You are watching: Determine the maximum tensile and compressive bending stress in the beam

EQUATIONSUSEDSOLUTIONUsing line b-b as referral, with positive direction downward, the y centroidal coordinate is established asBecause the beam is elastic, the netrual axis will certainly pass via itscross-sectional centroid.The minute that pressure P produces is around the horizontal axis, therefore,neutral axis corresponds via the horizontal centroidal axis in this problemas presented in the number listed below.The used load puts the percent above the NAin stress and anxiety and the one below in compression.The moment of inertia around the horizontal centroidal axis is the only onethat is required to be calculated right here, and is established as(a) The maximum compressive stress and anxiety will certainly occur at the farthest allude from the NA,on the compression side. At section n-n the bending moment isThe equivalent compressive stress and anxiety will be(b) The maximum tensile stress will occur at the farthest suggest from the NA on the stress side.(c) As was explained previously, the shear stress must be regarded as the ratio (V/I) times the ratio (Q/t). At a offered section alengthy the length of thebeam (V/I) is continuous. This suggests that the maximum shear anxiety will happen atthe point on the cross section which results in the biggest (Q/t) value.Without any type of exception, the initially moment of location, Q is constantly maximum at the NA. Because in this cross area the width t is likewise minimum at the NA, we deserve to conclude that the maximum shear tension will certainly happen tbelow.

**Side Note:**Based on the above conversation, it is rather feasible for the maximum shear stressto take place at a allude other than the NA. You should always be mindful of this factas soon as analyzing a beam under transverse shear loads.To determine Q, cut the cross area horizontally at the NA and also pick either the height or the bottom percent,it doesn"t matter which because the same answer will certainly result utilizing either percentage. Because of this, it makes sense to pick the percentage via thesimplest shape to calculate Q, in this case the bottom percentage.The maximum shear tension on the transverse plane n-n is uncovered to beThe shear tension circulation over the whole cross area was not asked forin the trouble statement, but it is presented below for completeness.

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Notice that the shear tension distribution mirrors only one component of sheartension (i.e., the vertical component). While this component is the leading one along the vertical flange, itis the smallest one along the horizontal flanges. To identify the dominantcomponent of shear tension alengthy the height flanges (i.e., horizontal component), we must make a vertical cutalengthy either the left or the ideal flange (as a result of symmetry) and calculate Q/talong the flange. The variation of shear anxiety will be direct via thelargest flange shear stress and anxiety occuring near the junction of the 3 flanges. We will certainly see this difference in later instance difficulties.However before, inthis trouble the all at once maximum shear tension at area n-n is the one discovered earlier.It is also necessary to allude out a pair of points about the shear stressdistribution presented. First, notice that the variation of shear tension isquadratic. This is governed solely by the equation for Q. 2nd, notification thediscontinuity at the junction. This is bereason as we go from height to bottomsurconfront, the thickness all of a sudden transforms at the junction. The sharp reduction inthickness is responsible for the sharp rise in the shear anxiety. To Section III.3 To Index Page of Transverse Shear Loading of Open Sections