For the idealized model of a structural component AD loaded as shown, determine the reactions that the supports exert on the component.
Include in your solution a check calculation. E.g. if you used one moment equation of equilibrium and two scalar force equations of equilibrium, use a second moment equation about a different point to re-calculate one of your previously calculated values. You may use one or both other previously calculated values in this second moment equation.
The load magnitudes are as follows:
PB = 5 + X/10 kN
MB = 3 + Y/10 kN·m
PC = 4 + Z/10 kN
Drawing is not to scale. Dimensions are in metres.
Where X, Y and Z are the last 3 digits of your student number.
Example: Student number 12345456 → X = 4, Y = 5, Z = 6
Thus PB = 5.4 kN, MB = 3.5 kN·m, PC = 4.6 kN
For the three loads shown, determine the equivalent single force and moment at the pin connection D.
Determine the equivalent single force and indicate where its line of action crosses a horizontal line that passes through the pin connection D.
Use the single equivalent force found in Part 2.2 to determine the reactions again to see that they are the same as the answers determined in Part 1, thus showing that the single force is equivalent to the original loading.
The idealized model of an initial design for a mechanism is shown below. The mechanism is attached to a rigid steel frame that may be considered to be ‘ground’. For the loading condition with vertical load PV and horizontal load PH, and the linear actuator ED holding it in the position shown, determine:
• The force required to be applied by the linear actuator
• The reaction forces in each of the pin connections at A, B, C, D, E and F
Load magnitudes:
PH = 3 + (X/10) kN
PV = 1 + (Y/10) kN
Where X and Y are the last 2 digits of your student number.
Example: Student number 12345452 → X = 5, Y = 2
PH = 3.5 kN
PV = 1.2 kN
Diagram not to scale. Dimensions in metres.
Disclaimer: The mechanism shown is a generic design for the purposes of the quiz. It is not meant to represent any specific practical mechanism.
An initial design for a beam is shown in the simplified drawing below. For the loading condition shown:
• Determine the reactions at the supports
• Use method of sections to determine the equations for the internal shear force (V) and bending moments (M)
• Determine and indicate the values of V and M at the ends of each section
• Determine and indicate the magnitude and location of any local maximum/minimum bending moment within each section
• Draw the shear and bending moment diagrams
• Determine the maximal normal stress in the beam if it has a rectangular hollow cross-section as shown below
Load magnitudes and dimensions are based on your student ID:
M = 50 + X N·m
P = 300 + Y×10 N
w = 1500 + Z×50 N/m
Example (ID 12345201):
X = 2 → M = 52 N·m
Y = 0 → P = 300 N
Z = 1 → w = 1550 N/m
Dimensions:
Beam lengths: 0.8Y and 0.3Z → 0.80 m and 0.31 m
Cross-section: 2X, 3Y and 2Z mm → 22 mm, 30 mm and 2.1 mm
Drawing not to scale.
Important Notes:
– You may use Excel or another graphing app to draw V and M diagrams and insert them as screenshots.
– If using Excel/graphing app, include input equations in your screenshot.
– Screenshots of V and M diagrams without supporting equations are not acceptable.
– Use sectioning method with FBDs to determine V and M equations.
– Read the assessment criteria in the rubric.
The solid 32 mm diameter shaft is used to transmit the torques applied to the pulley at A and gears at C, D and E. The shaft is supported in bearings at B and F.
You must:
• Determine the torque TA required to be applied by pulley A for the shaft to rotate at constant speed (equilibrium)
• Determine the internal torsional moments and draw the torque diagram for the entire shaft
• Determine the absolute maximum shear stress in the shaft (not including stress concentrations)
• If shaft is made of A-36 steel, determine the angle of twist (degrees) of end A relative to F
Loads and length:
TC = 100 + X×10 N·m
TD = 500 + Y×10 N·m
TE = 200 + X×10 N·m
L = 0.5 + Z/20 m
Example (ID 12345402 → X = 4, Y = 0, Z = 2):
TC = 140 N·m
TD = 500 N·m
TE = 220 N·m
L = 0.6 m
The W310×39 Wide-Flange beam (A-36 steel) is subjected to the loading shown.
Determine the slope (degrees) and deflection of end A of the cantilever beam:
• First using method of integration
• Then check answer by method of superposition and standard beam deflection tables
Loads and length:
w = 20 + X kN/m
M = 10 + Y kN·m
L = 2.5 + Z/10 m
Example (ID 12345452 → X = 4, Y = 5, Z = 2):
w = 24 kN/m
M = 15 kN·m
L = 2.7 m
The assessment requires application of statics, mechanics of materials and beam/shaft analysis across several problems. Key tasks:
• Part 1 (Statics — support reactions): compute support reactions for the idealised structural component AD using equilibrium; include an independent check using a second moment equation.
• Part 2 (Equivalent force & moment): reduce multiple loads to an equivalent single force + moment at pin D, locate its line of action, and show that using this equivalent force reproduces the Part 1 reactions.
• Mechanism problem: for the mechanism attached to ground, with given vertical/horizontal loads and actuator ED, determine actuator force and pin reaction forces at A, B, C, D, E, F.
• Beam problem: find support reactions; use method of sections to get V(x) and M(x); evaluate V and M at section ends; locate any internal extrema; draw shear and bending moment diagrams (screenshots from graphing app allowed but must include equations used); compute maximum normal stress for rectangular hollow cross-section.
• Torsion problem (shaft): find required pulley torque for equilibrium, internal torsional moments, torque diagram, absolute maximum shear stress, and angle of twist for A-36 steel.
• Beam deflection: for a W310×39 cantilever, compute slope and deflection at A by (a) integration and (b) superposition / standard tables and compare.
Additional requirements: use student-ID digits to generate load magnitudes, show free-body diagrams (FBDs) and section FBDs, include algebraic equations and check calculations, show any Excel/graph app inputs if used, follow rubric.
The mentor guided the student through a repeatable problem-solving workflow applied to each subtask. Below is the stepwise process and the brief explanation for each section.
Identify numerical inputs — extract X,Y,Z from the student ID and substitute into all load formulas. (Mentor emphasised doing this first to prevent propagation errors.)
Sketch and label — produce a clear, annotated sketch (dimensions, loads, support types, sign convention). Mentor required one neat FBD per major section.
Assumptions & sign conventions — state that beams/links are rigid bodies, small-deflection linear behaviour, material isotropic (A-36 when specified), and define positive directions for forces/moments and internal shear/moment sign conventions.
Free-body diagram of whole component — replace distributed/complex loads by their resultant(s) and show reaction forces/moments at supports.
Write equilibrium equations — ΣFx=0, ΣFy=0, ΣM_about_point=0. Mentor taught choosing a moment point that eliminates the largest unknown(s) to simplify algebra.
Solve algebraically — solve the three scalar equations for three unknown reaction components.
Check calculation (Part 1.2) — mentor required an independent moment equation about a different point (or an alternative equilibrium equation) that re-calculates one reaction value using previously computed values; confirm within tolerance. This ensures algebraic consistency.
Compute force resultant — vector sum of all applied forces to get equivalent single force R (magnitude and direction).
Compute resultant moment about D — sum moments of each original load about D to get equivalent couple M_D (include sign).
Locate line of action — find perpendicular offset a such that M_D = R × a (or use components and solve for intersection with horizontal through D).
Recompute support reactions using equivalent single force — treat the structure with only R and M_D acting at D and solve equilibrium — mentor insisted on showing that the same support reactions are obtained (numerical match to Part 1), which proves equivalence.
Isolate mechanism — draw FBD of mechanism with PH and PV applied, include linear actuator ED as unknown axial force along its line, and unknown reaction components at all pin joints.
Simplify via pins/links — use geometry to resolve reaction directions (pins give two unknown components each unless constrained). Mentor encouraged identifying any moments that can be summed to reduce unknowns.
Solve using equilibrium of the whole mechanism — ΣFx=0, ΣFy=0, ΣM_about_convenient_point=0 to get actuator force and some reactions.
Use sub-body equilibrium — take free-body of individual links/segments around pins to determine remaining reaction components. Mentor emphasised consistent sign usage and including the actuator orientation in component resolution.
Sanity checks — verify that action–reaction pairs at pin connections are equal and opposite, and units/signs are reasonable.
Support reactions — use equilibrium for the entire beam with point loads, moments and distributed loads substituted by resultants.
Method of sections — cut beam into the prescribed sections; for each section draw the FBD, write expressions for shear V(x) and bending moment M(x) (in terms of x) using equilibrium of the cut. Mentor required algebraic V(x), M(x) equations for each region.
Evaluate boundary values and local extrema — compute V and M at ends of each section; find dM/dx = 0 points inside each section to locate local maxima/minima and evaluate M there.
Shear & bending plots — plot V(x) and M(x) using Excel or plotting tool. Mentor required screenshots that include the input equations (not just the plots).
Stress calculation — compute section modulus for the given rectangular hollow cross-section, then maximal normal stress σ_max = M_max / S. Show units and conversion.
Checks — ensure discontinuities in V correspond to point load positions; ensure M is continuous except for applied moments.
Sign convention and FBD along shaft — draw shaft with applied torques at A, C, D, E and bearings at B, F.
Equilibrium for rotation — ΣT = 0 to compute unknown pulley torque TA required for equilibrium (TA balances applied torques and reactions). Mentor insisted on writing algebraic sums with sign convention.
Internal torque diagram — use method of sections to compute internal torque T(x) on each shaft segment; sketch torque diagram.
Max shear stress — use τ_max = (T*c)/J where J for solid circular shaft = π d^4 / 32, with d = 32 mm; compute τ_max numerically.
Angle of twist — compute θ = ∑(T L / (G J)) for each segment using G (A-36 steel: G ≈ 79–80 GPa — mentor required stating the source/assumption), convert to degrees.
Checks — ensure units correct and compare torque diagram extremes with stress locations.
Obtain M(x) for cantilever including w and applied M at tip: derive bending moment expression used in integration.
Integration method — integrate EI d^2v/dx^2 = M(x) → find slope θ(x) and deflection v(x), apply boundary conditions (at fixed root θ(0)=0, v(0)=0) and evaluate at tip A. Convert slope to degrees.
Superposition check — split loading into simpler cases (uniform load, tip moment), use standard deflection formulas/tables, sum contributions and compare with integration result. Mentor required numerical agreement (within rounding).
Documentation — show algebraic steps and numerical substitution for E, I (W310×39 section properties), and L.
• Substitution and algebra: all load magnitudes were substituted from student-ID digits before solving.
• Equilibrium and sections: global equilibrium gave support reactions; section FBDs produced V(x) and M(x) equations and located internal extrema.
• Equivalence demonstration: the resultant single force + moment at D reproduced Part 1 reactions numerically confirming correctness.
• Mechanism solution: actuator axial force and pin reactions at A–F were determined by whole-body and sub-body equilibrium, with reaction directions resolved by geometry.
• Torsion & stress: internal torque diagram drawn, required pulley torque found for equilibrium, τ_max computed for the 32 mm shaft, and total angle of twist between A and F computed using G for A-36.
• Deflection check: slope/deflection at cantilever tip computed by integration and confirmed by superposition/tables; both methods matched within rounding.
• Verification steps: second moment equilibrium checks, action–reaction checks at pins, continuity checks on V & M diagrams, and cross-method comparisons (integration vs superposition, Part1 vs Part2) were included and numerically consistent.
Apply static equilibrium (ΣFx, ΣFy, ΣM) to solve reactions for statically determinate structures.
Reduce distributed and multiple loads to equivalent resultant forces and moments and locate their lines of action.
Use method of sections and FBDs to derive shear V(x) and bending moment M(x) equations; draw shear and bending moment diagrams.
Locate internal maxima/minima of bending moment and identify critical sections.
Compute normal bending stress using section modulus for non-solid cross sections.
Perform torsion analysis: internal torque diagrams, shear stress from torsion, and compute angle of twist using material shear modulus.
Determine actuator and joint reaction forces for mechanisms using whole-body and sub-body equilibrium.
Calculate beam deflections and slopes by both direct integration and superposition; cross-verify results.
Practice good engineering documentation: clear FBDs, algebraic derivations, software graph inputs, and independent check calculations.
Translate student-ID based data correctly into numerical values and propagate units properly.
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