CIV 214:
Mechanics of
Materials
Chapter 1: Stress
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Mechanics of materials:
• Studies the internal effects of stress and strain in a
solid body that is subjected to an external loading.
• Study of the body’s stability when a body such as a
Introduction
column is subjected to compressive loading
• Many of the formulas and rules of design cited in
engineering codes are based upon the principles of
this subject.
Chapter Objectives
• Review key principles of statics
• Determine internal resultant loadings in a body
• Introduce normal stress and shear stress
• Apply analysis to members under axial load
• Apply analysis to members under direct shear
Equilibrium of a Deformable Body
External Loads
•Surface Forces
• Caused by direct contact between two bodies
• Distributed over the area of contact
• Can be idealized as:
• Concentrated force (applied at a point)
• Linear distributed load w(s)w(s)w(s)
•Body Forces
• Act without direct contact (e.g., gravity,
electromagnetic field)
• Represented as a single concentrated force
• For gravity → weight of the body acts through
center of gravity
Equilibrium of a Deformable Body
Support Reactions
• If a support prevents translation in a given direction → a reaction force
develops in that direction. If a support prevents rotation → a couple
moment must act on the member
Equilibrium of a Deformable Body
Equilibrium of a Deformable Body
Equations of Equilibrium.
• Equilibrium conditions:
Balance of forces → prevents translation
Balance of moments → prevents rotation
• Vector form:
• 3D Scalar form (six equations):
• 2D (coplanar) case (three equations):
Equilibrium of a Deformable Body
Internal Resultant Loadings.
• Statics is used to determine the internal loadings within a body.
• Free-Body Diagram (FBD):
• Cut a section through the body → isolate part of it .
• Draw a Free-Body Diagram (FBD) of one part
• Internal forces appear on the “exposed” section
Equilibrium of a Deformable Body
Internal Resultant Loadings.
• Internal Forces & Moments:
• Represent effects of material above acting on material below (or vice
versa).
• Exact distribution is unknown → replaced by resultant force 𝐹𝑅​ and
moment 𝑀𝑅𝑂 at a point O
Equilibrium of a Deformable Body
Internal Resultant Loadings.
• Three Dimensions:
• Four different types of resultant loadings can then be defined as follows.
• Normal force, N. This force acts perpendicular to the area. It is
developed whenever the external loads tend to push or pull on the two
segments of the body.
Equilibrium of a Deformable Body
Internal Resultant Loadings.
• Three Dimensions:
• Four different types of resultant loadings can then be defined as follows.
• Shear force, V. The shear force lies in the plane of the area and it is
developed when the external loads tend to cause the two segments of
the body to slide over one another.
Equilibrium of a Deformable Body
Internal Resultant Loadings.
• Three Dimensions:
• Four different types of resultant loadings can then be defined as follows.
• Torsional moment or torque, T. This effect is developed when the
external loads tend to twist one segment of the body with respect to
the other about an axis perpendicular to the area.
Equilibrium of a Deformable Body
Internal Resultant Loadings.
• Three Dimensions:
• Four different types of resultant loadings can then be defined as follows.
• Bending moment, M. The bending moment is caused by the external
loads that tend to bend the body about an axis lying within the plane
of the area
Equilibrium of a Deformable Body
Internal Resultant Loadings.
• Coplanar Loadings:
• Only three different types of resultant loadings can then be defined as
follows.
• Normal force , N. , Shear force, V. , and Bending moment, M.
EXAMPLE 1.1
Determine the resultant internal loadings acting on the cross section at C of the
cantilevered beam shown in the figure.
Solution:
Support Reactions. The support reactions at Ado not have to be determined if segment CB is considered.
EXAMPLE 1.1
Determine the resultant internal loadings acting on the cross section at C of the
cantilevered beam shown in the figure.
Solution:
Free body Diagram (FBD).
EXAMPLE 1.1
Determine the resultant internal loadings acting on the cross section at C of the
cantilevered beam shown in the figure.
Solution:
Equations of Equilibrium.
EXAMPLE 1.2
Determine the resultant internal loadings acting on the cross section at C of the
machine shaft shown in in the figure. The shaft is supported at A and B, as
shown, which provide only vertical reaction forces on the shaft.
Solution:
Support Reactions. The free-body diagram of the entire shaft is shown in Fig.1–5b.Since segment AC is to
be considered, only the reaction at Ahas to be determined. Why?
EXAMPLE 1.2
Determine the resultant internal loadings acting on the cross section at C of the
machine shaft shown in in the figure. The shaft is supported at A and B, as
shown, which provide only vertical reaction forces on the shaft.
Solution:
Equations of Equilibrium.
Free-Body Diagram.
NOTE: The negative signs for and indicate they act in the
opposite directions on the free-body diagram
EXAMPLE 1.3
The 500-kg engine is suspended from the crane boom in the figure. Determine
the resultant internal loadings acting on the cross section of the boom at point
E.
Solution:
Support Reactions. We will consider segment AE of the boom so
we must first determine the pin reactions at A. The free-body
diagram of the boom is shown in Fig.1–6b.
EXAMPLE 1.3
The 500-kg engine is suspended from the crane boom in the figure. Determine
the resultant internal loadings acting on the cross section of the boom at point
E.
Solution:
Support Reactions. Applying the equations of equilibrium,
EXAMPLE 1.3
The 500-kg engine is suspended from the crane boom in the figure. Determine
the resultant internal loadings acting on the cross section of the boom at point
E.
Solution:
Free body diagram. The free-body diagram of segment AE is shown,
Equations of Equilibrium.
EXAMPLE 1.4
Determine the resultant internal loadings acting on the cross section at G of the
beam shown in the figure. Each joint is pin connected.
Solution:
Support Reactions. Here we will consider segment AG. The
free-body diagram of the entire structure is shown in Fig.1–7b.
Verify the calculated reactions at E and C.
Since BAand BD are also two-force members, the freebody diagram of joint Bis shown in Fig.1–7c.
EXAMPLE 1.4
Determine the resultant internal loadings acting on the cross section at G of the
beam shown in the figure. Each joint is pin connected.
Solution:
Free-Body Diagram. Using the result for the free-body
diagram of segment AG is shown in Fig.1–7d.
Equations of Equilibrium.
EXAMPLE 1.5
Determine the resultant internal loadings acting on the cross section at B of the pipe shown
in the figure. The pipe has a mass of 2 kg/m and is subjected to both a vertical force of 50 N
and a couple moment of. 70 N.m at its end A. It is fixed to the wall at C.
Solution:
Support Reactions. The x, y, z axes are established at B and the
free-body diagram of segment AB is shown in Fig.1–8b.
The resultant force and moment components at the section are
assumed to act in the positive coordinate directions and to pass
through the centroid of the cross-sectional area at B.
The weight of each segment of pipe is calculated as follows.
EXAMPLE 1.5
Determine the resultant internal loadings acting on the cross section at B of the pipe shown in
the figure. The pipe has a mass of and is subjected to both a vertical force of 50 N and a
couple moment of. 70 N.m at its end A. It is fixed to the wall at C.
Solution:
Equations of Equilibrium. Applying the six scalar equations of
equilibrium, we have
Exercise 1.1
Determine the resultant internal loadings in the beam at cross sections through points D and
E. Point E is just to the right of the 3-kip load.
Solution:
Homework No. 1
Determine the resultant internal loadings acting on the cross section through point B of the
signpost. The post is fixed to the ground and a uniform pressure of 7 lb/ft2 acts perpendicular
to the face of the sign.
Solution:
Questions