Problem 4. = 1 1 = 1. = m 4. = m 4

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Myra Roberts
5 years ago
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1 Problem. (a) Calculate the stress in the bolt that connects steel plates and the wooden block as shown if the section is subjected to V kn. ssume the elastic moduli of steel as GPa and of wood as 1.5 GPa. The bolt used has a diameter of 16 mm and a spacing of s mm is used. Take steel as the reference material. Hence, E 1 = GPa, E.5 GPa n 1, n = E E 1 16 The transformed section will be the following Figure 7: Problem. Figure 75: Problem : The transformed section. The neutral axis will pass through the middle of the section. The second moment of inertia of part 1 about the neutral axis, I 1 1 n 1b 1 h 1 + n 1 1 d 1 1 (1) (.1 m) (. m) +(1) (.1 m) (. m) (.1 m) = m Similarly, For the full section, I 1 n b h 1 1 (.1 m) (. m) 16 = m I = I 1 + I = m Figure 76: Problem : The top steel plate.

2 To get the stress in the bolt we need to calculate the shear force at the bonded surface. Hence, we need the first moment of the steel pate about the neutral axis The shear flow is Q 1 = n 1 1 ȳ 1 =(1) (.1 m) (. m) (.11 m) = 1 6 m q = VQ 1 I = (1 1 N) ( 1 6 m ) m = N/m If the stress in the bolt is t b and the cross-sectional area of the bolt is b we can write t b b = F bolt = qs ) t b = qs = qs b pd b / = ( N/m) (.1 m) p (.16 m) / =.76 MPa (b) Instead if allowable shear stress is t all MPa determine the required spacing. We have s reqd = t all b q Hence, a spacing of 5 mm will be okay. =. m Problem 5. Calculate the shear stress in the bonded surface if the section is subjected to V kn. ssume the elastic moduli of steel as 1 GPa and of aluminum as 7 GPa. NOTE: In this problem, we need to know the centroid and the second moment of inertia of a semi-circular area. Please see the calculation at the end of this problem. Take the aluminum with the semi-circular hole in it as the reference material. Hence, E 1 = 7 GPa, E = 1 GPa Figure 77: Problem 5. n 1, n = E E 1 = Next, to locate its neutral axis

3 n i i (mm ) ȳ i (mm) n i i ȳ i (mm ) 1 (1) (1 1) luminum (1) p () 5 + p = 1.16 = 5.9 Steel () (1 5) S The neutral axis is located at a distance Ȳ from the bottom where Ȳ = Â i n i i ȳ i = Â i n i i mm Figure 79: Problem 5: The position of the N. Next, the calculate the second moment of inertia of the crosssectional area about the neutral axis. We will separately calculate for 1,, and, first. I 1 1 n 1b 1 h 1 + n 1 1 d 1 1 (1) (.1 m) (.1 m) +(1) (.1 m) (.1 m) (.5 m) = m I 1 n b h + n d 1 () (.1 m) (.5 m) +() (.1 m) (.5 m) (. m) m I 1 n 1 1 (1) pr pr + n 1 = m p (. m) d +(1) p (. m) (.5 m) Since I x = pr about the axis that passes through the center as shown in the figure. Detailed explanation is given at the end.

4 Hence, I = I 1 + I I = m To estimate the the shear stress we need to calculate the first moment Q of the cross-sectional area about the neutral axis and we will use the bottom steel part to do it. Q =Q = n ȳ =() (.1 m.5 m) (. m) = m Here, t mm mm = 6 mm =.6 m. Hence, the shear stress at the bonded surface t = VQ It = (1 1 N) (5 1 6 m ) ( m ) (.6 m).66 MPa Centroid and second moment of inertia of a semi-circular area Take a small area inside the semi-circular area as shown in the figure. The area of this element is d =(dr) (rdq) =rdrdq. Figure : Semi-circular area. The area of this semi-circular plate is = pr. Hence, if the distance to the centroid from the bottom is ȳ then

5 using the figure Z Z Z r Z p ȳ = yd = r sin(q)d = r sin(q) (rdrdq) pr Z r Z p ) ȳ = (sin(q)dq) (r dr) apple Z p apple Z r = sin(q)dq r dr r ) ȳ = = r pr = r p The second moment of inertia of the semi-circular area about the x axis Z I x = y d Z = [r sin(q)] d Z r Z p = r sin (q) (rdrdq) apple Z p apple Z r = sin (q)dq r dr = p = pr r

5. What is the moment of inertia about the x - x axis of the rectangular beam shown?

5. What is the moment of inertia about the x - x axis of the rectangular beam shown? 1 of 5 Continuing Education Course #274 What Every Engineer Should Know About Structures Part D - Bending Strength Of Materials NOTE: The following question was revised on 15 August 2018 1. The moment