Essentials in the theory of framed structures . FiG. 161. tinuous beam of two unequal spans h and h, supporting unequaluniform loads, wi and Wi per unit of length, will now be de-veloped in connection with Fig. 161. The M-diagram isdrawn as in the preceding problem. If the tangent to theelastic curve is drawn through C; andJi and h represent thetangential deviations at A and B respectively; then -=r-r)(i)e)+(^)(i-) 24 + Elk =tih = M = - 24 3 Will -\- WiH 8(/i + h) (12) 26o THEORY OF FRAMED STRUCTURES Chap. VI When the spans are equal in length I and the uniform load wper unit of length is the
Essentials in the theory of framed structures . FiG. 161. tinuous beam of two unequal spans h and h, supporting unequaluniform loads, wi and Wi per unit of length, will now be de-veloped in connection with Fig. 161. The M-diagram isdrawn as in the preceding problem. If the tangent to theelastic curve is drawn through C; andJi and h represent thetangential deviations at A and B respectively; then -=r-r)(i)e)+(^)(i-) 24 + Elk =tih = M = - 24 3 Will -\- WiH 8(/i + h) (12) 26o THEORY OF FRAMED STRUCTURES Chap. VI When the spans are equal in length I and the uniform load wper unit of length is the same in both spans, Eq. (12) reduces to M = -f (13) 165. When a continuous beam supports a combination ofuniform and concentrated loads, it will be found expedient tosketch the M-diagram in, parts as shown in Fig. 162. Theportion {a) is the M-diagram for the concentrated loads whenno continuity is considered at C; and the portion (&) is a similar. diagram for the uniform loads. The continuity is provided forby the portion (c). If the tangent to the elastic curve isdrawn through C; and h and h represent the tangential devia-tions at A and B; then Elh = (9,000 X 6 X 6) + (18,000 X 12 X - X 6) + If (6 X 8) = 1,188,000 + 48M ^ Elh = (6,000 X 3 X 4) + (6,000 X 6 X 9) + (6,000 X 3 X 14) + ^81,000 X 18 X - X 9) + M{g X 12) = 9,396,000 + 108M3/1 = —2/2M = —62,100 Sec. II RESTRAINED AND CONTINUOUS BEAMS 261 The reactions may now be determined by the principles ofstatics. The value of M may also be determined from Eqs. (5) and (12).Eq. (s) is applicable to the concentrated loads. For theload at Z); P = 3,000, k = }4, h = 12 and h = 18; hence 3,000 X 144 M = il-l) 2(12 + 18) = —2,700 For the load F; P = 1,000, k = -, h = 18 and h = 12;hence 1,000 X 324 ^3 27/ M = ,\ = -1,600 2(12 + 18) 2For the load at £; P = 1,000, k = -, h = i& and h = 12, hence 1,000 X 324(- ) M = ^3 27/ ^ _ 2(12 + 18)Hence the bending moment at C, due t
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