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Html http //wps. pearsoned.co. uk/emaukheharkermktgintro1/127/32609/8347930. cw/content/index. h tml ml 1 Companies must spend money and invest in transporting raw materials and finished goods materials handling warehousing storage and holding inventory and managing the distribution process including the cost of information technology. Basics of Logistics Management http //highered*mcgrawhill*com/sites/0073529931/studentview0/chapter16/multiplechoicequiz. Worldwide these activities and...

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In this problem we use the stiffness method to find member forces for a beam, and then we'll construct the shear diagram the key concepts we'll use are calculating deformations for a beam calculating member forces for a beam and constructing the shear diagram for a beam given member forces and loading and here's our beam fixed at node 1 rollers at nodes 2 & 3 it's a continuous beam he and I are constant along its length we are going to determine the internal shears at the ends of the members, and we'll use that information along with the loading to construct the shear diagram for the entire beam here's the approach we'll follow we'll begin by approximating the answer then we'll use that after we construct our shear diagram from the stiffness method results in order to evaluate the reasonableness of our answer then to begin the stiffness method we'll construct the structure stiffness matrix compiled the fixed end forces then we will find deformations at the unconstrained degrees of freedom then we'll find member forces specifically the shears at the ends of the members we use that to construct the shear diagram, and then we'll compare with our approximation let's begin, so we begin with step 1 let's approximate the answer to do that I'm going to assume that the distributed load all goes to the nearest vertical support that means since this is the only region with distributed load 1.5 meters worth goes to the reaction at node 3 and 1.5 meters worth goes to the reaction at node 2 so the reaction node 2 in the y-direction will be approximately the distributed load times the one and a half meters over which at X and that's 15 kilonewtons up the reaction at node 3 in the Y Direction will be approximately 10 kilonewtons per meter 1.5 meters so also 15 kilonewtons up that means that reaction 1 at node 1 in the Y direction will be about zero well we construct our approximate shear diagram in the units of kilonewtons we begin at zero and the shear diagram stays 0 because there is no applied load, so we have constant shear and since the magnitude was zero stays zero then at reaction to it jumps up a magnitude of 15 kilonewtons, and then we'll have a linear shear diagram because we have a constant distributed load, and it will decrease starting at a magnitude of 15 minus the distributed load acting over 3 meters gives us a minus 15 kilonewtons then reaction 3 gives us a jump, so we have the minus 15 plus the jump of 15 that equals 0 and so that checks out because the shear diagram should close to 0 now about this portion being 0 and this reaction being 0 if we sketch the anticipated elastic curve that is the deflected shape expect it to look something like this because it's fixed at this end the slope must be zero be concave up because of the downward distributed load but in order for this to this shape to occur we'll need a downward reaction, so I actually expect a small negative value here let's see now we're ready to begin the stiffness method we do that...

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