07-Str-A1 · December 2018

Question 2 of 8: Reactions, Shear and Bending Moment Diagrams for Three Structures

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Reactions, Shear and Bending Moment Diagrams for Three Structures (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.

Approach. All three are determinate. In (a) the internal hinge splits the beam into two free bodies and the hinge condition supplies the fourth equation; in (b) the hinge at the top-right knee lets the loaded right column be treated as a 4 m span simply supported between the hinge and its base pin; in (c) the roller plane is parallel to the inclined member, so its reaction is normal to the member and its lever arm about the base pin is the member length itself.

2(a) — continuous beam with an internal hinge and a loaded overhang

Given. Pin at the left end, roller 8 m along, an internal hinge 2 m past that roller, a second roller 4 m past the hinge and a free end 2 m beyond, giving overall stations 0, 8, 10, 14, 16 m. A downward UDL of $w = 6$ kN/m is drawn from the left end as far as the second roller (0 to 14 m), and a downward point load of 20 kN acts at the free tip.

Find. The three reactions, and the shear and bending moment diagrams with their maximum and minimum ordinates.

1. Confirm determinacy and pick the free body that contains only one unknown reaction. With $r = 2 + 1 + 1 = 4$ and one hinge, $r - 3 - c = 4 - 3 - 1 = 0$, so statics is sufficient. Cutting at the hinge, the right-hand free body (10 m to 16 m) carries the UDL over its first 4 m, the 20 kN tip load and the reaction $C_y$, and its moment about the hinge involves $C_y$ alone: $$4C_y = (6)(4)(2) + (20)(6) = 48 + 120 = 168 \qquad C_y = \boxed{42.0\ \text{kN}\ \uparrow}$$
2. Recover the force carried through the hinge. Vertical equilibrium of the same free body gives the force the left-hand part must hand over: $$V_H = (6)(4) + 20 - 42.0 = 2.0\ \text{kN}$$ so the left-hand part pushes up on the suspended length with 2.0 kN, and by Newton's third law it feels 2.0 kN downward at the hinge.
3. Balance the left-hand free body. That piece spans 0 to 10 m, carries $6 \times 10 = 60$ kN of UDL at its centroid and the 2.0 kN hinge force at its right end. Moments about the pin give $$8B_y = (60)(5) + (2.0)(10) = 320 \qquad B_y = \boxed{40.0\ \text{kN}\ \uparrow}$$ and vertical equilibrium then gives $$A_y = 60 + 2.0 - 40.0 = \boxed{22.0\ \text{kN}\ \uparrow}$$ with $A_x = 0$ because nothing acts horizontally. The arithmetic check is the total: $22.0 + 40.0 + 42.0 = 104.0$ kN, equal to $6 \times 14 + 20$.
4. Assemble the shear diagram, bay by bay. Starting from $+22.0$ kN at the pin and shedding 6 kN per metre, $$V(x) = 22.0 - 6x \quad (0 \le x < 8) \qquad V(8^-) = -26.0\ \text{kN}$$ The roller adds 40.0 kN, so $V(8^+) = +14.0$ kN; the UDL then carries it down to $V(10) = +2.0$ kN at the hinge (the value found in step 2 — a useful independent check) and to $V(14^-) = -22.0$ kN. The second roller adds 42.0 kN, leaving a constant $+20.0$ kN over the unloaded overhang, which the tip load closes to zero. Shear vanishes at $x = 22.0/6 = 3.667$ m and again at $x = 8 + 14.0/6 = 10.333$ m.
5. Integrate for the bending moments (sagging taken positive throughout). In the first bay $M = 22.0x - 3x^2$, giving the largest sagging moment at the point of zero shear, $$M_{\max} = 22.0(3.667) - 3(3.667)^2 = \boxed{+40.3\ \text{kN}\cdot\text{m}}$$ and $M(8) = 176 - 192 = -16.0$ kN·m over the roller. Past the roller the moment climbs back to exactly zero at the hinge (as it must), peaks at a trivial $+0.33$ kN·m at $x = 10.333$ m, and falls to $$M(14) = 22.0(14) - 3(14)^2 + 40.0(6) = -\,40.0\ \text{kN}\cdot\text{m}$$ which is also the overhang check $-20 \times 2 = -40.0$ kN·m. The tip moment is zero.

2(b) — portal frame with a hinge at the top-right knee

Given. An 8 m by 4 m portal, pinned at both bases, with a rigid top-left knee $B$ and an internal hinge at the top-right knee $C$. The beam carries a downward UDL of 6 kN/m over its full 8 m and the right-hand column carries a horizontal UDL of 6 kN/m over its full 4 m, acting from right to left. Resultants are therefore 48 kN vertically and 24 kN horizontally.

Find. The four reaction components and the shear and moment diagrams for all three members.

1. Check the count, then read the hinge as a release that isolates the right column. $m = 3$, $j = 4$, $r = 4$, $c = 1$, so $i = 3(3) + 4 - 3(4) - 1 = 0$ and the frame is determinate. Because member $C\!-\!D$ carries its own transverse load but is moment-free at both ends — hinge above, pin below — it behaves exactly like a 4 m simply supported beam under a uniform 24 kN, so each end takes half: $$D_x = A_x' = \tfrac{1}{2}(6)(4) = \boxed{12.0\ \text{kN}}$$ where the 12.0 kN at the hinge is delivered through the beam.
2. Take global moments about the left base pin for the vertical reactions. With the beam resultant 48 kN acting at $x = 4$ m and the column resultant 24 kN acting leftwards at $y = 2$ m, $$8D_y = (48)(4) - (24)(2) = 192 - 48 = 144 \qquad D_y = \boxed{18.0\ \text{kN}\ \uparrow}$$ $$A_y = 48 - 18.0 = \boxed{30.0\ \text{kN}\ \uparrow}$$ Horizontal equilibrium closes the set: $A_x = 24.0 - 12.0 = 12.0$ kN, so both bases push back to the right with 12.0 kN.
3. Work up the left column. It carries no transverse load, so the shear is the constant 12.0 kN and the moment grows linearly from zero at the pin to $$M_B = (12.0)(4) = \boxed{48.0\ \text{kN}\cdot\text{m}}$$ while the axial force is a uniform 30.0 kN compression.
4. Analyse the beam with its known end conditions. The 12.0 kN horizontal force delivered to the hinge travels along the beam, so the beam carries 12.0 kN of axial compression. Its bending moment must be $-48.0$ kN·m at $B$ (equilibrium of the rigid joint) and zero at the hinge $C$; with the shear starting at $+30.0$ kN, $$M(x) = -48.0 + 30.0x - 3x^2$$ which indeed gives $M(8) = -48 + 240 - 192 = 0$. Zero shear is at $x = 30.0/6 = 5.0$ m and there $$M_{\max} = -48.0 + 150.0 - 75.0 = \boxed{+27.0\ \text{kN}\cdot\text{m}}$$ The moment changes sign where $x^2 - 10x + 16 = 0$, i.e. at $x = 2.0$ m (and again at the hinge). The shear runs from $+30.0$ kN at $B$ to $-18.0$ kN at $C$.
5. Finish with the right column. Measuring $h$ upwards from the base pin, the moment is zero at both ends and $$M(h) = 12.0h - 3h^2 \qquad M_{\max} = M(2) = \boxed{+12.0\ \text{kN}\cdot\text{m}}$$ The shear varies linearly from $+12.0$ kN at the base to $-12.0$ kN at the hinge, vanishing at mid-height, and the axial force is 18.0 kN compression. The joint check at $B$ is that the beam hogging moment (48.0 kN·m) and the column moment (48.0 kN·m) are equal and opposite, which they are.

2(c) — straight inclined member on a pin and an inclined roller

Given. A straight member rising 5 m over a horizontal span of 12 m, pinned at its lower-left end and bearing at its upper-right end on a hatched plane drawn parallel to the member. Two vertical downward loads act on it, 39 kN at a horizontal distance of 4 m from the pin and 26 kN at 8 m; the span is dimensioned as three equal 4 m segments.

Find. The reactions and the shear and bending moment diagrams, with the maximum and minimum ordinates.

1. Exploit the geometry before writing any equation. The member length is $$L = \sqrt{12^2 + 5^2} = 13.0\ \text{m}$$ a 12-5-13 triangle, so the direction cosines are exactly $\hat{u} = (12/13,\ 5/13)$ along the member and $\hat{n} = (-5/13,\ 12/13)$ normal to it. A reaction normal to the member has the member length as its lever arm about the base pin, because the perpendicular from the pin to that reaction's line of action is the member itself rotated through a right angle.
2. Take moments about the pin for the roller reaction. Using that lever arm, $$13R = (39)(4) + (26)(8) = 156 + 208 = 364 \qquad R = \boxed{28.0\ \text{kN}}$$ directed normal to the member, i.e. with components $(-28 \times 5/13,\ 28 \times 12/13) = (-10.77,\ 25.85)$ kN. Force equilibrium then gives the pin reaction $$A = (10.77,\ 39.15)\ \text{kN} \qquad |A| = 40.6\ \text{kN}$$ The fact that $364/13$ is exactly 28 is the confirmation that the inclined-plane reading of the roller is the intended one.
3. Note why the alternative roller reading does not change the diagrams. Had the plane been horizontal, moments about the pin would give $R_v = 364/12 = 30.33$ kN and $A = (0,\ 34.67)$ kN. The difference between the two pin reactions is $(10.77,\ 4.49) = 0.897 \times (12,\ 5)$, a force acting purely along the member axis, which contributes neither shear nor bending moment anywhere. Only the reaction components and the axial force distinguish the two readings, so the diagrams below are unaffected.
4. Resolve the internal actions segment by segment. Projecting the resultant of all forces below each cut onto $\hat{n}$ and $\hat{u}$, and measuring distance $s$ along the member (each 4 m of horizontal span is $4 \times 13/12 = 4.333$ m of member), $$V = +32.0\ \text{kN} \ \ (0 < s < 4.333) \qquad V = -4.0\ \text{kN} \ \ (4.333 < s < 8.667) \qquad V = -28.0\ \text{kN} \ \ (8.667 < s < 13)$$ $$N = 25.0,\ 10.0,\ 0\ \text{kN compression respectively}$$ The last shear equals the roller reaction and the last axial force is exactly zero, both of which are automatic checks that the resolution is right; the fact that every one of these six numbers is an integer is the second confirmation of the roller reading.
5. Integrate the shear along the member for the moments. Because the shear is piecewise constant, $M$ is piecewise linear in $s$: $$M(4.333) = (32.0)(4.333) = \boxed{+138.7\ \text{kN}\cdot\text{m}}$$ $$M(8.667) = 138.7 - (4.0)(4.333) = +121.3\ \text{kN}\cdot\text{m}$$ $$M(13) = 121.3 - (28.0)(4.333) = 0$$ so the maximum moment sits under the 39 kN load, the minimum ordinate on the member itself is zero at each end, and the moment nowhere changes sign. Equivalently, in terms of horizontal distance, $M = 34.67x$ up to $x = 4$ m and the peak is $416/3 = 138.7$ kN·m.