- The Coefficients Are A 1 2 B 1 5 And C 1 1 1 (183.64 KiB) Viewed 15 times
The coefficients are: A= 1.2, B= 1.5 and C= 1.1
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The coefficients are: A= 1.2, B= 1.5 and C= 1.1
The coefficients are: A= 1.2, B= 1.5 and C= 1.1
Draw the distribution of internal forces M, N, V on a given structure. Determine the location and the magnitude of the maximal bending moment Mon intervals (k, a), (f, g) and (s, h). To check your results you will need to compute the reaction forces Kx, Kz, Rx, Rz, Sx, Sz, internal forces Nab, Nak, Nd, Ned, Nef, Nel, Nf, Ng, Nhg, Nhj, Nj, Nka, Nrl, Nsh, Vab, Vak, Vba, Vbc, Vcb, Vd, Ved, Vef, Vel, Vf, Vg, Vhg, Vhj, Vhs, Vjh, Vka, Vle, Vir, Vrl, Vsh, Mab, Mak, Mb, Mc, Md, Med, Mef, Mel, Mf, Mg, Mhg, Mhj, Mhs, Mj, MK, MI, Mr, Ms, and Xmax(k,a), Mmax(k,a), Xmax(f,g), Mmax(f.g), Xmax(s,h), Mmax(s,h) where Xmax is the distance of the maximal moment Mmax from the starting point of the particular interval (k, f, s). If there is no maximal moment on either of these intervals set the values of Xmax and Mmax for this interval equal to zero. The signs of checked forces M, N, V correspond to a recommended selection of bottom fibers (for vertical beams (columns) the bottom fibers are on the right). F1 = (20+10b) KN |F2 = (20+10c) KN = F3 = (20+10a) kN fi = (8+4b) kN/m b с 8 2 F4 = (20+4a) kN f2 = (12+4a) kN/m = 0.5 +c fo = (20+8a) kN/m Kx >Ar Rx Rz Sx Kz C of af b а 1.5 + b t b +2+a feki tit 2 +
Draw the distribution of internal forces M, N, V on a given structure. Determine the location and the magnitude of the maximal bending moment Mon intervals (k, a), (f, g) and (s, h). To check your results you will need to compute the reaction forces Kx, Kz, Rx, Rz, Sx, Sz, internal forces Nab, Nak, Nd, Ned, Nef, Nel, Nf, Ng, Nhg, Nhj, Nj, Nka, Nrl, Nsh, Vab, Vak, Vba, Vbc, Vcb, Vd, Ved, Vef, Vel, Vf, Vg, Vhg, Vhj, Vhs, Vjh, Vka, Vle, Vir, Vrl, Vsh, Mab, Mak, Mb, Mc, Md, Med, Mef, Mel, Mf, Mg, Mhg, Mhj, Mhs, Mj, MK, MI, Mr, Ms, and Xmax(k,a), Mmax(k,a), Xmax(f,g), Mmax(f.g), Xmax(s,h), Mmax(s,h) where Xmax is the distance of the maximal moment Mmax from the starting point of the particular interval (k, f, s). If there is no maximal moment on either of these intervals set the values of Xmax and Mmax for this interval equal to zero. The signs of checked forces M, N, V correspond to a recommended selection of bottom fibers (for vertical beams (columns) the bottom fibers are on the right). F1 = (20+10b) KN |F2 = (20+10c) KN = F3 = (20+10a) kN fi = (8+4b) kN/m b с 8 2 F4 = (20+4a) kN f2 = (12+4a) kN/m = 0.5 +c fo = (20+8a) kN/m Kx >Ar Rx Rz Sx Kz C of af b а 1.5 + b t b +2+a feki tit 2 +