Static Failure (pg 06) All material followed Hookeʹs law which states that strain is proportional to stress applied, until it exceed the proportional limits. It will reach and exceed the elastic limit to reach a well defined point known as the yield strength. The ultimate strength (maximum) lies beyond this point. Stress strain graph of several popular material NT-Failure G1.xmcd Pg 1 / 9
PRINCIPAL STRESSES stress acting on element in x direction stress acting on element in y direction shear stress acting on element in xy direction For long shaft or beam, the effect of transverse shear (shear along Z axis in diagram above) is very small (i.e. negligible) compared to the bending stress. Hence this analysis become a biaxial stress ( D) problem. The principal stresses and maximum shear stress are : = (Pg 77) σ = τ max = NT-Failure G1.xmcd Pg / 9
For round shaft, the effect of the stresses along y axis is relatively small compared to the stress along the x axis and hence for simplification, it can be ignored. The resulting principal stresses and maximum shear stress are : = σ = τ max = Failure Theories Ductile materials Maximum shear stress (MSS) Distortion energy (DE) Brittle materials Maximum normal stress (MNS) Modified Mohr (MM) NT-Failure G1.xmcd Pg 3 / 9
Maximum Shear Stress Hypothesis (MSS) (Pg 11) also called Tresca suitable for ductile material conservative Given > 0 > σ σ ve = ie. compressive stress. Yielding occurs when τ max or σ For MSS, Shear Yield strength, S sy = 0.5 Safety factor, = S sy τ max = σ Maximum Distortion Energy Hypothesis (DE) (Pg 13) valid for ductile materials popularly known as Von Mises Hencky moderately conservative Von Mises stress can be calt from Principal stresses as below: σ' = σ σ For biaxial stress state with axial, bending moment and torsional load : σ' = 3 For round shaft, 0 = σ' = 3 Yielding occurs if Safety factor ; = σ' σ' For DE, Shear Yield strength, S sy = 0.577 NT-Failure G1.xmcd Pg 4 / 9
Maximum Normal Stress Hypothesis (MNS) (Pg 6) earliest hypothesis by Rankine used for most brittle material (brittle or cast/forge) very conservative For d (biaxial stress state) analyses, if > 0 > σ S ut Failure occurs when where is the safety factor σ n.b: and σ are ( ve), ie compressive stresses. hence the safety factor S ut = or = whichever is smaller. σ Brittle-Coulomb-Mohr (BCM) (Pg 7) for brittle material not conservative If the vectors, σ 0 = S ut If the vectors, 0 σ σ S ut 1 If the vectors, 0 σ NT-Failure G1.xmcd Pg 5 / 9
= σ Modified Mohr (MM) (Pg 7) for brittle material moderately conservative If the vectors, σ 0 or 0 σ σ 1 = S ut If the vectors, 0 σ = σ If the vectors, 0 σ σ > 1 ( S ut ) S ut σ 1 NT-Failure G1.xmcd Pg 6 / 9
Failure Criterion Graph for brittle materials MNS stress line cont. line, prediction too conservative Modified Mohr blue dashed line, prediction moderately conservative. Coulomb Mohr red dashdot line, prediction not conservative Exercise 1 A boatʹs propeller weighing 70kg push the boat with a force of 1kN. The rotation of 3600 rpm create a torque of 500Nm on the shaft. A) If the shaft is 1.0m long and size 38mm dia. forged from steel with Ultimate tensile strength 340MPa and compressive strength 370MPa, determine the safety factor and state whether the shaft will hold up or fail? The recommended safety factor is.50. B) If the boat use a new shaft made of SAE 100 HR, will it support all acting loads find the safety factor. If the boat is anchored and the engine is off, find the safety factor? The recommended safety factor is 1.15. NT-Failure G1.xmcd Pg 7 / 9
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Exercise A shaft in a belt system is made of AISI 100 CD bar (see page 1016). With safety factor of 1.75, find the minimum size of the shaft using MNS and DE hypothesis. (Given: Belt tension, P1 = 1.75 P) NT-Failure G1.xmcd Pg 9 / 9