Question 2 of 8: Reactions, Shear Force and Bending Moment Diagrams

Question 2: Reactions, Shear Force and Bending Moment Diagrams (18 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.

(a) Gerber beam with one internal hinge

Given.

Find. The three reaction components, then the complete shear and bending moment diagrams with the maximum positive and negative ordinates labelled.

Approach. Four unknown reaction components against three equations of equilibrium plus one condition of zero moment at the hinge: cut the beam at the hinge, solve the short left segment first, then pass the hinge force onto the right segment.

1. Take moments about the hinge for the left segment. The segment from $x = 0$ to $x = 6$ carries only the reaction $A_y$ and the 30 kN load, and the hinge cannot transmit moment, so $$\sum M_{\text{hinge}} = A_y (6) - 30 (6-4) = 0 \;\Longrightarrow\; \boxed{A_y = 10.0\ \text{kN}\ \uparrow}$$ The pin also supplies a horizontal component, which is zero because no horizontal load is applied.
2. Extract the force carried through the hinge. Vertical equilibrium of the same segment gives the force the left half hands to the right half: $$V_{\text{hinge}} = 30 - A_y = 30 - 10 = 20.0\ \text{kN}$$ acting downward on the right segment at $x = 6\ \text{m}$.
3. Solve the right segment. It runs from $x = 6$ to $x = 16$, carries the 20.0 kN hinge force at its left end and a resultant $W = wL = 4(8) = 32\ \text{kN}$ acting at $x = 12\ \text{m}$, and is supported by the rollers at C and D. Taking moments about C, $$D(16-8) - 32(12-8) + 20(8-6) = 0 \;\Longrightarrow\; \boxed{D = 11.0\ \text{kN}\ \uparrow}$$ and vertical equilibrium of the same segment then gives $$\boxed{C = 20 + 32 - 11 = 41.0\ \text{kN}\ \uparrow}$$ As a check, $A_y + C + D = 10 + 41 + 11 = 62\ \text{kN}$, which is exactly the applied total $30 + 32$.
4. Build the shear diagram. Working from the left, $V = +10.0$ kN up to the point load, $V = 10 - 30 = -20.0$ kN from the load through the hinge to C, then C lifts it to $V = -20 + 41 = +21.0$ kN, after which the uniform load drives it down at $4\ \text{kN}/\text{m}$ to $V = 21 - 32 = -11.0$ kN at D. The maximum positive and negative ordinates are therefore $$\boxed{V_{\max}^{+} = +21.0\ \text{kN}\ \text{just right of C},\quad V_{\max}^{-} = -20.0\ \text{kN}\ \text{between the load and C}}$$
5. Integrate for the bending moments. The moment rises linearly to $M(4) = 10(4) = +40.0\ \text{kN}\cdot\text{m}$ under the point load, falls to zero at the hinge as it must, and continues down to $M(8) = -20(2) = -40.0\ \text{kN}\cdot\text{m}$ over C. Beyond C the shear is $V(x) = 21 - 4(x-8)$, which vanishes at $$x = 8 + \frac{21}{4} = 13.25\ \text{m}$$ and the sagging peak inside the right span is $$M(13.25) = -40 + 21(5.25) - 2(5.25)^2 = +15.125\ \text{kN}\cdot\text{m}$$ so that $$\boxed{M_{\max}^{+} = +40.0\ \text{kN}\cdot\text{m}\ \text{at }x=4\ \text{m}, \quad M_{\max}^{-} = -40.0\ \text{kN}\cdot\text{m}\ \text{at C}}$$ The moment returns to zero at D, confirming the arithmetic.

(b) Overhanging beam with a partial uniform load

Given. A 12 m beam with the origin at the free left end: pin B at $x = 2\ \text{m}$, roller D at $x = 12\ \text{m}$, a 44 kN downward point load at $x = 7\ \text{m}$, and $w = 12\ \text{kN}/\text{m}$ running from the free end to the point load, i.e. over $0 \le x \le 7\ \text{m}$. The load block on the drawing begins at the very tip of the overhang and stops exactly under the 44 kN arrow, so the loaded length is 7 m and not the 5 m span between the supports.

Find. Both reactions, and the shear and bending moment diagrams with their extreme ordinates.

Approach. The beam is simply supported with one overhang, so ordinary statics is enough; replace the distributed load by its resultant for the reaction calculation, then return to the distributed form to integrate the shear.

1. Replace the uniform load by its resultant. $$W = wL_w = 12(7) = 84.0\ \text{kN}\quad\text{acting at } \bar{x} = \tfrac{7}{2} = 3.5\ \text{m}$$
2. Take moments about the pin to find the far reaction. $$\sum M_B = D(12-2) - 84(3.5-2) - 44(7-2) = 0$$ $$10 D = 126 + 220 = 346 \;\Longrightarrow\; \boxed{D = 34.6\ \text{kN}\ \uparrow}$$
3. Close vertical equilibrium for the pin reaction. $$\boxed{B = 84 + 44 - 34.6 = 93.4\ \text{kN}\ \uparrow}$$ with no horizontal component.
4. Trace the shear. On the overhang the shear falls linearly to $V(2^-) = -12(2) = -24.0$ kN; the pin lifts it to $V(2^+) = -24 + 93.4 = +69.4$ kN, and the uniform load then reduces it to $V(7^-) = 69.4 - 12(5) = +9.4$ kN. Since the shear has not reached zero before the point load, there is no interior sagging peak in that stretch. The 44 kN load drops it to $V(7^+) = 9.4 - 44 = -34.6$ kN, which then holds constant to the roller. Hence $$\boxed{V_{\max}^{+} = +69.4\ \text{kN},\qquad V_{\max}^{-} = -34.6\ \text{kN}}$$
5. Integrate for the moments. Over the overhang $M(x) = -6x^2$, giving the hogging peak $M(2) = -24.0\ \text{kN}\cdot\text{m}$ at the pin. Between the supports the shear stays positive, so the moment climbs monotonically to its maximum under the point load: $$M(7) = -6(7)^2 + 93.4(7-2) = -294 + 467 = +173.0\ \text{kN}\cdot\text{m}$$ Beyond the load the shear is constant at $-34.6$ kN, so the moment runs straight back to $173 - 34.6(5) = 0$ at the roller, as it must. Therefore $$\boxed{M_{\max}^{+} = +173.0\ \text{kN}\cdot\text{m}\ \text{at } x = 7\ \text{m},\quad M_{\max}^{-} = -24.0\ \text{kN}\cdot\text{m}\ \text{at the pin}}$$

(c) Inclined member on a pin and an inclined roller

Given. A straight member running from a pin at $(0,\,12)$ down to a roller at $(16,\,0)$ — a 3:4 slope, marked as such on the drawing, so the member length is $L = \sqrt{16^2 + 12^2} = 20.0\ \text{m}$. Vertical point loads of 5 kN and 7 kN act at horizontal offsets of 6 m and 10 m from the pin, i.e. at 7.5 m and 12.5 m measured along the member. The roller hatching at the lower end is drawn parallel to the member, so its reaction is taken normal to the member axis.

Find. The reactions, and the shear and bending moment diagrams for the inclined member with the extreme ordinates.

Approach. Resolve along and normal to the member axis. The useful geometric fact is that a reaction normal to the member has the whole member length as its lever arm about the pin, which delivers the reaction in one line.

1. Take moments about the pin. With the reaction $R_n$ normal to the member, its lever arm about the pin is the member length itself, 20 m, while the applied loads act at horizontal offsets of 6 m and 10 m: $$R_n (20) = 5(6) + 7(10) = 100 \;\Longrightarrow\; \boxed{R_n = 5.0\ \text{kN}}$$
2. Resolve the roller reaction into components. The unit normal to a 3:4 member is $(0.6,\ 0.8)$, so $$\mathbf{R}_{\text{roller}} = 5.0\,(0.6,\ 0.8) = (3.0,\ 4.0)\ \text{kN}$$ i.e. 3.0 kN horizontal and 4.0 kN vertical.
3. Close global equilibrium at the pin. $$\begin{aligned} A_x &= -3.0\ \text{kN} \quad (3.0\ \text{kN acting to the left})\\ A_y &= 5 + 7 - 4.0 = 8.0\ \text{kN} \quad (\text{upward}) \end{aligned}$$ so $\boxed{\mathbf{A} = (-3.0,\ +8.0)\ \text{kN},\ |\mathbf{A}| = 8.544\ \text{kN}}$
4. Resolve the forces normal to the member for the shear. Taking $s$ along the member from the pin and dotting each force with the unit normal $(0.6,\ 0.8)$, $$\begin{aligned} V &= (-3.0)(0.6) + (8.0)(0.8) = +4.6\ \text{kN} && 0 < s < 7.5\ \text{m}\\ V &= 4.6 - 5(0.8) = +0.6\ \text{kN} && 7.5 < s < 12.5\ \text{m}\\ V &= 0.6 - 7(0.8) = -5.0\ \text{kN} && 12.5 < s < 20\ \text{m} \end{aligned}$$ so $\boxed{V_{\max}^{+} = +4.6\ \text{kN},\quad V_{\max}^{-} = -5.0\ \text{kN}}$
5. Integrate the shear along the member for the moment. The moment grows at $4.6\ \text{kN}$ per metre of member to $M = 34.5\ \text{kN}\cdot\text{m}$ under the 5 kN load, then at $0.6\ \text{kN}$ per metre to its peak under the 7 kN load, $$M_{\max} = 34.5 + 0.6(5.0) = +37.5\ \text{kN}\cdot\text{m} \quad\text{at } s = 12.5\ \text{m}$$ and finally falls at $5.0\ \text{kN}$ per metre back to zero at the roller, $37.5 - 5.0(7.5) = 0$. The member sags everywhere, so $$\boxed{M_{\max}^{+} = +37.5\ \text{kN}\cdot\text{m},\qquad M_{\max}^{-} = 0}$$
6. Record the axial force, which is what makes this member unusual. Dotting the same forces with the unit vector along the member, $(0.8,\ -0.6)$, the member carries $7.2\ \text{kN}$ tension above the 5 kN load, $4.2\ \text{kN}$ tension between the loads and zero below the 7 kN load — it hangs from the pin, and the normal roller adds no axial force at all.