Nội dung text 14 Shear and Bending Stress.pdf
PSAD 13: Shear and Bending Stress 1. Bending/Flexural Stress When a member such as a beam bends, fibers along its longitudinal plane experiences flexural stress. This stress can be calculated using the flexure formula: fb = Mc I (Eq. 13 − 1) fb = bending stress M = bending moment at section c = distance of fiber with respect to the neutral axis I = moment of inertia about neutral axis The neutral plane/axis is where the centroid of the member’s cross-section is found, and where the bending moment acts about. Equation 13-1 shows a linear relationship between distance from neutral axis, c, and bending stress, f. This implies that no bending stress occurs at the neutral plane (c = 0), and the farthest fiber (c = cmax) is most stressed. For a positive moment, fibers above the neutral axis contract causing them to experience compression, while fibers below the neutral axis are stretched longer due to tension. This behavior influences structural design, such as how to reinforce concrete with steel in consideration of concrete’s weak resistance to tension. 1.1 Computation Procedure for Bending/Flexural Stress The general procedure for solving the bending stress of a member’s section is provided below: Obtain the internal moment Locate which section’s internal moment is being asked. Typically, the section with the greatest moment is needed. This can be done by cutting the member to ‘release’ its internal reactions or through a shear and moment diagram. neutral plane neutral axis Positive Bending Negative Bending Section considered
Compute the moment of inertia The flexure formula requires the section’s moment of inertia, I, about the neutral axis. Great care must be taken for this step, especially for those with unusual geometry and for tapering members. See note below table for choosing the correct MOI. Identify location of fiber Usually, the outermost fiber is considered because it experiences the largest stress. However, for sections whose extreme fibers (top and bottom) are not equally distant from the neutral axis, it is crucial to know if the most stressed fiber must be in tension or compression. Choosing the correct moment of inertia is about imagining which axis of the cross-section moved or “translated” due to the applied force that caused the bending. This guide is particularly useful when solving combined stresses such as in the case of purlins that usually bend about both axes. The x-axis translated downwards due to the applied load USE MOI AT X-AXIS The y-axis translated to the right due to the applied load USE MOI AT Y-AXIS
1.2 Combined Axial and Bending Stress When an axial load is applied at a member with an eccentricity from the centroid, it causes both axial and bending stress. The general formula is given below. The positive and negative signs indicate tension (+) or compression (-). σ = ± P A ± Mc I (Eq. 13 − 2) At most times, the axial load is not only eccentric with respect to a one axis, but both axes of the cross-section. This condition is known as bi-axial bending. To demonstrate how bi-axial bending affects the stresses of a member’s components, consider a column loaded shown below: In statics, by transferring the location of the load to the centroid, an axial load is created, and moments are produced to induce the same stresses in the column as with the load in its original position. Mx = Pey My = Pex With the direction of the moment about the x-axis, Mx, all fibers at +y experience compression, while those at -y experience tension. As such, due to the direction of the moment about the y-axis, My, all fibers at +x experiences compression, while those at -x experience tension. The direction of the axial load causes compression over the entire area. Therefore, when solving for the stresses at the points specified in the image, the computation is as follows: σ1 = − P A − Mx (d/2) Ix + My(b/2) Iy σ2 = − P A − Mx (d/2) Ix − My(b/2) Iy σ3 = − P A + Mx (d/2) Ix − My(b/2) Iy σ4 = − P A + Mx (d/2) Ix + My(b/2) Iy
2. Flexural Shear Stress For the following cases: (a) load applied under/near the support of a beam, (b) load at the joint of two firmly glued members, or (c) shearing force experienced by bolts at member connections, using the average shear stress formula (τ = V A ) is sufficient. However, when the applied load causes significant bending, the stress caused by the transverse shear at the surface, which is numerically equal to the longitudinal shear, is calculated using the equation below. fv = VQ Ib (Eq. 13 − 3) fv = transverse/flexural shear stress V = internal shear force at section considered Q = moment of area about the neutral axis I = moment of inertia about neutral axis b = width of fiber considered Equation 13-3 is also referred to as the shear or flexural shear stress formula because it was derived using the flexure formula (Eq. 13-1). 2.1 Computation Procedure for Flexural Shearing Stress The general procedure for the flexural shearing stress is similar to the steps undertaken for the bending stress computation: (1) obtain internal shear, (2) compute the moment of inertia, and (3) identify location of fiber. When choosing the fiber to consider take note of the following: For uniform and symmetric cross- sectional areas, the maximum stress is located at the neutral axis. Example of other sections where this applies is a circle and a hollow circle. Longitudinal shear Transverse shear (c) (a) (b)
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