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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.