Question

Problem 12.34 Consider the frame shown in (Figure 1). The axial stress in a column is proportional to its distance from the centroid of the cross-sectional areas of the colum given floor level. In the special case of the columns having equal cross-sectional areas, the force in a column is also proportional to its distance from the centroid of the column areas. At a given floor level the shear at the interior column hinges is twice that at the exterior column hinges, since the frame considered to be a superposition of portals A hinge is placed at the center of each girder, since this is assumed to be a point of zero moment. Preview Answers Correct Part B Use the portal modo is and draw the moment diagram forgde FED Click on 'add vertical line off to add discontinuity lines. Then click on 'add segment" button to add functions between the Wines Note 1 - Make sure you place only one vertical line places that require a vertical line. If you inadvertently place we vertical lines at the same place, it will appear correct visually because the lines overlap, but the system will make it wrong Note 2 - You should not want extra vertical line at the point where the curve passes the axis Note 3. Be sure to indicate the correct types of the function between the lines in your answer the type of a function is near increasing slope for the function that actually has linear decreasing the answer will be graded as incorrect. Use the button change segment necessary 10k F E -20 ft- -20 ft Segment Attributes Segmenttype Increasing lines SO Figure 10k Lett points Rightpoint 30 . 2010 200 TU

Answer 1

# Beam Shears: SOF for end beams EF and ED; Let V x 10 = 37.5 k-fet 3.75k оо Shear force for Beam EF and CD = Axial force for each end colouma Axial force for each end calumn 3.75k. Axicel force for intermediate caloumn (BE) = 0 37.5k-ft 37.5k-ft F 37.5k ft E 75kft 1D 37.5k-fut 37.5 37.SKH A 3705 Kft B 75K-H 37-5kft moment diagram only for girder FED +37-5k-fa +37·5k-ful F - 37.5 Kft -37-5k-ft Answer.