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 2 2 2 2 y 2 4
A = bh I = 1 12 bh3 t xy = t yx = VQ Ib = V b2 h 2 4 y 2 1 12 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 2 4 - 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 = 5.27 10 6 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 = 9.09 10 6 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 = 14.36 10 6 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) = 102 10 6 m 3
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 = 8 103 N/m q all = V maxq I ) V max = q alli Q = (8 103 N/m) (14.36 10 6 m 4 ) 102 10 6 m 3 = 1.126 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) (119 10 6 m 3 ) (14.36 10 6 m 4 ) (0.02 m) = 414.35 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) = 51 10 6 m 4
Now, F nail ) = F nail q = q = VQ = (1 103 N) (51 10 6 m 3 ) I 14.36 10 6 m 4 = 3551.5 N/m 400 N = 3551.5 N/m = 0.113 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 = 3.27 10 6 m 4 Figure 68: Problem 2. I 2 = I 3 = 1 (0.02 m) (0.06 m)3 12 = 0.36 10 6 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 = 2 3.27 10 6 m 4 + 2 0.36 10 6 m 4 = 7.25 10 6 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) =80 10 6 m 3 2F nail = q = VQ I = (1 103 N) (80 10 6 m 3 ) 7.25 10 6 m 4 ) 2 400 N = 11034.5 N/m ) = 0.0725 m Hence, a pacing of 75 mm will be okay. (b) Calculate the maximum hear tre developed. At the neutral axi Q = 80 10 6 m 3 + 2 (0.03 m) (0.02 m) (0.015 m) = 98 10 6 m 3 Figure 69: Problem 2.
Maximum hear tre t = VQ = (1 103 N) (98 10 6 m 3 ) (7.25 10 6 m 4 ) (2 0.02 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 = 100 103 Nm 80 10 6 Pa = 1.25 10 3 m 3 = 1250 10 3 mm 3 Since the depth i limited chooe W250 80 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 = 126 10 6 m 4 apple 1 + 2 12 (0.254 m) (0.008 m)3 +(0.254 m) (0.008 m) (0.1325 m) 2 = 197.4 10 6 m 4 c = 136.5 mm S = I c = 1446 10 6 m 3 > S reqd Figure 72: Problem 3: Modified I ection.
Check for hear tre A (mm 2 ) ȳ (mm) Aȳ (mm 3 ) Plate 1 254 8 132.5 269.24 10 3 I-ection 2 254 15.6 120.7 478.26 10 3 3 112.9 9.4 56.45 59.91 10 3 S 807.41 10 3 Figure 73: Problem 3: Shear tre calculation. Q = Â Aȳ = 571.11 10 3 mm 3, t = 9.4 mm Hence, maximum hear tre i t max = V maxq = (20 103 N) (807.41 10 6 m 3 ) (197.4 10 6 m 4 ) (0.0094 m) = 8.7 MPa < t all (okay)