Solutions_2

# Solutions_2 - S.16.2 The bending moments half-way along the...

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Unformatted text preview: Solutions to Chapter 16 Problems 185 The second moments of area are then Ixx = 100 × 103 10 × 1153 + 100 × 10 × 33.42 + + 10 × 115 × 29.12 12 12 = 3.37 × 106 mm4 10 × 1003 115 × 103 + 10 × 100 × 24.12 + + 115 × 10 × 20.92 12 12 = 1.93 × 106 mm4 Iyy = Ixy = 100 × 10 × 33.4 × 24.1 + 115 × 10(−20.9)(−29.1) = 1.50 × 106 mm4 Substituting for Mx , My , Ixx , Iyy and Ixy in Eq. (16.18) gives σz = 0.27x + 0.65y (i) Since the coefﬁcients of x and y in Eq. (i) have the same sign the maximum value of direct stress will occur in either the ﬁrst or third quadrants. Then σz(A) = 0.27 × 74.1 + 0.65 × 38.4 = 45.0 N/mm2 (tension) (compression) σz(C) = 0.27 × (−25.9) + 0.65 × (−86.6) = −63.3 N/mm2 The maximum direct stress therefore occurs at C and is 63.3 N/mm2 compression. S.16.2 The bending moments half-way along the beam are Mx = −800 × 1000 = −800 000 N mm My = 400 × 1000 = 400 000 N mm By inspection the centroid of area (Fig. S.16.2) is midway between the ﬂanges. Its distance x from the vertical web is given by ¯ (40 × 2 + 100 × 2 + 80 × 1) x = 40 × 2 × 20 + 80 × 1 × 40 ¯ i.e. x = 13.33 mm ¯ The second moments of area of the cross-section are calculated using the approximations for thin-walled sections described in Section 16.4.5. Then 2 × 1003 = 5.67 × 105 mm4 Ixx = 40 × 2 × 502 + 80 × 1 × 502 + 12 2 × 403 1 × 803 Iyy = 100 × 2 × 13.332 + + 2 × 40 × 6.672 + 12 12 2 + 1 × 80 × 26.67 = 1.49 × 105 mm4 Ixy = 40 × 2(6.67)(50) + 80 × 1(26.67)(−50) = −0.8 × 105 mm4 186 Solutions Manual 40 mm y 2.0 mm 2.0 mm y C x 100 mm x 1.0 mm 80 mm Fig. S.16.2 The denominator in Eq. (16.18) is then (5.67 × 1.49 − 0.82 ) × 1010 = 7.81 × 1010 . From Eq. (16.18) σ= + i.e. σ = 2.08x − 1.12y and at the point A where x = 66.67 mm, y = −50 mm σ (A) = 194.7 N/mm2 (tension) 400 000 × 5.67 × 105 − 800 000 × 0.8 × 105 7.81 × 1010 x −800 000 × 1.49 × 105 + 400 000 × 0.8 × 105 7.81 × 1010 y S.16.3 Initially, the section properties are determined. By inspection the centroid of area, C, is a horizontal distance 2a from the point 2. Now referring to Fig. S.16.3 and taking moments of area about the ﬂange 23 (5a + 4a)t y = 5at (3a/2) ¯ from which y = 5a/6 ¯ 188 Solutions Manual S.16.4 Referring to Fig. S.16.4. W 2 2W 1 4 B D3 l/2 z A l/2 Fig. S.16.4 In DB Mx = −W (l − z) My = 0 In BA Mx = −W (l − z) M y = −2 W l −z 2 (ii) (iii) (i) Now referring to Fig. P.16.4 the centroid of area, C, of the beam cross-section is at the centre of antisymmetry. Then Ixx = 2 td Iyy = 2 td Ixy = td d 4 d 2 d 4 2 + 2 td 3 12 = 2td 3 3 d 4 2 + d 2 td 3 + td 12 d 4 = − d 2 5td 3 12 td 3 4 + td − = Substituting for Ixx , Iyy and Ixy in Eq. (16.18) gives σz = 1 [(3.10My − 1.16Mx )x + (1.94Mx − 1.16My )y] td 3 (iv) Solutions to Chapter 16 Problems 189 Along the edge l, x = 3d /4, y = d /2. Equation (iv) then becomes σz,1 = 1 (1.75My + 0.1Mx ) td 2 (v) Along the edge 2, x = −d /4, y = d /2. Equation (iv) then becomes σz,2 = From Eqs (i)–(iii), (v) and (vi) In DB σz,1 = − σz,2 = − In BA σz,1 = σz,2 = W td W 2 1 (−1.36My + 1.26Mx ) td 2 (vi) 0.1W 2 td 1.26W td 2 (1 − z) (1 − z) whence σz,1 (B) = − 0.05Wl td 2 0.63Wl td 2 1.85Wl td 2 0.1Wl td 2 whence σz,2 (B) = − (3.6z − 1.85l) (−1.46z + 0.1l) whence σz,1 (A) = − whence σz,2 (A) = td 2 S.16.5 By inspection the centroid of the section is at the mid-point of the web. Then Ixx = Iyy = Ixy h 10h3 t 2t (2h)3 (2t )h2 + ht h2 + = 2 12 3 th3 5h 3 t 2t (h/2)3 + = 3 3 12 h h = 2t − (h) + ht 2 4 h 3h 3 t (−h) = − 2 4 Since My = 0, Eq. (16.18) reduces to σz = −Mx Ixy Mx Iyy x+ y 2 2 Ixx Iyy − Ixy Ixx Iyy − Ixy (i) Substituting in Eq. (i) for Ixx , etc. Mx 5/12 3/4 x+ y 3 t (10/3)(5/12) − (3/4)2 h (10/3)(5/12) − (3/4)2 Mx σz = 3 (0.91x + 0.50y) ht σz = + i.e. (ii) 200 Solutions Manual Then 7W (−z3 + 4l 2 z) 113Ea3 t At the mid-span point where z = l , Eq. (v) gives u= u= Similarly v= 0.177Wl3 Ea3 t 0.186Wl3 Ea3 t...
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