Lecture 8 Bending & Shear Stresses on Beams

Transcription

1 Lecture 8 Bending & hear tresses on Beams Beams are almost always designed on the asis of ending stress and, to a lesser degree, shear stress. Each of these stresses will e discussed in detail as follows. A) Bending tresses A ending stress is NOT considered to e a simple stress. In other words, it is not load divided y area. The formula for ending stress,, is as follows: y I where: moment acting on eam from moment diagram (kip-in or l-in) y distance from neutral axis to extreme edge of memer (in) I moment of inertia aout the axis (in 4 ) Recalling that I y, the ending stress formula could e re-written as: where: section modulus aout the axis (in 3 ) Bending stress is distriuted through a eam as seen in the diagram elow: o, in reality, ending stresses are tensile or compressive stresses in the eam! A simply-supported eam always has tensile stresses at the ottom of the eam and compressive stresses at the top of the eam. Lecture 8 - Page 1 of 9

2 Example 1 GIVEN: A nominal 2x10 (actual dims. 1½ x 9¼ ) is used as a simply-supported eam with loading as shown. The allowale ending stress is 1200 psi. REQUIRED: a) Determine the maximum moment on the eam. ) Determine the maximum actual ending stress on the eam c) Determine if the eam is acceptale ased upon allowale ending stress. w 140 PLF (includes eam weight) 11-0 The maximum ending moment, max, on a simply-supported, uniformly loaded eam is: max wl 8 2 The actual ending stress is: 2 (140 PLF)(11') max 8 max l-ft l - ft(12"/ft) in PI ee textook appendix ince the actual ending stress of PI is less than the allowale ending stress of 1200 PI, THE BEA I ACCEPTABLE. Lecture 8 - Page 2 of 9

3 Example 2 GIVEN: A W14x30 steel eam experiences a moment of 95 kip-feet. The allowale ending stress for the steel eam is 24 KI. REQUIRED: a) Determine the maximum ALLOWABLE moment ased on the allowale ending stress (leave answer in units of kip-ft). ) Determine if the eam is acceptale or not ased upon allowale ending moment. Using the ending stress formula aove, re-write it to solve for moment: ustituting 24 KI for and using 42.0 in 3 (from textook appendix) gives: all 24 KI(42.0 in 3 ) 1008 kip-in Dividing y 12 to get into units of kip-ft: all 84 kip-ft ince the actual ending moment of 95 kip-ft is more than the allowale ending moment of 84 kip-ft, THE BEA I UNACCEPTABLE. Lecture 8 - Page 3 of 9

4 Example 3 GIVEN: The eam prolem in Example 2. REQUIRED: Determine the lightest weight W12 eam that can carry a moment of 95-kip-feet ased upon allowale ending stress of 24 KI. Using the ending stress formula aove, re-write it to solve for section modulus: ustituting 24 KI for and 95 kip-ft for gives: 95 kip - ft(12"/ft) 24 KI 47.5 in 3 Looking at the textook appendix, the lightest weight W12 that has a section modulus of at least 47.5 in 3 is a W12x40 ( x 51.9 in 3 > 47.5 in 3 ). Lecture 8 - Page 4 of 9

5 B) hear tresses It is easy to imagine vertical shear on a eam that was made up of concrete locks: Load This type of shear is called transverse shear, and occurs if there is no ending stresses present. The transverse shear stress A V However, almost all real eams have ending stresses present. In this case, eams are more like a deck of cards and ending produces sliding along the horizontal planes at the interfaces of the cards as shown elow: Load This type of shear is called longitudinal or horizontal shear. The formula used for determining the maximum longitudinal shear stress, f v, is as follows: VQ f v I where: V vertical shear, usually from shear diagram (l. or kip) Q first moment of area Ay A area of shape aove or elow the neutral axis (in 2 ) y distance from neutral axis to centroid of area A (in) I moment of inertia of shape (in 4 ) width of area A (in) Lecture 8 - Page 5 of 9

6 Example 4 GIVEN: The nominal 2x10 wood eam from Example 1. The allowale horizontal shear stress is 95 PI. REQUIRED: a) Determine the maximum horizontal shear stress on the eam. ) Determine if the eam is acceptale ased upon allowale horizontal shear stress. w 140 PLF (includes eam weight) 11-0 End reaction ½(140 PLF)(11 ) 770 ls. 770 ls. 0 0 hear diagram -770 ls. 9 ¼ N.A Area A shown shaded y 2.31 N.A ½ 1½ Beam X-ection Lecture 8 - Page 6 of 9

7 Using the horizontal shear stress formula: and sustituting in the values: VQ f v I V 770 ls. (from shear diagram) Q Ay where: A (4.625 )(1.5 ) 6.94 in 2 y ½(4.625 ) 2.31 Q (6.94 in 2 )(2.31 ) in 3 I in 4 (from textook appendix, or calculate 3 h I ) 12 1½ 3 (770ls)(16.03in ) f v 4 (98.93in )(1.5") f v 83.2 PI ince the actual horizontal shear stress of 83.2 PI is less than the allowale horizontal shear stress of 95 PI, THE BEA I ACCEPTABLE. Lecture 8 - Page 7 of 9

8 Example 5 GIVEN: A wood uilt-up eam is constructed using a nominal 2X8 we and 2X6 flanges as shown elow. The moment of inertia aout the strong axis, I, in 4. The maximum vertical shear is shown elow. REQUIRED: Determine the spacing of 12d nails connecting the top & ottom flanges to the we. Assume the allowale shear strength of each nail is 140 ls. w 200 PLF (includes eam weight) 11-0 End reaction ½(200 PLF)(11 ) 1100 ls ls. 0 0 hear diagram ls. Beam Cross-ection Lecture 8 - Page 8 of 9

9 The shear flow, q can e used to determine the horizontal shear along the length of the eam in terms of force per unit length. In particular, VQ q I Where: V aximum vertical shear (from shear diagram) 1100 ls. Q A p y A p Area of piece to e fastened, in 2 (1½ x 5½ ) 8.25 in 2 y Distance from eam N.A. to centroid of piece, in Q (8.25in 2 )(4.375 ) 36.1 in 3 ustituting: VQ q I 3 (1100ls.)(36.1in ) q in ls inch The allowale shear capacity per nail 140 ls., o then the nail spacing can e determined as follows: pacing Allowale. hear _ per _ Nail q 140ls._ per _ nail ls inch pacing 1.29 inches use 1¼ pace the 12d nails at 1¼ apart along entire length of eam. Lecture 8 - Page 9 of 9