The maximum shear stress (MPa) in the stepped beam shown below is most nearly: 4

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Explanation / Answer

Unable to put the diagram, try to understand the solution. you willget it. at B consider the equilibrium of AB, M=0 =>MB= 50*1.5= 75 kN-m {in clockwise sense at B}         and vertical shear forcedeveloped in AB= 50 kN consider the equilibrium of AC, M=0 => MC=50*1.5 + 50*3 - 75 = 150kN-m{in clockwise sense at C}       and vertical shear force developedin BC= 100 kN consider the quilibrium of AD, M= 0 => MD=50*1.5 + 50*3 + 50*4.5 - 75 -150 =225 kN-m{in clockwise sense}      and vertical shear force developed in CD=150 kN Now, for the section AB maximum moment will act at B, againat C and D respectively so let us calculate axial and shear stress at each of thispoint, B= 50*103/(10*12*10-4) =4.16 MPa cross sectional moment of inertia about centroid =(12*103/12) cm4= 1* 10-5m4 {I= wh3/12} so, B= 75*103*5* 10-2/(1*10-5) section= 375 MPa { maximum shear stress at B =((/2)2+2 ) = 187.54MPa It is coming irrelevent with the option given, something wrong withthe data. But, the procedure will remain same. calculate maximum shear at C and D and then finally whichever isthe maximum will be the answer. calculating shear stress developed in part BC=100*103/(20*12*10-4)= 4.16 MPa                shear stress developed in part CD=150*103/(30*12*10-4)= 4.16 MPa or, may be it is asking about the maximum direct shear acting alongvertical plane, not about plane at any angle. in that case theanswer will be (a) 4.16 MPa which is coming same everywhere.

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