H2 kNK2 kNMNwwwRAYRAXAGBCDEF3 m↑RGY4mFill in the multiple blanks.Figure Q19Finding the vertical reactions is the starting point which can be done by taking moments at A and G but since this is symmetrical loading case the vertical reactions can simply be calculated by halving the total loading 4 kN.Ideally, we can solve the problem using the Method ofcutting through the members JK, DJ andIt would be sensible to select the left-hand side of the diagram as there are less full members and only one force from the reaction at node A.This will expose the internal forces which can be labelled with the names of the members themselves.Since we are required to find JK, examining the framework shows it is not a straight-forward matter, and we will require finding all three unknown internal forces.The easiest internal force to find isNext, we can take moments at node, as we can resolve forces in the vertical direction.in order to find the internal force JK and find that it has value ofKN (use the convention that compressive internal forces are negative values).

Method of: The correct answer is sections, as the problem involves using the Method of Sections to…

Answered: H 2 kN K 2 kN M N www RAY RAX A G B C D…

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter9: Deflections Of Beams

Section: Chapter Questions

Problem 9.5.16P: Repeat Problem 9,5-15 for the anti-symmetric loading shown in the figure.

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The inclined beam represents a ladder with the Following applied loads: the weight (W) of the house painter and the distributed weight (u) of the ladder itself. Find support reactions at A and B: then plot axial force (N), shear (V), and moment (M) diagrams. Label all critical N, V, and M values and also the distance to points where any critical ordmates are zero. Plot N, V, and M diagrams normal to the inclined ladder. Repeat part (a) for the case of the ladder suspended from a pin at B and traveling on a roller support perpendicular to the floor at A.
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