Bending Moments and Shears: Equal Span Coefficients

For two span Beam

Using the three moment equation

M_AL + 2M_B(L+L) + M_CL = (wL³/4) + (wL³/4)

Since M_A = M_B = 0

4M_BL = wL³/2

M_B = wL²/8 = 0.125wL²………………………………….......…..1

Reaction R_AB

Note: S. S. R. means simple suport reaction
          C. C. means continuity correction

S.S.R =wl/2

C.C = (|M_A| - |M_B|)/L = (-M_B)/L = wl/8                  

 Reaction R_AB = ∑R = S.S.R + C.C.

                        = wL/2-wL/8 =3wL/8

            R_AB =3wL/8 = 0.375wL…………………………………2

Reaction R_BA

S.S.R, = wL/2

C.C = (|M_B|-|M_A|)/L = (M_B)/L  

Reaction R_AB = ∑R = S.S.R + C.C. = wL/2 + wL/8 = 5wL/8

R_BA =5wL/8 = 0.625wL................................................................3

Due to symmetry, R_AB =R_CB = 3wL/8

R_BA = R_BC =5wL/8

Maximum Span Moment on Span A-B

M_x = R_AB - wx²/2...........................................................................4

At maximum moment

dM_x/d_x = R_AB - wx = 0

x = R_AB/w = 3L/8 = 0.375L

Substituting for x = 3L/8 in equ (4)

M_max = (3wL/8).(3L/8) - (w/2). (3L/8)²

              = 9wL²/64( 1-1/2) = 9wL²/128

 M_max = 9wL²/128 = 0.0703wL²...................................................5

Alternatively,

M_x = R_BA - wx²/2 - M_B

At maximum moment

dM_x/d_x = R_BA - wx = 0

x = R_BA/w = 5L/8

Substituting for x = 5L/8 in equ (4)

M_max = (5wL/8).(5L/8) - (w/2). (5L/8)² -wL²/8

= 9wL²/128

M_max = 9wL²/128 = 0.0703wL².................................................6

For three span Beam

Using the three moment equation

SPANS A-C

M_AL + 2M_B(L+L) + M_CL = wL³/4 + wL³/4

M_A = 0

4M_BL + M_CL = wL³/2………………………………………………7

SPANS B-D

M_BL + 2M_C(L+L) + M_DL = wL³/4 + wL³/4

M_D = 0

M_BL + 4M_CL = wL³/2………………………………………………8
Equating 7 and 8

4M_BL + M_CL = M_BL + 4M_CL

4M_B + M_C  = M_B + 4M_C

3M_B = 3M_C

M_B = M_C

Substituting for M_B = M_C in equation 7
4M_CL + M_CL = wL³/2

5M_CL = wL³/2

M_C = wL²/10 = 0.1wL²....……………………………………………9

Since M_B = M_C

M_B = M_C = wL²/10 = 0.1wL²

S.S.R = wL/2

C.C = (|M_A|-|M_B|)/L = (- M_B)/L = wL/10
R_AB = ∑R = S.S.R + C.C.  = wL²/2-wL/10

R_AB = 4wL/10 = 0.4wL......................................................................10
Similarly

R_BA = wL/2 + wL/10 = 6wL/10

R_BA = 6wL/10 = 0.6wL……………………………………………..11
S.S.R. = wL/2
C.C = (|M_B|-|M_C|)/L = 0

R_BC =R_CB = wL/2 = 0.5wL.................................................................12

Maximum Span Moment on Span A-B

M_x = R_AB - wx²/2...........................................................................13

At maximum moment

dM_x/d_x = R_AB - wx = 0

x = R_AB/w = 4L/10 = 0.4L

Substituting for x = 4L/10 in equ (13)

M_max = (4wL/10).(4L/10) - (w/2). (4L/10)²

              = 16wL²/100( 1-1/2) = 16wL²/200

 M_max = 16wL²/200 = 0.08wL²......................................................14

Maximum Span Moment on Span B-C

M_x = R_BC - wx²/2 - |M_B| ...............................................................15

At maximum moment

dM_x/d_x = R_BC - wx = 0

x = R_BC/w = 5L/10

Substituting for x = 5L/10 in equ (15)

M_max = (5wL/10).(5L/10) - (w/2). (5L/10)² -wL²/10

= 25wL²/200 - wL²/10

M_max = wL²/40 = 0.025wL²............................................................16