Given :M1 = 400 kN-mM2 = 500 kN-mM3 = 600 kN-mP = 150 kN w1= 60 kN-m w2 = 120 kN/m a = b = d = 6 mc = 4 m Determine the following:a. The maximum positive shear (in kN).b. The maximum negative shear (in kN).c. The maximum positive moment (in kN-m).d. The maximum negative moment (in kN-m).e. The location of maximum shear from the left end of the beam (in m).f. The location of the maximum moment from the left end of the beam (in m). Kindly Draw the Shear and Moment Diagram. Compute for all of the necessary elements, Include the proper units, use the proper formula and round-off all the answers and final answers to 2 decimal places. PW2WiWiM3M2ad Mechanics of Deformable BodiesI =bh3SHEAR STRESS FORMULA12McMcorBending StressIIArea = A'MOVING LOADSFor more than one moving Load:1. The bending moment under a particular load is amaximum when the center of the beam is midwaybetween that load and the resultant of all loads on they'Aspan of the beam.2. The maximum shear occurs at the reaction where theresultant load is nearest. Usually, it happens if thebiggest load is over that support and as many as possibleof the remaining loads are still on the span.PROCEDURE:1. Determine the location of the Resultant relativeto the given loads.2. Put the center of the load and the resultant tocoincide with the beam.3. Calculate supports.4. Section/cut the beam where the load inconsideration is located.The shear stress in the%3DVQT =Itmember at the point located at adistanceThis stress is assumed to beconstant and therefore averagedy from the neutral axis.across the width t of the member.MOST ECONOMIC SECTIONSV= the shear forceI = the moment of inertia of the ENTIRE cross-sectional area calculated about the neutral axis.When a structural section is selected to be used as abeam, the section modulus must be equal to or greaterthan the section modulus determined by the flexureequation;t = the width of the member's cross section measuredat the point where the shear stress is to be determined.|M\maxOallowThis equation indicates that the section modulus of theselected beam must be equal to or greater than the ratioof the bending moment to the working stress.A design engineer is often required to select the lighteststandard structural shape (such as a W-shape) that cancarry a given loading in addition to the weight of thebeam. Following is an outline of the selection process:1. Determine the maximum shear and moment ofthe (loaded) beam. A V-M diagram is the fastestway to do it. Neglect beam weight at this point.2. Determine the minimum allowable sectionQ = 2ÿ'\A'j + y½A'½modulus, Smin3. Choose the lightest shape from the list ofstructural shapes where S Smin4. Calculate the maximum bending stress in theselected beam caused by the loading PLUS thebeam self weight. If ơmax

Question

Given :M1 = 400 kN-mM2 = 500 kN-mM3 = 600 kN-mP = 150 kN w1= 60 kN-m w2 = 120 kN/m a = b = d = 6 mc = 4 m Determine the following:a. The maximum positive shear (in kN).b. The maximum negative shear (in kN).c. The maximum positive moment (in kN-m).d. The maximum negative moment (in kN-m).e. The location of maximum shear from the left end of the beam (in m).f. The location of the maximum moment from the left end of the beam (in m). Kindly Draw the Shear and Moment Diagram. Compute for all of the necessary elements, Include the proper units, use the proper formula and round-off all the answers and final answers to 2 decimal places. PW2WiWiM3M2ad Mechanics of Deformable BodiesI =bh3SHEAR STRESS FORMULA12McMcorBending StressIIArea = A'MOVING LOADSFor more than one moving Load:1. The bending moment under a particular load is amaximum when the center of the beam is midwaybetween that load and the resultant of all loads on they'Aspan of the beam.2. The maximum shear occurs at the reaction where theresultant load is nearest. Usually, it happens if thebiggest load is over that support and as many as possibleof the remaining loads are still on the span.PROCEDURE:1. Determine the location of the Resultant relativeto the given loads.2. Put the center of the load and the resultant tocoincide with the beam.3. Calculate supports.4. Section/cut the beam where the load inconsideration is located.The shear stress in the%3DVQT =Itmember at the point located at adistanceThis stress is assumed to beconstant and therefore averagedy from the neutral axis.across the width t of the member.MOST ECONOMIC SECTIONSV= the shear forceI = the moment of inertia of the ENTIRE cross-sectional area calculated about the neutral axis.When a structural section is selected to be used as abeam, the section modulus must be equal to or greaterthan the section modulus determined by the flexureequation;t = the width of the member's cross section measuredat the point where the shear stress is to be determined.|M\maxOallowThis equation indicates that the section modulus of theselected beam must be equal to or greater than the ratioof the bending moment to the working stress.A design engineer is often required to select the lighteststandard structural shape (such as a W-shape) that cancarry a given loading in addition to the weight of thebeam. Following is an outline of the selection process:1. Determine the maximum shear and moment ofthe (loaded) beam. A V-M diagram is the fastestway to do it. Neglect beam weight at this point.2. Determine the minimum allowable sectionQ = 2ÿ'\A'j + y½A'½modulus, Smin3. Choose the lightest shape from the list ofstructural shapes where S> Smin4. Calculate the maximum bending stress in theselected beam caused by the loading PLUS thebeam self weight. If ơmax < Oallow then theselection is finished. If it is not satisfied, choosethe second lightest shape.5. Repeat the selection process until a satisfactory,economic shape is found.Q = j'A'Q = j'A'Q = j' A', where A' is the area of the top (or bottom)portion of the member's cross-section, above (or below)the section plane where t is measured and y' is thedistance from the neutral axis to the centroid of A'. orA'= Area of the section above/below t.j' = Distance of the centroid of A' to Neutral Axis(N.A.)4rE(А)(Location of Neutral Axis) = y =ΣΑ(I about the Neutral Axis) I = Iy'+ Ad,?

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