierrez, Georgia
● Leviste, CJ
● Lopiga, Hersha Mhele
● Macayanan, Russel Marie
● Serrano, Dara Czarina