Question 1 of 9: Schematic Shear and Bending Moment Diagrams for Three Structures

Question 1: Schematic Shear and Bending Moment Diagrams for Three Structures (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three schematic structures, all members of constant $EI$ and axially inextensible. Only structure (a) carries printed dimensions (in multiples of $L$ and $w$); structures (b) and (c) are dimensioned symbolically, so their ordinates are reported as closed forms in $P$, $a$, $b$ (clear span), $h$ (column height) and the storey heights $h_1,h_2$.

Find. The shear force and bending moment diagrams of each structure, with the governing ordinates, the points of contraflexure and the qualitative shape stated.

Approach. Count the degree of static indeterminacy, exploit symmetry or anti-symmetry to collapse the unknowns, solve the reduced compatibility (structure a) or slope-deflection (structures b and c) equations, then plot $V$ and $M$ from statics.

1. Structure (a) — classify and choose redundants. The beam carries two roller reactions and a built-in end, so $r=1+1+3=5$ against three equations of equilibrium: the beam is $5-3=\boxed{2}$ degrees statically indeterminate. Taking the two roller reactions $R_B$ and $R_C$ as redundants leaves a cantilever fixed at D as the primary structure, and the compatibility conditions are simply that the deflection vanishes at B and at C.
2. Solve the two compatibility equations. With $f_{ij}=\int m_i m_j\,\mathrm{d}x/EI$ and $\Delta_{i0}=\int M_0 m_i\,\mathrm{d}x/EI$ evaluated over the full $7L/3$ length, the flexibility equations $f_{ij}R_j+\Delta_{i0}=0$ give $$R_B=\frac{265}{378}\,wL=0.7011\,wL,\qquad R_C=\frac{125}{378}\,wL=0.3307\,wL.$$ Vertical equilibrium of the whole beam, whose total load is $wL$, then fixes the reaction at the built-in end, $$R_D=wL-R_B-R_C=-\tfrac{2}{63}\,wL=\boxed{-0.0317\,wL},$$ a hold-down: the last span is so lightly loaded that the built-in end must pull the beam down.
3. Bending moment ordinates for (a). Working from the free end with $M(x)=\sum R_i\langle x-x_i\rangle-\sum w\langle\cdot\rangle^{2}/2$, the governing sagging ordinates are $$M_B=-\frac{wL^{2}}{18}=-0.0556\,wL^{2},\qquad M_C=-\frac{4wL^{2}}{189}=-0.0212\,wL^{2},\qquad M_D=+\frac{2wL^{2}}{189}=+0.0106\,wL^{2}.$$ Inside the loaded strip of span BC the shear vanishes at $x=1.0344L$ from the free end, where the moment reaches its largest sagging value $\boxed{M_{\max}=+0.0235\,wL^{2}}$. The diagram crosses zero at $x=0.546L$, $1.251L$ and exactly $2.000L$ — the last of these is the point of contraflexure in the unloaded span CD, whose moment varies linearly because the shear there is constant.
4. Shear ordinates for (a). The overhang shear grows linearly to $-wL/3=-0.333\,wL$ just left of B, jumps by $R_B$ to $+0.368\,wL$, falls linearly to $+0.034\,wL$ where the first load block ends, stays constant to $x=L$, falls again to $-0.299\,wL$ just left of C, jumps by $R_C$ to $+0.032\,wL$ and stays constant to D. The constant shear over the whole of CD is the numerical statement that span CD carries no load at all.

Structure (b) is the classic anti-symmetry problem. An upward $P$ at the left tip mirrors into a downward $P$ at the right tip, so the loading is anti-symmetric about the centre line and the response must be anti-symmetric too: the two joints rotate through the same angle $\theta$, both columns sway through the same $\Delta$, and neither column top can translate vertically because the columns are inextensible.

1. Prove the column shear is zero. There is no net horizontal load, so the two base shears sum to zero; anti-symmetry makes them equal, hence each is zero and $M_{12}+M_{21}=0$ in every column. Substituting the slope-deflection expressions for a fixed-base column of height $h$, $$\frac{2EI}{h}\left(3\theta-6\psi\right)=0\;\Longrightarrow\; \psi=\frac{\theta}{2},\qquad M_{21}=\frac{EI\theta}{h}=-M_{12}.$$
2. Joint equilibrium. The cantilevered tips are determinate: each delivers a shear $P$ and a couple $Pa$ to its joint. With $\theta_2=\theta_3=\theta$ and no chord rotation in the clear span, $M_{23}=M_{32}=6EI\theta/b$, and moment equilibrium of joint 2 gives $$\frac{EI\theta}{h}+\frac{6EI\theta}{b}=-Pa \;\Longrightarrow\;EI\theta=-\frac{Pabh}{b+6h}.$$
3. Ordinates for (b). Hence $$\boxed{M_{\text{column}}=\frac{Pab}{b+6h}}\quad\text{(equal and opposite at the two ends of each column)},\qquad \boxed{M_{\text{span end}}=\frac{6Pah}{b+6h}},$$ with $M=\pm Pa$ where the overhangs meet the joints and a linear span diagram that passes through zero at midspan. The overhang shear is $P$, the clear-span shear is the constant $12Pah/[b(b+6h)]$, the column shear and the horizontal reactions are exactly zero, and the base moments are $Pab/(b+6h)$. As a check the two contributions add to the applied couple: $Pab/(b+6h)+6Pah/(b+6h)=Pa$.

1. Structure (c) — kill the sway first. Structure and loading are both symmetric, so the horizontal translation of every joint is zero and all chord rotations vanish. The joint rotations are equal and opposite, $\theta_3=-\theta_2$, which reduces the intermediate beam to the modified stiffness $2EI/L$ and leaves a single unknown $\theta_2$.
2. Assemble the one joint equation. The lower column contributes $4EI/h_1$ (fixed base), the upper column contributes the pinned-end value $3EI/h_2$, the beam contributes $2EI/L$ and the midspan load supplies $\mathrm{FEM}=PL/8$: $$\theta_2=-\frac{PL/8}{\dfrac{4EI}{h_1}+\dfrac{3EI}{h_2}+\dfrac{2EI}{L}}.$$ Taking the drawing's near-square proportions $h_1=h_2=L=h$ as a representative case gives $EI\theta_2=-Ph^{2}/72$ and the ordinates $$\boxed{M_{\text{beam end}}=-\tfrac{7}{72}Ph},\qquad M_{\text{beam midspan}}=+\tfrac{11}{72}Ph,\qquad M_{\text{lower col, top}}=-\tfrac{1}{18}Ph,\qquad M_{\text{lower col, base}}=+\tfrac{1}{36}Ph.$$
3. The upper storey is the informative part. Because the top beam is pin-connected at both ends it is simply supported: its moment diagram is the plain triangle peaking at $\boxed{PL/4}$ and it delivers only $P/2$ vertically to each column top, with no moment. The upper column therefore carries a moment that falls linearly from $3EI\theta_2/h_2=-Ph/24$ at the intermediate joint to exactly zero at the pin, and a constant shear $P/24$ per unit $h$; the two upper column shears are equal and opposite and are equilibrated by axial force in the top beam. The lower columns carry $P/12$ of shear, again equal and opposite.