The A-frame shown stands on a pin at A and a roller at B, 2.4 m apart. The legs meet at the pin C, 3 m high, and the horizontal tie DE is 1.2 m above the ground. A load of 5 kN acts at C. Find the magnitude of the force C that the pin C exerts on leg BC.
P = 5 kNAB = 2.4 mheight of C = 3 mheight of DE = 1.2 m
C = kN
8.Frames & machinesHard
A uniform horizontal boom AB of weight 2 kN is pinned to a wall at A and held by a cable from B to the point C on the wall, 2 m above A. A load of 15 kN hangs at B. Find the magnitude of the pin reaction A.
W = 15 kNWb = 2 kNAB = 6 mAC = 2 m
A = kN
9.Shear force & bending momentMedium
The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.
35 kN @ 4 m4 kN/m, 4–12 m
1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)
10.Shear force & bending momentHard
Find the bending moment Mx at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 8 mP = 60 kN at 3 mw = 12 kN/m from 0 to 8 mx = 2.5 m
Mx = kN·m
11.Centroid & second moment of areaMedium
Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.
bf = 220 mmtf = 15 mmtw = 12 mmhw = 160 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A symmetric I-section has flanges 130 mm wide and 14 mm thick, an overall depth of 200 mm and a web 12 mm thick. It carries a bending moment of 80 kN·m about the axis x. Find the size of the bending stress σj at the junction of the web and the top flange.
Dot product:a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.
Dot product equals zero
a·b = 3·0 + (−2)·k + 1·(−4) = 0
−4 + (−2)·k = 0
Solve for k
k = −(−4)/(−2) = −2
Answer
k = −2
2.Vectors: dot & cross product
Answer: n̂y = 0.5571
Cross product:a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.
Cross product
(a × b)x = aybz − azby = (−2)·3 − 1·(−2) = −4
(a × b)y = azbx − axbz = 1·(−3) − 1·3 = −6
(a × b)z = axby − aybx = 1·(−2) − (−2)·(−3) = −8
Normalise
The z-component of a × b is negative, so use −(a × b).
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.
Free-body diagram
Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
ΣMA = 0: RB·4 = 10·1.5 + 50·3
RB = 165/4 = 41.25 kN
ΣFy = 0: RA = 60 − 41.25 = 18.75 kN
Answer
RA = 18.75 kN
4.Beam support reactions
Answer: RB = 23 kN
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.
Free-body diagram
Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
UDL resultant: W = w·ℓ = 10·2 = 20 kN, acting at x̄ = 3 m
ΣMA = 0: RB·5 = 55·1 + 10·2·3
RB = 115/5 = 23 kN
ΣFy = 0: RA = 75 − 23 = 52 kN
Answer
RB = 23 kN
5.Truss: method of joints
Answer: member CG
Zero-force members: at a joint, the forces must balance in every direction. If a joint has no load and no support, and two of its three members are in line, then the balance across that line involves only the third member, so its force is zero. If only two members meet at an unloaded joint and they are not in line, both are zero.
Joint C
BC and CD lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.
ΣF⊥ = 0 → FCG = 0
Check with the method of joints
Solving all joints gives these member forces (T = tension, C = compression).
Zero-force members: at a joint, the forces must balance in every direction. If a joint has no load and no support, and two of its three members are in line, then the balance across that line involves only the third member, so its force is zero. If only two members meet at an unloaded joint and they are not in line, both are zero.
Joint B
AB and BC lie on one straight line and the joint carries no load. Balancing forces across that line leaves only BF.
ΣF⊥ = 0 → FBF = 0
Check with the method of joints
Solving all joints gives these member forces (T = tension, C = compression).
Frames and machines: unlike a truss, a frame contains at least one member that carries forces at more than two points, or a load between its pins. First look for two-force members: a member loaded only at two pins has its force along the line through them, so it adds one unknown, not two. Then use the whole structure for the support reactions when there are no more than three unknowns. Finally take the frame apart: draw each member separately, with the pin forces equal and opposite on the two members they connect, and write ΣFx = 0, ΣFy = 0 and ΣM = 0 for it. Taking moments about a pin removes both components of that pin force from the equation.
Tie force from leg BC
By = 2.5 kN
T = 2.5·1.2/1.8 = 1.667 kN
Components of the pin force on BC
Cx = T = 1.667 kN, Cy = −By = −2.5 kN
Magnitude
C = √(1.667² + 2.5²) = 3.005 kN
Answer
C = 3.005 kN
8.Frames & machines
Answer: A = 48.01 kN
Frames and machines: unlike a truss, a frame contains at least one member that carries forces at more than two points, or a load between its pins. First look for two-force members: a member loaded only at two pins has its force along the line through them, so it adds one unknown, not two. Then use the whole structure for the support reactions when there are no more than three unknowns. Finally take the frame apart: draw each member separately, with the pin forces equal and opposite on the two members they connect, and write ΣFx = 0, ΣFy = 0 and ΣM = 0 for it. Taking moments about a pin removes both components of that pin force from the equation.
Cable tension
T = (15 + 2/2)/0.3162 = 50.6 kN
Components at A
Ax = T·cos β = 48 kN
Ay = 15 + 2 − T·sin β = 1 kN
Magnitude
A = √(48² + 1²) = 48.01 kN
Answer
A = 48.01 kN
9.Shear force & bending moment
Answer: C (x = 4 m)
Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.
Support reactions
Take moments about one support, then use the vertical force balance.
R1 = 34 kN, R2 = 33 kN
Bending moment at every mark
Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.
MA = 0, MB = 68, MC = 136 kN·m, MD = 126, ME = 100, MF = 58, MG = 0
Answer
Mark C: MC = 136 kN·m
10.Shear force & bending moment
Answer: Mx = 176.3 kN·m
Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut.
Support reactions
Equilibrium of the whole beam.
UDL resultant: W = w·ℓ = 12·8 = 96 kN, acting at x̄ = 4 m
ΣMA = 0: RB·8 = 60·3 + 12·8·4
RB = 564/8 = 70.5 kN
ΣFy = 0: RA = 156 − 70.5 = 85.5 kN
Cut at x = 2.5 m and keep the left part
Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.
M = RA·x − (each load left of the cut × its distance to the cut)
Mx = 85.5·2.5 − 12·2.5·1.25 = 176.3 kN·m
Answer
Mx = 176.3 kN·m
11.Centroid & second moment of area
Answer: Ix = 13.45 × 10⁶ mm⁴
Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.
Split into two rectangles
Measure every y from the bottom edge.
flange: Af = 220·15 = 3300 mm², yf = 160 + 15/2 = 167.5 mm
Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.
Outer rectangle minus the two side cut-outs
Both cut-outs together are (B − tw) wide and hw = H − 2tf high, and share the axis x.