Question 20.9: Construct influence lines for the reaction at B and for the ...
Construct influence lines for the reaction at \mathrm{B} and for the shear force and bending moment at \mathrm{D} in the two-span continuous beam shown in Fig. 20.21(a).
The shape of each influence line may be drawn using the Mueller-Breslau principle as shown in Fig. 20.21(b), (c) and (d). However, before they can be of direct use in determining maximum values, say, of the various functions due to the passage of loading systems, the ordinates must be calculated; for this, since the influence lines are comprised of curved segments, we need to derive their equations.
However, once the influence line for a support reaction, R_{\mathrm{B}} in this case, has been established, the remaining influence lines follow from statical equilibrium.
R_{B} influence line
Suppose initially that a unit load is a distance x_{1} from \mathrm{A}, between \mathrm{A} and \mathrm{B}. To determine R_{\mathrm{B}} we may use the flexibility method described in Section 16.4. Thus we remove the support at B (point 2) and calculate the displacement, a_{21}, at B due to the unit load at x_{1} (point 1). We then calculate the displacement, a_{22}, at \mathrm{B} due to a vertically downward unit load at \mathrm{B}. The total displacement at \mathrm{B} due to the unit load at x_{1} and the reaction R_{\mathrm{B}} is then
a_{21}-a_{22} R_{\mathrm{B}}=0 (i)
since the support at \mathrm{B} is not displaced. In Eq. (i) the term a_{22} R_{\mathrm{B}} is negative since R_{\mathrm{B}} is in the opposite direction to the applied unit load at B.
Both the flexibility coefficients in Eq. (i) may be obtained from a single unit load application since, from the reciprocal theorem (Section 15.4), the displacement at B due to a unit load at x_{1} is equal to the displacement at x_{1} due to a unit load at \mathrm{B}. Therefore we apply a vertically downward unit load at B.
The equation for the displaced shape of the beam is that for a simply supported beam carrying a central concentrated load. Therefore, from Eq. (iv) of Ex. 13.5
v=\frac{1}{48 E I}\left(4 x^{3}-3 L^{2} x\right) (ii)
or, for the beam of Fig. 20.21(a)
v=\frac{x}{12 E I}\left(x^{2}-48\right) (iii)
At \mathrm{B}, when x=4 \mathrm{~m}
v_{\mathrm{B}}=-\frac{32}{3 E I}=a_{22} (iv)
Furthermore, the displacement at B due to the unit load at x_{1} (=displacement at x_{1} due to a unit load at B) is from Eq. (iii)
v_{x_{1}}=\frac{x_{1}}{12 E I}\left(x_{1}^{2}-48\right)=a_{21} (v)
Substituting for a_{22} and a_{21} in Eq. (i) we have
\frac{x_{1}}{12 E I}\left(x_{1}^{2}-48\right)+\frac{32}{3 E I} R_{\mathrm{B}}=0
from which
R_{\mathrm{B}}=-\frac{x_{1}}{128}\left(x_{1}^{2}-48\right) \quad\left(0 \leq x_{1} \leq 4.0 \mathrm{~m}\right) (vi)
Equation (vi) gives the influence line for R_{\mathrm{B}} with the unit load between \mathrm{A} and \mathrm{B}; the remainder of the influence line follows from symmetry. Eq. (vi) may be checked since we know the value of R_{\mathrm{B}} with the unit load at \mathrm{A} and \mathrm{B}. Thus from Eq. (vi), when x_{1}=0, R_{\mathrm{B}}=0 and when x_{1}=4.0 \mathrm{~m}, R_{\mathrm{B}}=1 as expected.
If the support at \mathrm{B} were not symmetrically positioned, the above procedure would be repeated for the unit load on the span BC. In this case the equations for the deflected shape of \mathrm{AB} and \mathrm{BC} would be Eqs (xiv) and (xv) in Ex. 13.6. In this example we require the S_{\mathrm{D}} influence line so that we shall, in fact, need to consider the value of R_{\mathrm{B}} with the unit load on the span BC. Therefore from Eq. (xv) in Ex. 13.6
v_{x_{1}}=-\frac{1}{12 E I}\left(x_{1}^{3}-24 x_{1}^{2}+144 x_{1}-128\right) \quad\left(4.0 \mathrm{~m} \leq x_{1} \leq 8.0 \mathrm{~m}\right) (vii)
Hence from Eq. (i)
R_{\mathrm{B}}=\frac{1}{128}\left(x_{1}^{3}-24 x_{1}^{2}+144 x_{1}-128\right) \quad\left(4.0 \mathrm{~m} \leq x_{1} \leq 8.0 \mathrm{~m}\right) (viii)
A check on Eq. (viii) shows that when x_{1}=4.0 \mathrm{~m}, R_{\mathrm{B}}=1 and when x_{1}=8.0 \mathrm{~m}, R_{\mathrm{B}}=0.
S_{D} influence line
With the unit load to the left of \mathrm{D}, the shear force, S_{\mathrm{D}}, at \mathrm{D} is most simply given by
S_{\mathrm{D}}=-R_{\mathrm{A}}+1 (ix)
where, by taking moments about \mathrm{C}, we have
R_{\mathrm{A}} \times 8-1\left(8-x_{1}\right)+R_{\mathrm{B}} \times 4=0 (x)
Substituting in Eq. (x) for R_{\mathrm{B}} from Eq. (vi) and rearranging gives
R_{\mathrm{A}}=\frac{1}{256}\left(x_{1}^{3}-80 x_{1}+256\right) (xi)
whence, from Eq. (ix)
S_{\mathrm{D}}=-\frac{1}{256}\left(x_{1}^{3}-80 x_{1}\right) \quad\left(0 \leq x_{1} \leq 2.0 \mathrm{~m}\right) (xii)
Therefore, when x_{1}=0, S_{\mathrm{D}}=0 and when x_{1}=2.0 \mathrm{~m}, S_{\mathrm{D}}=0.59, the ordinate \mathrm{d}_{1} \,\mathrm{g} in the S_{\mathrm{D}} influence line in Fig. 20.21(c).
With the unit load between D and B
S_{\mathrm{D}}=-R_{\mathrm{A}}
so that, substituting for R_{\mathrm{A}} from Eq. (xi)
S_{\mathrm{D}}=-\frac{1}{256}\left(x_{1}^{3}-80 x_{1}+256\right) \quad\left(2.0 \mathrm{~m} \leq x_{1} \leq 4.0 \mathrm{~m}\right)
Thus, when x_{1}=2.0 \mathrm{~m}, S_{\mathrm{D}}=-0.41, the ordinate \mathrm{d}_{1} \mathrm{f} in Fig. 20.21(c) and when x_{1}=4.0 \mathrm{~m}, S_{\mathrm{D}}=0.
Now consider the unit load between B and C. Again
S_{\mathrm{D}}=-R_{\mathrm{A}}
but in this case, \mathrm{R}_{\mathrm{B}} in Eq. (x) is given by Eq. (viii). Substituting for \mathrm{R}_{\mathrm{B}} from Eq. (viii) in Eq. (x) we obtain
R_{\mathrm{A}}=-S_{\mathrm{D}}=-\frac{1}{256}\left(x_{1}^{3}-24 x_{1}^{2}+176 x_{1}-384\right) \quad\left(4.0 \mathrm{~m} \leq x_{1} \leq 8.0 \mathrm{~m}\right) (xiv)
Therefore the S_{\mathrm{D}} influence line consists of three segments, \mathrm{a}_{1} \mathrm{~g}, \mathrm{fb}_{1} and \mathrm{b}_{1} \mathrm{c}_{1}.
M_{\mathrm{D}} influence line
With the unit load between A and D
M_{\mathrm{D}}=R_{\mathrm{A}} \times 2-1\left(2-x_{1}\right) (xv)
Substituting for R_{\mathrm{A}} from Eq. (xi) in Eq. (xv) and simplifying, we obtain
M_{\mathrm{D}}=\frac{1}{128}\left(x_{1}^{3}+48 x_{1}\right) \quad\left(0 \leq x_{1} \leq 2.0 \mathrm{~m}\right) (xvi)
When x_{1}=0, M_{\mathrm{D}}=0 and when x_{1}=2.0 \mathrm{~m}, M_{\mathrm{D}}=0.81, the ordinate \mathrm{d}_{2} \mathrm{~h} in the M_{\mathrm{D}} influence line in Fig. 20.21(d).
Now with the unit load between \mathrm{D} and \mathrm{B}
M_{\mathrm{D}}=R_{\mathrm{A}} \times 2 (xvii)
Therefore, substituting for R_{\mathrm{A}} from Eq. (xi) we have
M_{\mathrm{D}}=\frac{1}{128}\left(x_{1}^{3}-80 x_{1}+256\right) \quad\left(2.0 \mathrm{~m} \leq x_{1} \leq 4.0 \mathrm{~m}\right) (xvii)
From Eq. (xviii) we see that when x_{1}=2.0 \mathrm{~m}, M_{\mathrm{D}}=0.81, again the ordinate \mathrm{d}_{2} \mathrm{~h} in Fig. 20.21(d). Also, when x_{1}=4.0 \mathrm{~m}, M_{\mathrm{D}}=0.
Finally, with the unit load between \mathrm{B} and \mathrm{C}, M_{\mathrm{D}} is again given by Eq. (xvii) but in which R_{\mathrm{A}} is given by Eq. (xiv). Hence
M_{\mathrm{D}}=-\frac{1}{128}\left(x_{1}^{3}-24 x_{1}^{2}+176 x_{1}-384\right) \quad\left(4.0 \mathrm{~m} \leq x_{1} \leq 8.0 \mathrm{~m}\right) (xix)
The maximum ordinates in the S_{\mathrm{D}} and M_{\mathrm{D}} influence lines for the span BC may be found by differentiating Eqs (xiv) and (xix) with respect to x_{1}, equating to zero and then substituting the resulting values of x_{1} back in the equations. Thus, for example, from Eq. (xiv)
\frac{\mathrm{d} S_{\mathrm{D}}}{\mathrm{d} x_{1}}=\frac{1}{256}\left(3 x_{1}^{2}-48 x_{1}+176\right)=0
from which x_{1}=5.7 \mathrm{~m}. Hence
S_{\mathrm{D}}(\max )=0.1
Similarly M_{\mathrm{D}}(\max )=-0.2 at x_{1}=5.7 \mathrm{~m}.
In this chapter we have constructed influence lines for beams, trusses and continuous beams. Clearly influence lines can be drawn for a wide variety of structures that carry moving loads. Their construction, whatever the structure, is based on considering the passage of a unit load across the structure.