Study Companion: S20 — Beams: External Effects
A beam is the most common structural member there is: a shelf, a floor joist, a bridge girder, a crane jib. It is long and straight, and it is loaded across its length, often by a load spread along it rather than by a few point forces. This lesson does the first half of a beam analysis, the external half: replace each distributed load by its resultant, using the centroids of S18–S19, and find the support reactions by the equilibrium of S11. The second half, the forces inside the beam, is S21.
Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 5/6 ("Beams—External Effects"), with Sample Problems 5/11–5/12. Harper, Solving Statics Problems in MATLAB, "Beams-External Effects" (Problem 5/105, which is Problem 5/99 in Meriam's 9th edition). Meriam's Sample Problem 5/11 and Harper's problem are in US units, converted here: 1 lbf = 4.448\,\mathrm N, 1 ft = 0.3048\,\mathrm m, so 1 lb/ft = 14.59\,\mathrm{N/m}. Axes: x along the beam to the right, y up; moments counter-clockwise positive on every free-body diagram.
How to use this companion
Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.
The puzzle
A shelf plank 1.5\,\mathrm m long rests on a bracket at each end. Books are stacked on it more and more heavily toward the right, so that the load rises steadily from nothing at the left bracket to 800\,\mathrm{N/m} at the right one.
How much does each bracket carry? The plausible guess is "the total load, split equally between the two brackets". The answer is in Nugget 03.
Nugget 01: A Beam Is Determinate Only If Statics Alone Fixes Its Reactions
Claim. A beam whose support reactions can be found from the equilibrium equations alone is statically determinate. One with more supports than equilibrium needs is statically indeterminate: its reactions depend on how the beam deflects, and statics cannot find them. Count the unknowns before solving.
The picture. Meriam's Fig. 5/18.
A pin and a roller give three unknowns (A_x, A_y, B), a fixed end three (A_x, A_y, M_A): three equations solve either. A combination beam, a cantilever and a simple span joined by a pin, has four unknowns, but the pin adds a fourth equation (no moment can pass through it). Add one more support to any of them and the count exceeds the equations.
Check with numbers. Unknown reactions against the three equations of a plane body: simple 2 + 1 = 3; cantilever 3; continuous over three supports 2 + 1 + 1 = 4; end-supported cantilever 3 + 1 = 4; fixed at both ends 3 + 3 = 6. The last three exceed 3.
The idea. Meriam (5/6): beams are structural members that resist bending under loads applied, usually, normal to their axes. Analysing one has two parts. First, the equilibrium of the beam as a whole and of any portion of it, which is statics. Second, the relation between those forces and the beam's internal resistance, which needs the strength of the material and belongs to mechanics of materials. This article treats the external loads and reactions; Article 5/7 (S21) the internal force and moment along the beam. Statically determinate beams (simple, cantilever, combination) are solved by statics alone; statically indeterminate beams (continuous, end-supported cantilever, fixed) need their load–deformation properties as well, and only determinate beams are treated here. This is S11's count of unknowns against equations, applied to beams.
The equation (physical law: S11's planar equilibrium of the whole beam):
Worked number. A footbridge plank on three trestles carrying 6\,\mathrm{kN}: four unknowns, three equations. The middle trestle's share depends on how much the plank sags, which statics cannot see.
Real example (one you can see). A garden plank bridging a ditch on two supports is determinate; lay it on three bricks and one brick often rocks loose, carrying nothing: the indeterminate beam has settled its own reactions by deflection.
Plausible wrong answer. "Three supports share the load equally." Equilibrium cannot say so; the shares depend on stiffness and on the supports' exact heights.
Check. A cantilever has a roller added under its free end. Determinate? (No: four unknowns, three equations.)
Narration spine. Six beams in two columns. Reaction arrows appear on each, with a counter: 3, 3, 4 (plus a pin equation), then 4, 4, 6 in red. The red column's beams flex slightly, showing that deflection is what fixes their reactions.
Bridge. The loads on beams are often spread out. They need a description.
Nugget 02: A Distributed Load Is Described by Its Intensity Along the Beam
Claim. A load spread along a beam is described by its intensity w(x), force per unit length (N/m). It may be constant or vary, and it may jump: either w itself changes abruptly, or only its slope dw/dx does.
The picture. Meriam's Fig. 5/19.
From A to C the intensity grows; from C to D it is constant; at D it drops at once; from D to B it falls to zero.
Check with numbers. A 200\,\mathrm{mm} thick concrete floor slab, 2400\,\mathrm{kg/m^3}, 3\,\mathrm m wide, resting on one beam: per metre of beam it weighs 2400(9.81)(0.2)(3) = 14.1\times10^{3}\,\mathrm{N}, so w = 14.1\,\mathrm{kN/m}, constant. Where a partition wall sits on part of the span, w jumps there.
The idea. Meriam (5/6): beams may be classified by their loading as well as by their supports. The intensity w of a distributed load is force per unit length of beam. It may be constant or variable, continuous or discontinuous. In Fig. 5/19 the intensity is constant from C to D and variable from A to C and from D to B; it is discontinuous at D. At C the intensity itself is continuous, but its rate of change dw/dx is not. (S21 shows why these distinctions matter: they shape the shear and moment diagrams.)
The equation (definition of load intensity):
Worked number. A uniform w = 14.1\,\mathrm{kN/m} over a 5\,\mathrm m span is 70.6\,\mathrm{kN} in total.
Real example. Snow on a flat roof, the weight of the beam itself, a row of stacked pallets: each is a distributed load, and building codes specify design loads in kN/m² of floor, which become kN/m on each beam.
Plausible wrong answer. "A load with a kink at C is discontinuous there." Only its slope is; w is continuous at C.
Check. Water standing 0.1\,\mathrm m deep on a flat roof 2\,\mathrm m wide per beam. What is w on the beam? (1000(9.81)(0.1)(2) = 1962\,\mathrm{N/m}.)
Narration spine. Arrows rain onto a beam, crowding where the load is heavier; their tips trace the curve w(x). At C the curve bends sharply; at D it steps down; both points are marked with what changes.
Bridge. For the reactions, all those arrows can be replaced by one.
Nugget 03: The Resultant Is the Area of the Load Diagram, Through Its Centroid
Claim. The resultant of a distributed load equals the area under its load diagram w(x), and acts through the centroid of that area. A uniform load w over L gives wL at L/2; a triangular one gives \frac12wL at \frac23L from the zero end.
The picture. Meriam's Fig. 5/20.
A trapezoid is split into a rectangle and a triangle, each replaced by its own resultant; a single combined resultant is rarely needed.
Check with numbers (the puzzle). The books' load is a triangle, 0 to 800\,\mathrm{N/m} over 1.5\,\mathrm m: R = \frac12(800)(1.5) = 600\,\mathrm N, at \frac23(1.5) = 1.0\,\mathrm m from the left bracket. Moments about A, counter-clockwise positive: R_B(1.5) - 600(1.0) = 0, so R_B = 400\,\mathrm N; \Sigma F_y: R_A = 600 - 400 = 200\,\mathrm N. The right bracket carries twice the left one, not 300\,\mathrm N each.
The idea. Meriam (5/6): constant or linearly varying loads are easily handled (Fig. 5/20). For (a) and (b) the resultant R is represented by the area formed by the intensity w and the length L, and passes through the centroid of that area. For (c), the trapezoid is divided into a rectangle, R_1 = w_1L at L/2, and a triangle, R_2 = \frac12(w_2 - w_1)L at \frac23L; one resultant could be found by the composite method of 5/4, but it is usually unnecessary. For a general distribution (Fig. 5/21) start from dR = w\,dx: R = \int w\,dx, and the principle of moments, R\bar x = \int xw\,dx, places it at the centroid of the area under w(x). Its vertical coordinate is never needed. The load diagram's area and centroid are exactly the area and centroid of S18–S19: the "area" just carries units of newtons.
The equation (mathematical identity: S06's resultant of parallel forces, by the principle of moments):
Assumption: the resultant replaces the load for the external effects (reactions) only; inside the beam it does not (S21).
Worked number. A trapezoidal load from 2 to 5\,\mathrm{kN/m} on a 6\,\mathrm m simple beam: R_1 = 2(6) = 12\,\mathrm{kN} at 3\,\mathrm m, R_2 = \frac12(3)(6) = 9\,\mathrm{kN} at 4\,\mathrm m. R_B = [12(3) + 9(4)]/6 = 12\,\mathrm{kN}, R_A = 21 - 12 = 9\,\mathrm{kN}.
Real example (one you can see). Pile books along a shelf heavier at one end and press on the brackets: the bracket under the heavy end is the one that creaks. A builder loading a trailer puts the heavy end of the load toward the axle for the same reason.
Plausible wrong answer. "The two brackets share the books equally." Only if the load diagram is symmetric about the middle; here its centroid is at 1.0\,\mathrm m, and the shares are 200 and 400\,\mathrm N.
Check. Where does a triangular load act if it is largest at the left end? (A third of the span from the left: \frac23L from its zero end.)
Narration spine. The books' load arrows merge: the triangle fills with colour, its area counts up to 600\,\mathrm N, and one arrow drops through its centroid at 1.0\,\mathrm m. The brackets' reaction arrows grow to 200 and 400\,\mathrm N as a balance beam levels.
Bridge. With every load replaced, the beam is an ordinary rigid body.
Nugget 04: Replace the Loads, Then Solve the Beam Like Any Rigid Body
Claim. Once the distributed loads are replaced by their equivalent concentrated loads, the reactions follow from S11's equilibrium equations. Take moments about each support in turn so each equation has one unknown, and check the sign of every reaction against what its support can actually supply.
The picture. Meriam's Sample Problem 5/11, in SI.
The 3.048\,\mathrm m (10 ft) beam carries 1751\,\mathrm{N/m} (120 lb/ft) everywhere, plus a triangle growing over the last 1.829\,\mathrm m (6 ft) to a total of 4086\,\mathrm{N/m} (280 lb/ft) at B.
Check with numbers. The rectangle: R_1 = 1751(3.048) = 5338\,\mathrm N (1200 lb) at 1.524\,\mathrm m. The triangle, height 2335\,\mathrm{N/m} (160 lb/ft): R_2 = \frac12(2335)(1.829) = 2135\,\mathrm N (480 lb) at 1.219 + \frac23(1.829) = 2.438\,\mathrm m from A.
Check: 3096 + 4377 = 7473\,\mathrm N = 5338 + 2135.
The idea. Meriam (5/6): once the distributed loads have been reduced to equivalent concentrated loads, the external reactions follow by the static analysis of Chapter 3. Meriam's hint to SP 5/11: it is usually unnecessary to reduce a distributed load to a single concentrated load. (Here the single resultant would be 7473\,\mathrm N at 1.785\,\mathrm m, which gives the same reactions with no saving.)
One-sided supports. A roller, a bracket or a beam resting on a wall can only push. If a reaction comes out negative, the beam would lift off that support, and the support must be a tie or a hold-down. A beam on supports at 0 and 3\,\mathrm m with 2\,\mathrm{kN/m} on an overhang from 3 to 4\,\mathrm m gives R_B = 2(1)(3.5)/3 = 2.333\,\mathrm{kN} and R_A = 2 - 2.333 = -0.333\,\mathrm{kN}: end A must be held down, or it lifts.
The equation (physical law: equilibrium of the beam with the equivalent loads):
Worked number. The single-resultant route for SP 5/11: R = 7473\,\mathrm N at \bar x = [5338(1.524) + 2135(2.438)]/7473 = 1.785\,\mathrm m; R_B = 7473(1.785)/3.048 = 4377\,\mathrm N, the same.
Real example. A diving board is a beam on two supports with its load on the overhang: the back support is pulled up and is bolted down, while the front fulcrum is pushed, carrying more than the diver's weight.
Plausible wrong answer. "A negative reaction just means the arrow was drawn the wrong way." For a pin it does; for a roller or a beam resting on a wall it means the support cannot do the job.
Check. In SP 5/11, why use \Sigma M_B = 0 for R_A rather than \Sigma F_y = 0? (Each moment equation has one unknown; \Sigma F_y is then a free check.)
Narration spine. The load diagram splits into a blue rectangle and an orange triangle; each collapses to one arrow at its centroid. The beam lifts off its supports into a free-body diagram; moment arms sweep from A and from B, and the reactions fill in at 3096 and 4377\,\mathrm N.
Bridge. Not every load is a rectangle or a triangle. Some are formulas.
Nugget 05: A Load Given by a Formula Is Integrated Directly
Claim. When w(x) is a function, R = \int w\,dx and \bar x = \int xw\,dx/R are evaluated as integrals; then the reactions follow as before. Mind the units of the constants in w.
The picture. Meriam's Sample Problem 5/12: an 8\,\mathrm m cantilever under w = w_0 + kx^3.
At the wall w = 1000\,\mathrm{N/m}; at the free end 2024\,\mathrm{N/m}. The wall supplies a vertical force and a couple.
Check with numbers. w(0) = w_0 = 1000\,\mathrm{N/m} and w(8) = 1000 + k(512) = 2024 give k = 2\,\mathrm{N/m^4}.
On the cantilever, counter-clockwise positive: \Sigma M_A: M_A - 10\,048(4.49) = 0, M_A = 45\,100\,\mathrm{N\,m} (counter-clockwise); \Sigma F_y: A_y = 10\,048\,\mathrm N; A_x = 0 by inspection. (Meriam rounds R to 10\,050\,\mathrm N; \bar x and M_A are the same to three figures.)
The idea. Meriam's hints to SP 5/12: use caution with the units of the constants, w_0 in N/m and k in N/m⁴, so that kx^3 is in N/m; and recognise that computing R and \bar x is simply the centroid calculation of Article 5/3 applied to the area under w(x).
The equation (mathematical identity: the resultant of a load given by a function):
Worked number. For a cantilever the moment at the wall is simply the first moment of the load: M_A = \int_0^8x(1000 + 2x^3)\,dx = 45\,107\,\mathrm{N\,m}, without finding \bar x at all.
Real example. The wind on a tall sign or a flagpole grows with height roughly as a power of the height; the base moment is the integral of x times that load, exactly as here.
Plausible wrong answer. "The load acts at mid-span, 4\,\mathrm m, since the beam is 8\,\mathrm m long." The load is heavier near the tip, so \bar x = 4.49\,\mathrm m, and M_A is 12\,\% more than 10\,048(4).
Check. What are the units of k in w = w_0 + kx^3? (N/m⁴.)
Narration spine. The load curve grows along the cantilever, its arrows lengthening toward the tip. An integral sign sweeps along it summing to 10\,048\,\mathrm N; the single arrow settles at 4.49\,\mathrm m; a curved arrow at the wall grows to 45\,100\,\mathrm{N\,m}.
Bridge. Harper keeps the load as a parameter and lets the reactions follow.
Nugget 06: Keep the Load as a Parameter and the Reactions Become Functions
Claim. If the load's shape depends on a parameter, the reactions are functions of it. For Harper's beam they are linear in the end load w_0, and the plot shows at once which support is more sensitive to it.
The picture. Harper's Problem 5/105, in SI: a 3.048\,\mathrm m beam whose load is w = k_1x - k_2x^2.
The load starts at zero, rises at 2394\,\mathrm{N/m} per metre (50 lb/ft per foot) at A, and reaches w_0 at B. With w_0 = 0 it is a symmetric arch; as w_0 grows the load leans toward B.
Check with numbers. The two conditions fix the constants: dw/dx(0) = k_1 = 2394\,\mathrm{N/m^2}, and w(L) = k_1L - k_2L^2 = w_0 gives k_2 = (k_1L - w_0)/L^2. Moments about A and the sum of forces:
With k_1L^2/12 = 2394(3.048)^2/12 = 1853.4\,\mathrm N: R_A = 1853.4 + 0.254w_0 and R_B = 1853.4 + 0.762w_0 (N, with w_0 in N/m). At w_0 = 0 both are 1853.4\,\mathrm N (416.7 lb); at w_0 = 4378\,\mathrm{N/m} (300 lb/ft), R_A = 2965.5\,\mathrm N (667 lb) and R_B = 5189.6\,\mathrm N (1167 lb), Meriam's answers to its Problem 5/99.
The idea. Harper (5.2) builds w(x) from its boundary conditions, takes moments about A with the load discretised into forces dF = w\,dx at distance x, integrates symbolically, and plots w(x) for several w_0 and the reactions against w_0. Both reactions rise linearly; R_B three times as fast as R_A, because the extra load w_0 adds a triangle-like piece whose centroid is near B.
The equation (mathematical identity: reactions as functions of the load parameter):
Worked number. At w_0 = 2\,\mathrm{kN/m}: R_A = 1853.4 + 0.254(2000) = 2361.4\,\mathrm N and R_B = 1853.4 + 0.762(2000) = 3377.4\,\mathrm N.
Real example. A road bridge is checked for every position and size of its traffic load; engineers tabulate each bearing's reaction as a function of the load, as here, and design each bearing for its own maximum.
Plausible wrong answer. "Raising w_0 adds equally to both supports." It adds three times as much to B as to A.
Check. For which w_0 is the load linear (k_2 = 0)? (w_0 = k_1L = 7297\,\mathrm{N/m}, 500 lb/ft.)
Narration spine. The load arch on the beam; a slider for w_0 lifts its right end and the arch leans over. Beside it two reaction bars grow, the right one three times faster, while the straight lines of the right-hand plot draw themselves.
Bridge. The reactions hold the beam up. What happens inside it, section by section, is S21.
Synthesis
The shelf's brackets carry 200\,\mathrm N and 400\,\mathrm N, not 300\,\mathrm N each. The books' load was a triangle; its area gave the total, 600\,\mathrm N, and its centroid, two-thirds of the way toward the heavy end, gave where that total acts. One moment equation did the rest.
That is all an external beam analysis is. Classify the beam and count: statics alone can solve it only if it is determinate. Describe the load by its intensity w(x). Replace it by areas acting at centroids, piece by piece or by integration. Then the beam is a rigid body under point loads, solved by S11's three equations, with each reaction checked against what its support can supply. A distributed load on a beam acts, for the reactions, as its load diagram's area through the diagram's centroid.
- Central idea: resultant of w(x) = area under it, at its centroid; then ordinary equilibrium.
- Key relationship: R = \int w\,dx, R\bar x = \int xw\,dx; reactions from \Sigma M_A, \Sigma M_B.
- Most important assumption: a statically determinate beam, and the resultant used only for external effects.
- Open question: the resultant is not how the beam itself feels the load. What force and moment act at each cross-section inside the beam? (S21.)
Computed views
The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Harper's programs; the Python is a translation, and both print the same numbers.
Python (scipy). S20_beam_loads.py uses one function for
every simply supported beam. It integrates w and xw with quad and returns R, \bar x,
R_A and R_B, for the trapezoid, Sample Problem 5/11 and the cubic cantilever load. It then
tabulates Harper's reactions against the end load w_0.
Part 1: resultants and reactions by quadrature
trapezoid 2 to 5 kN/m on 6 m: R = 21.0 kN at 3.429 m, R_A = 9.0, R_B = 12.0 kN
SP 5/11: R = 7472 N at 1.785 m, R_A = 3096 N, R_B = 4377 N
cantilever 1000 + 2x^3 on 8 m: R = 10048 N at 4.49 m, M_A = 45107 N m
Part 2: Harper's P 5/105, w = k1 x - k2 x^2 with w(L) = w0
w0 (N/m) R_A (N) R_B (N)
0 1853.4 1853.4
1460 2224.3 2965.9
2000 2361.4 3377.4
2919 2594.8 4077.7
4378 2965.4 5189.5
slopes: L/12 = 0.254, L/4 = 0.762; k1 L^2 / 12 = 1853.4 N
Sample Problem 5/11's R = 7472\,\mathrm N is the exact 5337.0 + 2135.4; Nugget 04 adds the parts after rounding them, giving 7473. At w_0 = 4378\,\mathrm{N/m} the exact slopes give 2965.4 and 5189.5\,\mathrm N, a tenth below the rounded slopes' 2965.5 and 5189.6.
MATLAB (Octave-compatible). S20_beam_loads.m does the
same with integral.
SP 5/11: R = 7472 N at 1.785 m, R_A = 3096 N, R_B = 4377 N
cantilever: R = 10048 N, M_A = 45107 N m
w0 (N/m) R_A (N) R_B (N)
0 1853.4 1853.4
1460 2224.3 2965.9
2000 2361.4 3377.4
2919 2594.8 4077.7
4378 2965.4 5189.5
Traps
- Solving an indeterminate beam with three equations. Count unknowns first. (From Nugget 01.)
- Calling a kink in w(x) a jump. w is continuous where only its slope changes. (From Nugget 02.)
- Putting a triangular load's resultant at mid-span. It acts at \frac23L from the zero end. (From Nugget 03.)
- Using the peak intensity times the length. The resultant is the area, \frac12wL for a triangle. (From Nugget 03.)
- Accepting a negative reaction at a roller. A roller only pushes; the beam lifts. (From Nugget 04.)
- Using the resultant to find forces inside the beam. It replaces the load for external effects only. (From Nugget 03.)
- Mixing units in a load formula. In w_0 + kx^3, k is in N/m⁴. (From Nugget 05.)
- Assuming a parameter changes both reactions alike. Plot them. (From Nugget 06.)
Self-check
- Count the unknown reactions of each beam in Fig. 5/18 and say which are determinate.
- A 4\,\mathrm m simple beam carries 3\,\mathrm{kN/m} along its whole length. Find the reactions.
- A 3\,\mathrm m simple beam carries a load rising from 0 at A to 6\,\mathrm{kN/m} at B. Find the resultant, its position and the reactions.
- A 6\,\mathrm m simple beam carries a load rising linearly from 2 to 5\,\mathrm{kN/m}. Find the reactions.
- Find the single resultant of SP 5/11's load in SI, and use it to check the reactions.
- SP 5/12's cantilever shortened to 6\,\mathrm m, same w = 1000 + 2x^3 (N/m). Find R, \bar x and M_A.
- In w = w_0 + kx^3, with w in N/m and x in m, what are the units of w_0 and k?
- Harper's beam with w_0 = 2\,\mathrm{kN/m}: find R_A and R_B.
- For Harper's beam, find the w_0 that makes the load linear, in N/m and lb/ft.
- A beam on supports at 0 (a roller) and 3\,\mathrm m carries 2\,\mathrm{kN/m} on an overhang from 3 to 4\,\mathrm m. Find the reactions, say whether the roller can provide them, and find the smallest uniform load on the span 0–3\,\mathrm m that keeps the beam down. Then find the reactions with 0.5\,\mathrm{kN/m} on that span.
- A 3\,\mathrm m cantilever carries 2\,\mathrm{kN/m} and a 5\,\mathrm{kN} load at its free end. Find the reactions at the wall.
- A load is 8\,\mathrm{kN/m} on the left half of a beam and 4\,\mathrm{kN/m} on the right half. Is w continuous at mid-span? Is dw/dx?
Fully worked solutions
1. Simple 3, cantilever 3: determinate. Combination 4 unknowns, 4 equations (three plus no moment at the pin): determinate. Continuous 4, end-supported cantilever 4, fixed 6: indeterminate.
2. R = 3(4) = 12\,\mathrm{kN} at mid-span: R_A = R_B = 6\,\mathrm{kN}.
3. R = \frac12(6)(3) = 9\,\mathrm{kN} at \frac23(3) = 2\,\mathrm m from A. \Sigma M_A: R_B(3) = 9(2), R_B = 6\,\mathrm{kN}; R_A = 3\,\mathrm{kN}.
4. Rectangle 12\,\mathrm{kN} at 3\,\mathrm m, triangle 9\,\mathrm{kN} at 4\,\mathrm m. R_B = (36 + 36)/6 = 12\,\mathrm{kN}, R_A = 21 - 12 = 9\,\mathrm{kN}.
5. R = 5338 + 2135 = 7473\,\mathrm N at \bar x = [5338(1.524) + 2135(2.438)]/7473 = 1.785\,\mathrm m. R_B = 7473(1.785)/3.048 = 4377\,\mathrm N, R_A = 7473 - 4377 = 3096\,\mathrm N.
6. R = 1000(6) + 6^4/2 = 6000 + 648 = 6648\,\mathrm N. \int_0^6x\,w\,dx = 500(36) + \frac25(7776) = 18\,000 + 3110.4 = 21\,110\,\mathrm{N\,m}. \bar x = 21\,110/6648 = 3.175\,\mathrm m; M_A = 21\,110\,\mathrm{N\,m} counter-clockwise, A_y = 6648\,\mathrm N.
7. w_0: N/m. kx^3 must be N/m, so k is N/m⁴.
8. R_A = 1853.4 + 0.254(2000) = 2361.4\,\mathrm N; R_B = 1853.4 + 0.762(2000) = 3377.4\,\mathrm N.
9. k_2 = 0 when w_0 = k_1L = 2394(3.048) = 7297\,\mathrm{N/m}; in Harper's units 50(10) = 500\,\mathrm{lb/ft}.
10. The overhang load is 2\,\mathrm{kN} at 3.5\,\mathrm m. \Sigma M_A: 3R_B = 2(3.5), R_B = 2.333\,\mathrm{kN}; R_A = 2 - 2.333 = -0.333\,\mathrm{kN}: A must pull down, which a roller cannot. A load w_1 on 0–3\,\mathrm m adds 1.5w_1 to R_A, so 1.5w_1 \ge 0.333, w_1 \ge 0.222\,\mathrm{kN/m}. With w_1 = 0.5\,\mathrm{kN/m} (1.5\,\mathrm{kN} at 1.5\,\mathrm m): R_B = [1.5(1.5) + 2(3.5)]/3 = 3.083\,\mathrm{kN}, R_A = 3.5 - 3.083 = 0.417\,\mathrm{kN}, upward: the roller works.
11. A_y = 2(3) + 5 = 11\,\mathrm{kN}; M_A = 6(1.5) + 5(3) = 24\,\mathrm{kN\,m} counter-clockwise; A_x = 0.
12. No: w jumps from 8 to 4\,\mathrm{kN/m}. Its slope is zero on both sides, so dw/dx is continuous except at the jump itself, where it is undefined.