# Design procedure used in beam design

In the design of beams, the principal stresses and absolute maximum shear stresses are not calculated. Instead, we calculate maximum bending stress using flexure formula and equate it to allowable stress to get dimensions of cross-section and then using these dimensions we calculate maximum shear stress on cross-section. If maximum shear stress on cross-section comes out to be less than allowable stress, then design is safe. But there is a possibility that maximum shear stress in the beam (whose location is at the top and bottom of cross-section; Absolute max shear stress= (max. bending stress)/2) )exceeds allowable shear stress, then the beam will fail. So, why do we use this design procedure if this gives unsafe design?

• Generally beams are selected from standard tables for the requirements of the application. Trusses made from beams would need to be designed. Nov 9 '21 at 17:01

The design process isn't exactly as you describe it. More specifically you write:

But there is a possibility that maximum shear stress in the beam (whose location is at the top and bottom of cross-section; Absolute max shear stress= (max. bending stress)/2) )exceeds allowable shear stress, then the beam will fail.

(Although it isn't very clear at first glance) it shows that there is an understanding of Mohr's circle for a simple loading case (I am not considering biaxial or complex loading) and that the max shear stress will always be smaller than the maximum principal stress.

In the case of the outer layers, like you said there are only normal stresses, and therefore the max shear stress is half the maximum normal stress.

However, regarding the allowable shear stress it seems as though you are making the assumption that is independent from the maximum tensile stress. In the general anisotropic case, that might be true, however when common isotropic engineering materials are considered then there is a relationship between the shear and tensile values. Depending on which yield failure theory you are considering the allowable shear $$\tau_{all, s}$$ is at least 0.5 (or 0.57) times the allowable tensile stress ($$\sigma_{all}$$.

Please note that, as allowable value I am considering the yield values, in order to limit the deformation to the elastic range.

• Regarding your second point, absolute maximum shear stresses will still occur on the outer edges (since we already had a conversation where you supported this claim by saying that its on the edges basically where the yield begins in a ductile material) but there is no need to include the value of this absolute maximum shear stress within any equation to size the beam, since its effect is already taken into account by the principal stresses in the Von-Mises equation. Nov 9 '21 at 16:52
• That only occurs for torsion though, and this question is about bending. Nov 9 '21 at 16:54
• I meant the absolute maximum shear stresses coming from the Mohr's cirlce for an element (ofcourse, at a different orientation), which are bound to occur at the outer surfaces of a beam subjected to bending, since that is where the bending stresses are the maximum. Nov 9 '21 at 17:05
• You are right. I did not read properly the question. I will either modify it or delete all together when I get home. Nov 9 '21 at 18:05

We don't use the maximum bending stress and equate it to allowable stress, and we do use the principal stresses.

We calculate the Von-Mises stress (if the material is ductile) and equate it to the allowable stress. Von-Mises (or equivalent stress) consists of both the components i.e. axial (coming from bending) and shear. The formula below shows the calculation of Von-Mises stress.

As it can clearly be observed that it includes the contribution from the shear stress as well, since principal stresses depend on the value of shear stress when an element within a body is subjected to a combination of bending stress and shear stress.

Tresca theory can also be used to calculate the stress and equate it to the maximum allowable stress, and it also depends on the value of shear stress that the element is subjected to since it involves principal stresses within its formula as well. But Von-Mises is usually preferred since it agrees with the reality more than Tresca.

Failure usually refers to plastic failure, and the allowable stress is a fraction of the yield strength of the material. I have not heard if there is something like allowable shear stress which might cause a failure.

• But at the extreme ends at which bending stress is maximum shear stress is zero there, so how will the von mises stress include the effect of maximum shear stress on cross-section? I think you are calculating von mises stress considering critical point on the extreme end and there uniaxial state of stress is developed because only bending stress is present
– Max
Nov 9 '21 at 20:13
• @Max, well you are right. The stress state of the elements located at the top and bottom of the cross section of a beam shouldn't have a shear stress associated with it since it is equal to zero. Which means that the right hand side of the equation has influence only from max compressive or tensile stress (coming from bending). Well, in these cases, what could be done is that you can calculate the equivalent Von-Mises stress at some intervals of the cross section, starting from the middle and going to the top (and bottom of the cross section). Nov 10 '21 at 7:24
• In the middle, there is no bending stress and only max shear stress. On the top/bottom, there is no shear stress and only max bending stress. So the max Von-Mises stress should lie somewhere in between these limits. Making some intervals between these two limits allows you to calculate a range of Von-Mises stress from where the max can be chosen. Nov 10 '21 at 7:29
• Other way is to actual use the max shear stress (in the middle) and max bending stress (at the top/bottom), apply it on an element together and then create a Mohr's Circle from it, calculate the principal stresses from this stress state and find the Von-Mises, equate it to allowable stress and size the beam, but it would oversize it most probabaly since the max shear stress and max bending stress (in a cross section) won't occur on an element at the same time. Nov 10 '21 at 7:29

Because the allowable stress is always lower than the maximum shear stress possible.

For a rectangular beam, the maximum shear can be calculated as $$f_{v_{max}} = {2P_{act}}/{A}$$. Now let's set the maximum shear stress to its limit, $$f_{v_{max}} = {2P_{act}}/{A} \leq f_y$$, then $$P_{act} = 0.5f_yA$$. However, the "allowable shear stress" is limited to be $$f_a \leq 0.4f_y$$, then for the same cross-section $$P_a = f_aA = 0.4f_yA$$. So, compare $$P_{act}$$ and $$P_a$$, we conclude that the design is always safe as $$P_a$$ (allowable shear force) $$\lt P_{act}$$ (maximum shear force) with a 25% safety margin, as $$P_{act} = 1.25P_a$$.

The above proof can be held for the bending stress of rectangle beams using the generalized bending stress formula ($$f_b = P/A \pm M_xy/I_x \pm M_yy/I_y$$), however, for other shapes, it pays to check the principal stresses especially when the cross-section is subjected to complex loadings. In the US, AISC has made the check simple by providing necessary parameters about the principal axes.

The proof was based on the ASD (allowable strength design method), but the same holds for the USD (ultimate strength design method) and LSD (limit state design method).

Final note - the maximum shear occurs at the centroid of a cross-section, at which the static moment of area $$Q$$ is the maximum ($$\tau = VQ/Ib$$).

ADD - Figure below shows stresses in a beam of rectangular cross-section at various depths:

(a) show points in the cross-section.

(b) normal and shear stresses acting on horizontal and vertical planes at each point.

(c) principal stresses at each point.

(d) the maximum shear stresses at each point.

Ref: "Mechanics of Materials", Gere & Timoshenko, 2nd.