A Numerical Introduction to the Transverse Shear Stress Formula

Transcription

1 A Numerical Introduction to the Transverse Shear Stress Formula Brian Swartz 1 Abstract The transverse shear stress equation is particularly challenging for students to understand and fully comprehend. Careful consideration of the students learning process and the sequence of instruction which supports learning is critical on this topic. The paper details a numerical exercise and a simplified derivation that mirrors the initial exercise and an appropriate means of instruction for transverse shear stress. Keywords: Transverse Shear Stress BACKGROUND The equation for transverse shear stress (τ=vq/it) often poses significant challenges for sophomore mechanics students. Even students who learn to compute the correct numerical result struggle to fully understand the concept and become comfortable with the subject. It seems that a different, softer, sequence of the instruction might be necessary in this case. There exists a tendency among faculty and textbook authors to introduce topics through deeply theoretical derivations. Such is often the case for the shear stress formula. But students are more comfortable with numerical values and values give the students a better feel for the topic at hand. They tend to be more comfortable with concrete values than algebraic variables[1]. Furthermore, students must learn material contextually if for no other reason than it helps to trigger the brain s recall mechanism[2]. In other words, the human brain does not learn things in a vacuum. Instead it is attempting to connect new knowledge with previously held concepts. It is crucial that our teaching practices reflect the brain s own learning process. Therefore, it is important when introducing the concept of transverse shear stress that the proper context be established and that students have enough concrete information to be comfortable. The simple exercise described next provides a helpful primer for student learning in this subject. NUMERICAL EXERCISE A beam problem is devised here with an intentionally unusual cross-section for the purposes of the exercise. The beam (shown in Figure 1) is simply supported over 20 feet and loaded at midspan with 10 kips. The cross-section is 2 inches wide and 12 inches deep. 1 University of Hartford, 200 Bloomfield Avenue West Hartford, CT 06078, bswartz@hartford.edu

2 Figure 1. Span arrangement and cross section of the example beam. The problem statement asks students to sketch the stress state on each of the elements indicated in Figure 1. The elements are 2 inch cube-shaped pieces of the beam. A stack of six elements make up the full beam section. You could think of this as a very crude finite element model. Students should start with what they know about stresses in beams first. Presumably they have just been through a unit related to flexural stresses ( ). So they have the skills to determine the flexural stress distribution through the cross section and determine the normal stresses on each beam element. The choice of finite dimensions for the beam elements is very intentional, and an important first discovery should occur as students attempt to calculate the flexural stresses the moment is not the same on the two sides of the element (see Figure 2). Encourage them to plot the flexural stresses actually felt on the two sides of the element the differential is one of the biggest parts of the learning adventure. Figure 2. Shear and moment diagrams for the example beam problem.

3 Students should be able to produce the flexural stress distribution over the entire cross-section, but focus on the top element first. It will be helpful to think of this action both in terms of stress and also the resultant force of that stress. After calculating the stress distribution, students should calculate the magnitude and location of that resultant force. The results of the calculation are shown in Figure 3. Figure 3. Flexural stresses and resultant forces acting on the second element. Attentive mechanics students should soon recognize that the element diagram they have drawn fails a basic equilibrium check. That acknowledgement is, in fact, precisely the point of this exercise. Their observations from reality (a simple foam beam demo is helpful at this point) should reinforce the idea that each element does indeed satisfy equilibrium; the elements are not moving around in the beam. The students should start to think about what must be missing from the diagram drawn. So far they have considered only that adjacent elements push against each other and resist pulling away from one another. They should slowly rationalize that elements do not want to slide past adjacent elements either. They can hypothesize that the element below the current element resists the sliding that would occur if the free-body diagram in Figure 3 were accurate and determine the magnitude of force necessary at that interface to hold force equilibrium (shown in Figure 4). Dividing by the area of the face one can then back-calculate the shear stress present on that face. Figure 4. Flexural stresses and resultant force on the top element, in addition to the shear force and stress that must be developed on the lower face for force equilibrium. Students will likely dismiss the force diagram in Figure 4, assuming it satisfies equilibrium. But after some additional prodding and a few Are you sure? questions they will realize that it still fails rotational equilibrium. What holds the element in place and prevents it from rotating? Maybe a little brainstorming is in order and they eventually realize that the adjacent elements must be playing a role, also preventing a sliding action and building up a shear stress against this element. Summing moments about some point in the element allows calculation of the force and then stress which must be present on those vertical faces (see Figure 5). Note that the stress on the top face has to be zero simply because it is not touching anything.

4 Figure 5. Flexural and shear stresses and resultant forces on the top element When students advance to the second element (Figure 6) they should achieve another objective of this exercise. The equal and opposite stress from the bottom of element 1 is felt on the top of element 2. Furthermore, the flexural stress differential on element 2 causes additional shear stress. That is, the shear stress is accumulating as the point of interest moves deeper in the section (that is an important lesson understanding Q in the shear stress formula later). Figure 6. Stress and resultant forces on the first two elements Students can work through the remainder of the elements fairly easily as the pattern becomes apparent. The remaining key point to absorb is recognizing the change when passing through the neutral axis. When working from the top of the section towards the neutral axis shear stress is accumulating. But after passing through the neutral axis shear stress begins to dissipate until it reaches a zero value at the bottom face (see Figure 7).

5 Stress (KSI) Force (KIPS) A B " " " " " " Figure 7. Stress and resultant force on all elements in the example cross section. The numerical results of the example problem can be plotted to yield the general shape of the shear stress distribution, shown in Figure 8.

6 Figure 8. Shear stress distribution It is important to note that, as the size of the blocks gets smaller, the values for the shear stress on all four faces will approach the same value. Therefore, if a differential element is examined, it can be assumed that the shear stress is the same on all faces. Therefore, only the shear stress on one face needs to be calculated. THE SHEAR STRESS EQUATION: BASIC DERIVATION Now we set the problem up for a more generic case to derive an equation that can be used to calculate shear stress at any point. Consider all the stresses (and resulting forces) acting on the stack of elements from the previous example. Refer to Figure 9.

7 Figure 9. Summary of all resultant forces (flexural compression/tension and shear) on the series of elements from the previous example Now we will isolate one of those blocks. To make the derivation more general, one of the interior blocks will be considered as shown in Figure 10. M left M I y A M left I y A Figure 10. Generic beam element enduring bending and shear forces.

8 The force due to flexural stresses on the generic element can be found as the flexural stress multiplied by the area over which it acts (ΔA). The force due to flexural stresses on the opposite face is slightly different because the value of the applied moment changes by an increment (ΔM). The total length of the element is given by Δx, with a thickness, t. The total shear force is equal to the shear stress multiplied by the area (t*δa). As noted previously, once the shear stress (τ) on one face of the block has been found, all other values will be known if the element is infinitesimally small. Let s develop an equation for the shear stress on the bottom of the block. Recalling the accumulation of shear stress for elements deeper in the beam, and referring to Figure 10, it can be said that the shear stress on the top and bottom of the element are equal except for a small change in shear stress (Δτ) that occurs over the depth of the element: In terms of stress: Δ OR, in terms of force: Δ Δ ΔΔ Equation 1 Considering equilibrium in the generic element of Figure 10: From Equation 1, it can be seen that: Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Equation 2 ΔΔ Δ Δ Equation 3 Rearranging Equation 3 yields: Δ Δ Δ 1 Δ Equation 4 The slope of the moment diagram Δ is equal to the internal shear: Δ Δ Δ

9 Equation 4 can then be simplified to say: Δ Δ Equation 5 If the element is very small, it can be given differential dimensions and integrated to solve for the shear stress at any location, y 2 : If a variable Q is used to represent y 2 : Equation 6, then equation 6 can be rewritten for shear stress any location, Equation 7 In practice, Q can be determined as (Refer to Figure 11) Figure 11. Definition of variables used in the generic shear stress equation. CONCLUSION Student learning improves when concrete examples are used to introduce concepts and new ideas are clearly tied to previous learning. The approach demonstrated in this paper achieves both of those objectives in introducing the

10 concept of transverse shear stress and the general equation for calculating its value. This is a topic often difficult for students to grasp, so much care should be taken when introducing the subject. REFERENCES [1] R.M. Felder, Reaching the Second Tier: Learning and Teaching Styles in College Science Education. Journal of College Science Teaching, 23(5), (1993). [2] J.M. Haile, Toward Technical Understanding. (i) Part 1. Brain Structure and Function. Chem. Engr. Education, 31(3), (1997). (ii) Part 2. Elementary Levels. Chem. Engr. Education, 31(4), (1997). (iii) Part 3. Advanced Levels. Chem. Engr. Education, 32(1), (1998). Brian Swartz, PhD, PE Brian Swartz is Assistant Professor of Civil Engineering at University of Hartford. He holds a PhD in Civil Engineering from The Pennsylvania State University.