Normal stress & Shear Stress
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Transcript of Normal stress & Shear Stress
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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.
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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
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Support Reactions• Surface forces developed at the supports/points of
contact between bodies
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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
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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
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Example 1.1• Determine the resultant internal loading acting on
the cross section at C of the cantilevered beam shown in the Fig.
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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.
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Stress• Normal Stress σ : Force per unit area
acting normal
• Shear Stress τ: Force per unit area acting tangent
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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.
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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
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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?
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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.
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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
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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.
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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
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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.