Introduction• Mechanics of materials is a study of the

relationship between the external loads on a body and the intensity of the internal loads within the body.

• This subject also involves the deformations and stability of a body when subjected to external forces.

Equilibrium of a Deformable Body

External Forces:1. Surface Forces: caused by direct contact of other

body’s surface2. Body Forces: other body exerts a force without

contact

Support Reactions• Surface forces developed at the supports/points of

contact between bodies

Equilibrium of a Deformable Body

Equations of Equilibrium• Equilibrium of a body requires a balance of

forces and a balance of moments

• For a body with x, y, z coordinate system with origin O (3D Space)

• Best way to account for these forces is to draw the body’s free-body diagram (FBD).

∑ 𝐹=0 ∑ 𝑀𝑜=0

Equilibrium of a Deformable Body

Internal Resultant Loadings• Objective of FBD is to determine

the resultant force and moment acting within a body.

• In general, there are 4 different types of resultant

• loadings:• a) Normal force, N• b) Shear force, V• c) Torsional moment or torque,

T• d) Bending moment, M

Example 1.1• Determine the resultant internal loading acting on

the cross section at C of the cantilevered beam shown in the Fig.

Example 1.3• Determine the resultant internal loading acting

on the cross sectional at G of the beam shown in the Fig. each joint is pin connected.

Stress• Normal Stress σ : Force per unit area

acting normal

• Shear Stress τ: Force per unit area acting tangent

Average Normal Stress in an Axially Loaded Bar

When a cross-sectional area bar is subjected to axial force through the centroid, it is only subjected to normal stress. Stress is assumed to be averaged over the area.

Normal StressAverage Normal Stress Distribution • When A bar is subjected to constant deformation,

then

P : Resultant Normal Force (N)σ : Normal Stress (Pa)A: Cross Sectional Area (m2)

∫𝑑𝐹=∫𝜎 𝑑𝐴𝑃=𝜎 𝐴𝜎=

𝑃𝐴

Example 1.5• The bar shown in the Fig., has a constant width

of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown?

Example 1.6• The 80 kg lamp is supported by two rods AB and

BC as shown in the Fig. If the AB has a diameter of 10 mm and BC has a diameter of 8 mm. Determine the average normal stress in each rod.

Average Shear Stress • The average shear stress distributed over each

sectional that develops a shear force. τ: Average Shear Stress V: Resultant Internal Shear Force A: Area at the Section• Different types of Shear

1. Single Shear 2. Double Shear

𝜏𝑎𝑣𝑔=𝑉𝐴

Example 1.9• Determine the average shear stress in the 20 mm

diameter pin at A and the 30 mm diameter at B that supports the beam in the Fig.

Allowable Stress • Many unknown factors that influence the actual

stress in a member. • A factor of safety is needed to obtained allowable

load. • The factor of safety (F.S.) is a ratio of the failure

load divided by the allowable load

𝐹 .𝑆=𝐹 𝑓𝑎𝑖𝑙

𝐹 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

𝐹 .𝑆=𝜎 𝑓𝑎𝑖𝑙

𝜎 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

𝐹 .𝑆=𝜏 𝑓𝑎𝑖𝑙

𝜏𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

Example 1.12• The control arm is subjected to the loading

shown in the Fig., determine the required diameter of the steel pin at C if the allowable shear stress for the steel is 55 MPa.