Category: Shear for steel beams

All posts include shear data for steel beams. Based on the 2010 and 2016 codes, how can we get the shear strength for a given W section?

  • 20S- A guide to solved problem 2-22 for shear stress-part-A

    20S- A guide to solved problem 2-22 for shear stress-part-A

    A Solved Problem 2-22 For Shear Stress Part-A.

    The next slide shows the shear stress distribution for a rectangular section. The shear stress formula can be written as τ max =V*Q/Ixx*b, where V is the statical first moment of area about the neutral axis for the portion enclosed by the upper edge and the neutral axis. Q is the acting force.

    The Ix is the moment of inertia of the section at the x-axis, and b is the breadth of the section.

    Derive the expression for the shear stress for a rectangular section.

    We know the expression of the moment of inertia about the x-axis can be written as Ix= b*h^3/12, where b is the breadth of the rectangle while h is the height of the section.

    The first moment of area is the area enclosed between the NA and the upper edge multiplied by the distance from Cg to the N.A. The enclosed area is (b*h/2), and the distance from CG is h/4.

    The statical first moment of area =1*0.50(b*h)*(h/4)=b*h^2/8. The maximum shear stress at the neutral axis of the rectangular section can be estimated as the applied force multiplied by Q divided by the product of ix by b; the expression can be simplified to be τ max =Q*(b*h^2/8)/(b*h^3/12)*b)=(3/2)*Q/A. A is the area of the rectangle.

    What is the shear stress for a rectangle?


    How do you get the value for the first moment of area for the I section?

    If we use the expression V as an indication for the first moment, the portion included between the upper edge and the neutral axis is the T-section or half of the complete I section.

    We can find that the first moment of the area at the top of the flange=0; we can derive the expression for the first moment of the area at the bottom of the flange as shown in the next slide image.

    The first moment of area for I section at the bottom         of flange.

    Then we have two items. The first moment of area for the I section consists of the first moment of area for Flange, and we will add the first moment of the web at the N. A.

    The final first moment of area value equals the area of flange*y bar=bf*tf* (d/2-tf/2). The term d is the overall height of the I section, while the bf is the flange breadth.

    The first moment of area for the V web at the N>A will be (1/2*hw*tw)*hw/4=1/8*(b*h2w). By adding these values together, we can get the final first moment of area for the I beam. Ix for the same section is shown on the next slide.

    The first moment of area  for I section.

    Two methods can be used to find the moment of inertia Ix for the I section. Please refer to the next slide image for more details.

    The first expression for Ix for I section.

    This is the second method for expressing the moment of inertia for Ix.

    The second expression for Ix for I section.

    The next image will show the shear stress shape for an I section.

    Shear stress shape for an I section.

    Part A of the solved problem 2.22 for shear stress.

    We have solved problem 2.22, for which the maximum shearing stress for the following sections when the external shear force V=75.0 kips must be estimated.

    For part A, when the shear force is given as V=75 kips for a rectangular section, the expression for the maximum shear stress can be written as τ max =3/2*(Q/A).

    Q is the applied force, and A is the area of the rectangle. The rectangle has a dimension of (4 inches) by (12 inches). The area will be equal to (4*12)=48 inch2.

    We will substitute as τ max =3/2*(75/(48)=2.344 ksi

    In the next post, we will solve the same problem for the other shapes.

    A solved problem 2-22 for shear stress-part A.
    A solved problem 2-22 for shear stress-part A.

    This is the next post, A-solved problem- 2-22 for shear stress.
    For a good external site to estimate the shear stress for various sections.

  • 19S- Review of shear stresses for steel beams.

    19S- Review of shear stresses for steel beams.

    Review Of Shear Stresses for steel beams.

    Introduction to shear stress for beams and proof that both vertical and horizontal shear stresses are equal.

    If we take two sections at a beam apart by a distance, at elevation =y, there will be two compression forces due to M and M+dM due to the moment difference.

    The stress developed from the known formula f=M*y/I will be multiplied by dA, creating two compressive forces. These forces are not equal due to the current difference, which is=dM. To create a balance, a shear force acting horizontally will be created.

    As the sum of Fx =0, the shear force is developed to create a balance; this force=shear stress* area=τ*dx*b=σ’*dA-σ*dA, but (σ’-σ)*dA=(dM)*y/Ix, then τ*dx*t=dM*y/I.

    How do we derive the expression for shear stress?

    Making integration, we get an expression for the stress for shear τ=∫dMydA/(dx*t)=V*Q/I*T, where V is the shear.

    Q is the first moment of the area at the point of interest, I is the moment of inertia for the section, and finally,t is the breadth of the section.

    Derivation of shear formula.

    Horizontal shear stress for beams. 

    For the balance of a beam element, the horizontal stress of shear should be accompanied by vertical stress due to shear, as shown in the next slide.

    Derivation of shear formula.

    It is proven that the value of τh=τv, where τv is the vertical stress due to shear, while τh is the horizontal stress due to shear by multiplying by area and taking a moment at the edge.

    The proof of the relationship between vertical and horizontal stress for a beam.

    The following post will contain a solved problem for stress estimation for different shapes.
    For more detailed shear in bending, please read link-Shear in Bending
    This is a link for the next post, A solved problem for shear stress.