† Muddy Card Responses Lecture M6 2/12/2004 Synopsis. Continued derivation of equations of simple beam theory. Obtained three keyequations: Moment-curvature, Moment-Stress and shear force-shear stress. Also introduced second moment of area as the property of the beam cross-section that defines the stiffness ofthe beam, in conjunction with the Young’s modulus of the material. You went way to fast on the last part. I know, but we will review more slowly tomorrow. Last part of the lecture a bit confusing. What is Q? Q is the first moment of area of that partof the cross-section above the plane defined by z, the location at which we are evaluating theshear stress. We will review tomorrow. Not quite keen on what “Q” is. Fine, we will discuss further. I have no idea what actually happened for the last five minutes of lecture. OK, see above. Real question: We’ve only been talking about pure moments on beams, what happenswhen they’re loaded elsewhere? Good question. We will discuss this tomorrow. More explanation of the shear force/stress stuff at the end of lecture. Example would bevery helpful. Will do. Can you review the last part of the lecture tomorrow. It hit me a little quickly and I didn’tunderstand. Will do. Moved to fast, even through the first parts. Thankyou for the feedback. We will recover this material over the next lecture or two – I hope that this will consolidate things for you. A bit too fast. Understood. Beams must have a large length relative to their thickness to be considered as a beam, butwhen we analyze individual cross-sections the dimensional relationship is switched. How does this affect the assumptions we make during analysis? The assumption of slendernessallows us to make the assumption of plane-sections remaining plane, from which everythingelse flows. We will examine non-slender structures in the next couple of lectures. If we vary the cross-section of a beam along z, and the stress sxx remains constant, then does the beam still bend in a circular arc? It doesn’t make sense logically that it would,but if there is some way to explain it does, please do. Thanks. This is a perceptivequestion. The beam no longer has the same cross-section, and therefore although themoment is the same at any section the stiffness of the beam is not. The curvature of the beam will therefore vary with position along the beam and so we will not obtain a circular arc. We can still integrate up the moment-curvature relationship to obtain the deformed shape of thebeam.† † † I find it counter-intuitive that a sloping beam (varying cross-section with x) can beanalyzed in the same way we did for the uniform one. If you apply similar moments at thetwo ends will it still form part of a circular arc? No – see explanation above. Mud: In PRS #1 why doesn’t the shape of the object affect the stress equation. Because Mz s= -applies at any given cross-section. If I is a function of x, then that is OK, theIequation will take care of it. Remember this stems from plane sections remaining plane. Anyand every cross-section will remain plane so this equation applies everywhere, so long as thebeam is made of an homogenous material (no layers of dissimilar materials). Could you please review the reasoning behind the concept question? Are you saying that the stress is not linear through the thickness of the beam because your equation (4) e= -sxx is based upon the assumption that the strain and stress both varyxx Exlinearly? You mentioned strain is linear though stress is not, this was confusing. You are on the right track. See explanation above. The strain varies linearly, consistent with planesections remaining planar and perpendicular to the neutral axis of the beam. However, if there are different materials through the cross-section, the different materials experiencingthe same strains will experience different stresses – so a linear strain distribution will notproduce a linear stress distribution. Can you explain conceptual differences between I and Q again? Physical meanings. In future please give me something to work with here – what do you understand? I is a property of the whole cross-section. It reflects how material is distributed away from thecentroid of the cross-section. A higher value of I will produce a stiffer beam. I is termed the second moment of area. Q is a special case of the first moment of area. It is the first moment of area (integral of z, not z2) for that part of the cross-section more than a distance “z” fromthe centroid of the cross-section. It allows us to calculate the shear stress at that z location. Still don’t understand negative signs for F, S and M integral equations. There is no negative sign in F. The negative signs in the expressions for S and M arise due to the signconventions for positive shear stresses and tensile stresses acting in the opposite direction tothe sign convention for positive S and M. Don’t understand PRS 1. See above. Lost last 10 minutes. I will attempt to recover next lecture. In beams F=0, why is there sxx ? Where is the force for the xx stress coming from? The moment? This is absolutely correct. You could think of the bending moment on a particularcross-section as a couple made up of a net compressive force on the top half of the cross-section and a net tensile force on the lower half of the cross-section. The two forces are equalin magnitude and opposite in direction. Thus, they have no net force, but because they areoff-set from each other, they do exert a moment.† † † † MzIs Iyy the same moment as that in sxx = -. Thinking yes. Yes, you are correct, Iyy is the I second moment of area about the y axis (i.e. the one we are interested in). It should technically be written as Iyy since it is a second order tensor quantity. You will meet this if you take 16.20. Since we are only considering bending about one axis in Unified I will oftenjust use “I”. Is there any reason why you refer to I as the second moment of area? My roommate thecivil engineer likes to talk about just “the second moment” Having often heard tell aboutthis mythical second moment, I was just wondering is this it? Presumably yes. I would encourage you to avoid calling I simply a moment – since it should not be confused with amoment due to a force acting with a moment arm (i.e P x L = M), also confusion with adynamic moment of inertia can become confusing
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