Shear Stress. Horizontal Shear in Beams. Average Shear Stress Across the Width. Maximum Transverse Shear Stress. = b h

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Georgina Rogers
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1 Shear Stre Due to the preence of the hear force in beam and the fact that t xy = t yx a horizontal hear force exit in the beam that tend to force the beam fiber to lide. Horizontal Shear in Beam The horizontal hear per unit length i given by q = VQ I where V = the hear force at that ection; Q = the firt moment of the portion of the area (above the horizontal line where the hear i being calculated) about the neutral axi; and I = moment of inertia of the cro-ectional area of the beam. The quantity q i alo known a the hear flow. Average Shear Stre Acro the Width Average hear tre acro the width i defined a t ave = VQ where t = width of the ection at that horizontal line. For a narrow rectangular beam with t = b apple h/4, the hear tre varie acro the width by le than 80% of t ave. Maximum Tranvere Shear Stre For a narrow rectangular ection we can work with the equation t = VQ to calculate hear tre at any vertical point in the cro ection. Hence, the hear tre at a ditance y from the neutral axi apple h Q = b y y + h/2 y = b h y 2 4

2 A = bh I = 1 12 bh3 t xy = t yx = VQ Ib = V b2 h 2 4 y bh3 b = 3V(h2 4y 2 ) 2bh 3 = 3V 2A 4y 2 1 h 2 OR t xy = t yx = V h 2I 2 y a parabolic ditribution of tre. Hence, the maximum tre in a rectangular beam ection i at y = 0 and t max = 3V 2A In cae of a wide flanged beam like the one hown here the maximum hear tre i at the web and can be approximated a t max = V A web Problem 1. (a) Uing the wooden T ection a hown below and ued in the previou clae find the maximum hear it can take where the nail have a capacity of 800 N againt hear load and the pacing between the nail i 50 mm. Uing the parallel axe theorem, I 1 = 1 12 bh3 + Ad 2 = 1 12 (0.1 m) (0.02 m)3 +(0.1 m) (0.02 m) (0.051 m) 2 = m 4 I 2 = 1 12 bh3 + Ad 2 = 1 12 (0.02 m) (0.15 m)3 +(0.02 m) (0.15 m) (0.034 m) 2 = m 4

Figure 65: Problem 1: cro-ection. @ @ Ą 9 " ' Hence, the moment of inertia of the T ection about the centroidal axi x 0 I = I 1 + I 2 = 14.

3 Figure 65: Problem Ą 9 " ' Hence, the moment of inertia of the T ection about the centroidal axi x 0 I = I 1 + I 2 = m 4 Figure 66: Problem 1: pacing of nail. The firt moment of the cro-ectional area i Q = A 1 ȳ 1 =(0.1 m) (0.02 m) (0.051 m) = m 3

4 The nail have F nail = 400 N. If q all i the allowable hear per unit length and i the pacing between the nail then Hence, F nail = q all ) q all = F nail = 400 N 0.05 m = N/m q all = V maxq I ) V max = q alli Q = (8 103 N/m) ( m 4 ) m 3 = kn (b) If V = 1 kn and etimate the maximum hear tre. Maximum hear tre occur at the neutral axi t max = VQ = (1 103 N) ( m 3 ) ( m 4 ) (0.02 m) = kpa (c) Intead of two wooden plank a hown before if four wooden plank, two horizontal nail, and a ingle vertical nail are ued a hown below. etimate the pacing required for the two horizontal nail for V = 1 kn and F nail = 400 N. Figure 67: Problem 1: four plank are ued. In thi cae, the hear at the joint of 1t and the 2nd part need to be etimated. For thi Q = A 1 ȳ 1 =(0.05 m) (0.02 m) (0.051 m) = m 4

5 Now, F nail ) = F nail q = q = VQ = (1 103 N) ( m 3 ) I m 4 = N/m 400 N = N/m = m Hence, a pacing of 100 mm will be okay. Problem 2. (a) For the box ection hown here etimate the nail pacing required if V = 1 kn and F nail = 400 N. I 1 = I 4 = 1 12 (0.1 m) (0.02 m)3 +(0.1 m) (0.02 m) (0.04 m) 2 = m 4 Figure 68: Problem 2. I 2 = I 3 = 1 (0.02 m) (0.06 m)3 12 = m 4 The econd moment of inertia of the cro-ectional area about the neutral axi I = I 1 + I 2 + I 3 + I 4 = m m 4 = m 4 The firt moment of the top part about the neutral axi i Q = A 1 ȳ 1 The hear flow here =(0.1 m) (0.02 m) (0.04 m) = m 3 2F nail = q = VQ I = (1 103 N) ( m 3 ) m 4 ) N = N/m ) = m Hence, a pacing of 75 mm will be okay. (b) Calculate the maximum hear tre developed. At the neutral axi Q = m (0.03 m) (0.02 m) (0.015 m) = m 3 Figure 69: Problem 2.

6 Maximum hear tre t = VQ = (1 103 N) ( m 3 ) ( m 4 ) ( m) = 338 kpa Problem 3. Deign the beam a hown below for all = 80 MPa and t all = 10 MPa. The depth of the beam i limited to 275 mm. Ue tandard rolled teel ection. The hear force and bending moment diagram are drawn firt. From the diagram, V max = 20 kn and M max = 100 knm. Figure 70: Problem 3. Deign for bending tre Hence, ection modulu required S reqd = M max all = Nm Pa = m 3 = mm 3 Since the depth i limited chooe W and add two 8 mm thick plate at the top and bottom. Total depth = 273 mm < 275 mm (okay). The modified I ection ha a econd moment of inertia about the neutral axi Figure 71: Problem 3: SFD, BMD. I = I beam + 2I plate = m 4 apple (0.254 m) (0.008 m)3 +(0.254 m) (0.008 m) ( m) 2 = m 4 c = mm S = I c = m 3 > S reqd Figure 72: Problem 3: Modified I ection.

7 Check for hear tre A (mm 2 ) ȳ (mm) Aȳ (mm 3 ) Plate I-ection S Figure 73: Problem 3: Shear tre calculation. Q = Â Aȳ = mm 3, t = 9.4 mm Hence, maximum hear tre i t max = V maxq = ( N) ( m 3 ) ( m 4 ) ( m) = 8.7 MPa < t all (okay)

Bending Stress. Sign convention. Centroid of an area

Bending Stress. Sign convention. Centroid of an area Bending Stress Sign convention The positive shear force and bending moments are as shown in the figure. Centroid of an area Figure 40: Sign convention followed. If the area can be divided into n parts