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Unformatted text preview: Lecture 6 1 Lecture 6 1 Outline: Combined bending and axial loadComposite sections under bending 1. Combined bending and axial load Recall that under axial (normal) force the stress distribution is uniform across the crosssection, i.e. N σ Fig. 1 Beam under axial (normal) force The stress is proportional with the axial force N and inversely proportional with the crosssectional area A , i.e. N A σ = Recall that under pure bending moment the stress distribution is linear across the crosssection, i.e. σ M Fig. 2 Beam under bending moment The stress is proportional with the bending moment M and the distance from the neutral axis y and inversely proportional with the moment of inertia of the section about the neutral axis I , i.e. M y I σ = − When both bending moment and axial force act together we can use superposition principle to find the stress distribution across the section 1 Mechanics of Solids Dr Emre Erkmen Lecture 6 2 M N Fig. 3 Beam under combined bending moment and axial force Stress at a point y from the centroidal axis in this case can be written as N M y A I σ = − Neutral axis by definition is where the strains (and hence stresses) are zero. Under combined bending moment and axial force the centroidal axis does not coincide with the neutral axis as shown in the figure below. x N My A I σ = − − y Centroidal axis Neutral axis not passing from centroid M N Fig. 4 Shifting of the neutral axis due to combined bending moment and axial force This is because the normal stresses due to axial force shifts the axis of zero stresses. We will be locating our axes at the centroid. Thus y is measured from the centroid not the neutral axis. Eccentrically applied force When an axial force P is applied at a point that is different than the centroid of the section it causes bending action due to eccentricity e . This is because the eccentric axial force is equal to an axial force acting at the centroid of the section plus a force couple causing moment. Case 1 If the load is tensile (hence positive) and application point of the force P is below the centroid it causes positive bending moment (according to our sign convention that bottom fibres are in tension) as shown in the figure below. x e P x P = M=Pe x e P = + P P y Fig. 5 Eccentrically applied tensile force below the centroid Lecture 6 3 The stresses due to this load can be calculated as P M P P e y y A I A I σ × = − = − Note that the sign in front of the first term is positive because the force is tensile and the second term is negative because a positive bending moment causes compression above the centroid according to our bending formula (lecture 4). The stress distribution and the location of the neutral axis are as shown in the figure below....
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 Three '11
 BROWN
 Force, Stress, Second moment of area, yn, neutral axis

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