# Stress in a perfect bar

## Main Question or Discussion Point

I understand that if you glue together two pieces of a bar together at an incline, and apply an axial tensile force (normal) to the bar, that as far as the layer of glue is concerned, the force applied to it is partly normal (normal to the surface of the glue layer) and shear (perpendicular to the normal force).

Okay, now let's just consider a bar that is perfectly homogeneous and prismatic. Are there any shear forces within the bar? I know bars often fracture at an angle, that is, not exactly perpendicular to the length of the bar, but in a perfect bar there should be no shear forces inside the bar, at least that is what I understand.

But I read "To obtain a complete picture of the stresses in a bar, we must consider the stresses acting on an "inclined" (as opposed to a "normal") section of the bar". That sort of seems to hint that even in a perfect bar there are shear forces. But, that seems contrary to what I've previously believed.

So, are there only normal stresses in a bar that is only subject to normal stress? Just a plain straight bar. Not glued together or whatever. Thanks.

P.S. Only if there is some defect in a bar, can I imagine a shearing stress. in a straight bar entirely under uniaxial normal stress.

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PhanthomJay
Homework Helper
Gold Member
Under axial loads only, there are still shear stresses on cross section planes that are inclined to the longitudinal axis. That is to say, the resultant internal load on this section which in the longitudinal axial direction can be broken up into its components normal and parallel to that inclined section. The parallel component causes the shear stress. So this is true of the bar whether it is glued at an incline or not, but it needs a close look with glues on inclines, some of which can have rather low shear strengths,.

Another way to get at what I'm asking is. Let's say that a uniaxial force of 100 MPa normal stress was applied to a homogeneous and prismatic round bar.

Say the bar will yield with a normal stress at 200 Mpa. Also the bar will begin to yield at a shear stress of 50 Mpa.

As far as I understand it, the bar will take the 100 MPa normal stress. It won't yield at 50 MPa along an incline. Because no shear stresses will be set up in the bar.

But if I were to cut the bar at an incline, put some glue in, put the bar back together, the bar will yield at 50 MPa, if the glue has a 50 MPa stress limit. The glue having a critical stress limit of 50 MPa yeilds, but the rest of the metal, with the same critical shear stress limit will not yield. My understanding.

P.S. If you chose a piece of metal bar with a tensile strength of 200 Mpa, I presume that means the bar will take a normal stress of 200 MPa, that it won't yield at 50 MPa - by way of shear stress, when put under tensile stress. Unless yield stress results are not a strictly a matter of normal stress. I am assuming a tensile test produces a stress yield that is the result of normal cleavage fracture - well, at any rate, not a sliding fracture (shear). Although really I'm supposed to be dealing with things occurring in the elastic zone.

.I'm actually trying to stay within happenings occurring within an elastic limit.

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Looks like if you take a bar and subject it to purely tensile force, shearing takes place and thus a shearing force develops in the bar. . Involving a point of maximum shear stress. But, I don't understand why. I only get it, I only understand how you get shear, if you cut an incline in the bar and glue it back together. A tensile force acting at normal, should only produce elongation at all points. In the elastic zone.

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Chestermiller
Mentor
I agree with PhanthomJay. In the case that you are considering, the axial tensile stress is one of the three principal stresses, and there is no shear stress on planes normal to the three principal directions of stresses. When you said
But, that seems contrary to what I've previously believed.
that belief was incorrect. The way the stress tensor works is that you choose a plane of any arbitrary orientation within the material, and take the dot product of the stress tensor with the normal to that plane. This gives you the stress vector (with both normal and shear components) acting on that plane. Unless the plane is oriented perpendicular to one of the principal directions of stress, you will calculate a shear stress on that plane.

Are you saying that in a uncut perfect bar, if I chose to orientate the infinite element at 45 degrees, that shows there is a shear force in the bar acting at 45 degrees (τ) in my diagram? And that a normal stress is at 45 degrees to the face of that element?

I must admit, I don't understand why there would be a shear stress in an uncut bar. I can imagine the stress as fictitious, in the sense of if you cut the bar at 45 degrees there would be a sliding force generated, but if the bar is not actually cut, I cannot see why there would be a sliding force in the bar.. Although we do have the Poisson Effect.

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Chestermiller
Mentor
Are you saying that in a uncut perfect bar, if I chose to orientate the infinite element at 45 degrees, that shows there is a shear force in the bar acting at 45 degrees (τ) in my diagram? And that a normal stress is at 45 degrees to the face of that element?

I must admit, I don't understand why there would be a shear stress in an uncut bar. I can imagine the stress as fictitious, in the sense of if you cut the bar at 45 degrees there would be a sliding force generated, but if the bar is not actually cut, I cannot see why there would be a sliding force in the bar.. Although we do have the Poisson Effect.
Yes. The picture is correct, except that the two $\sigma_n$'s are not the same. The stress vector on an internal element of surface oriented in an arbitrary direction is not fictitious. It represents the force exerted by the material on one side of the surface on the material on the other side of the surface. Have you ever learned about the Cauchy stress relationship, or have you ever learned about Mohr circles? The derivation of the equation for a stress vector on an arbitrary element of internal surface within a material in terms of the principal stresses (or the components of the stress tensor expressed with respect to any specified set of coordinate axes) is in every book on Strength of Materials or Solid Mechanics.

Of course, my questions are not principally an issue of mathematics. Math is really little to do with answering my questions.. I'm posing a question of what is real and what is imagined. If the applied force is normal to the bar and the bar is not sliced at an angle, is their really a force of strain acting on an element in the bar, as a shear force? (tangentially to a face of the element). I can see that if the bar is cut at an angle, the two halves will want to slide over each other. Fine, that's showing the existence of a shearing force. But in an uncut bar does an element within the bar really experience a force of shear ? This does not principally call for a math answer. It's simpler than that. If the answer is yes, there is a real force of shear in the bar, you can then go on to express mathematically how to compute that force. But, the first thing is to say whether there is a real shearing force on an element, in an uncut bar. This has not been answered yet in terms of just saying yes there is a real shearing force acing on an element in an uncut bar or, no, a shearing force only manifests itself when the bar is cut. I'm still unclear what the answer is.

Can anyone else please chime in and not focus on math as a means of seeking to answer my question. I think the answer to what I'm asking is best in the yes/no form. And then if necessary backed up with math, if the math explains why that shearing force in a uncut bar exists. At the moment I cannot see how shearing stress arises on an infinite element. Why should we see such a shearing stress if I draw the element at an angle on a piece of paper? Of course, messing about with an element on paper has nothing to do with what stresses are really within a bar, it's just messing around on paper. Stresses that really exist depend on what is really the case per the bar. And in this matter, the uncut bar is under normal stress at it's end, with no other stresses applied. I draw an element at 36 degrees. So what? Does it mean I can use my little drawing to show a shear stress in an uncut bar, or is it only good if the bar were to be actually cut at 36 degrees?

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Chestermiller
Mentor
OK. I see that you want yes-no, so here goes.

Suppose that, rather than having an actual cut, you just have a surface of interest within the bar, and the surface is perpendicular to the axis of the bar (rather than being at some other angle). Do you accept the idea that there is a real force exerted by the material to the right on the material to the left, equal to the tensile force T you apply at the left end of the bar? Do you accept the idea that this force T exerted by the material on the right is equal to the tensile stress times the cross sectional area of the bar, and the stress vector on the surface is strictly normal to the surface, so that there is no shear component?

After you respond to these yes-no questions, we can continue.

Chet