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Stress describes how a body responds to an external force. Or, i.e., stress quantifies the internal force that resists the applied force to maintain a state of equilibrium. So stress, as a vector quantity, should point in the opposite direction to any externally applied force.

In the drawing below, if my hand applies an external uniform force pointing into the wall (or pushes against area A of the rod), the rod would compress and store potential energy to bounce back to its uncompressed state if my hand is off the rod. Therefore, the direction of stress intrinsic to the rod itself should point in the opposite direction, or out of the rod, under compression.

In textbooks or tutorials I found they always portray the directional arrows of "negative compressive stress" to be pointing into the rod), and "positive tensile stress" pointing out of the rod, which make absolutely no sense to me. To cause compression, external force must points into the rod which results in a stress pointing out of the rod. Vice versa for tensile force, if it pulls or stretches the rod, the force will induce a stress pointing into (the center of) the rod.

[Edit]
Let's ignore gravitational force here as that will bend the rod causing both tensile and compression stress and unnecessarily complicate the concept here.

stress v force textbook

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  • $\begingroup$ what do you basically want to know ? $\endgroup$
    – Ankit
    Oct 21, 2020 at 10:45
  • $\begingroup$ @Ankit I want to know why the stress arrows shown in textbooks point in the same direction as the force externally applies as seen in the image attached here. The object elongates or stretches if an external force or pressure is applied away from either end of the object (or pulling it to stretch). Internal tensile stress should then point inwards as each atomic bond is trying to pull itself back to its originate state like a spring. So for tensile stress, the arrow should point into the object, whereas for compressive stress, outwards instead. $\endgroup$
    – KMC
    Oct 21, 2020 at 12:26

4 Answers 4

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Your diagrams are showing the directions of the external applied compressive and tensile stresses. In order to show the directions of the resulting internal compressive and tensile stresses you need to draw free body diagrams of cut sections of the beams. Those diagrams will slow the direction of the internal stresses as opposite to the direction external stresses as you say they should be.

Hope this helps

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  • $\begingroup$ That's is what makes the concept confusing or contradictory. One CANNOT apply stress - you can only apply force. Stress is internal - if the rod is made up of two atom then the stress is the spring or hooke force countering the applied force. If spring force goes in opposite to applied force, so does stress. Then the arrows drawn in the many textbooks are simply incorrect $\endgroup$
    – KMC
    Oct 21, 2020 at 5:57
  • $\begingroup$ One applies a force F and assuming it is uniformly distributed over an area A then the external applied stress is F/A. If you don’t like the term external stress call it external pressure. My point is the internal stress is, in my experience, always shown in a FBD of a cut section of the beam in equilibrium. That’s the way it is normally done $\endgroup$
    – Bob D
    Oct 21, 2020 at 6:12
  • $\begingroup$ The external applied pressure (not stress) is F/A. Pressure is a scalar hence I prefer relating force to stress here. Stress is essentially a spring force - it is a property of the material. If I compress a spring, the spring will exert a force equal and opposite to applied compressive force. Same goes for stress! If I draw a FBD according to modern textbooks, then all forces will not sum to zero. I'm questioning if we have been taught the wrong concept all along. $\endgroup$
    – KMC
    Oct 21, 2020 at 6:21
  • $\begingroup$ Technically stress is neither a vector nor scalar but a tensor. The external applied stress is usually shown as a load P (mechanical engineering term for force) perpendicular to an area A. The sigma symbol is to indicate P/A or F/A. So fine, use F or P. But to say stress is “a property of the material” is flat out wrong. The modulus of elasticity, E, is a property of a material in the linear elastic region. I don’t know what you’re talking about when you say the forces don’t sum to zero. The last statement is just silly. This discussion is going nowhere so I’m done $\endgroup$
    – Bob D
    Oct 21, 2020 at 6:47
  • $\begingroup$ I believed you've misunderstood my question. If you compress a spring to the left, the spring exerts a force to the right and bounces to the right if it's set free. One goes left, the other goes right, and the two forces sum to zero. That's the concept. Same if you compress a rod to the left, the rod will experience a stress to the right - out of the rod, not into the rod like every textbook says. Tell me, if the applied force and the compressed force both point to the left, how could that sums to zero? The arrow is drawn incorrectly for the axial direction of stress $\endgroup$
    – KMC
    Oct 21, 2020 at 8:19
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Stress is actually a 2nd order tensor, which, by its very nature is bi-directional. The usual sign convention is that compressive stress is negative and tensile stress is positive. If you want a more precise description of how this all works, Google "Cauchy stress relationship," which provides the mathematical framework for mapping the stress tensor into the traction vector on a surface of arbitrary orientation within a material (or at its surface).

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There is actually a very simple reason that compressive stress components are negative and tensile stress components are positive. For an elastic material the stress is proportional to the strain. This is Hooke’s law. A negative strain is the object getting smaller which happens in compression. So compressive stress components must also be negative so that the stress can be proportional to the strain

Now, regarding the direction of the force. It is important to understand that stress is not a vector, it is a tensor. A tensor is an object that if you give it a vector it returns another vector. In the case of stress, you give it a vector representing a small area and it returns the force on that area. The direction of the area vector is normal to the area and the length of the vector is the size of the area.

Now, at a given point in the object you can look at forces on areas pointing in any direction, but it is sufficient to consider the forces on the six faces of an infinitesimal cube. The thing is that each face of the cube has two sides, the inside and the outside. In principle, either could be chosen as the direction of the area vector.

Making tensile stress components positive corresponds to choosing the outside face as the direction of the area vector. With that, a positive stress component corresponds to an outward force which causes a positive strain. This is the force acting on the infinitesimal cube, not the 3rd law reaction force that the cube exerts on the rest of the material. It is a force internal to the object, but it is the force on the infinitesimal cube, not the reaction force as you described.

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Thats simply the way compressive and tensile stress is defined. Stress isn't strictly a vector, having the same dimentions as pressure. So 'stress produced when the body is compressed' is called compressive stress, and 'stress produced when the body is stretched' is called tensile stress, even though as you said, the actual reaction forces are opposite.

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  • $\begingroup$ So you're implying that the arrow denoting the direction of stress pointing into the object under compression is just an engineering convention. When the object is under tensile stress, the arrows are drawn to point out the direction of elongation or direction of external force, even though the reaction force per unit of cross-section area (i.e. stress) is pointing in the opposite direction. That's a misleading convention .... $\endgroup$
    – KMC
    Oct 21, 2020 at 9:55
  • $\begingroup$ I think this will answer your question: physics.stackexchange.com/questions/499018/…. $\endgroup$
    – dnaik
    Oct 21, 2020 at 12:24
  • $\begingroup$ The reaction force is pointing in the opposite direction, not the stress. The direction of stress is taken as the direction of deformation by convention. $\endgroup$
    – dnaik
    Oct 21, 2020 at 12:29

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