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 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 m
C = kN
8.Frames & machinesHard
A uniform horizontal boom AB of weight 1.5 kN is pinned to a wall at A and held by a cable from B to the point C on the wall, 2.5 m above A. A load of 6 kN hangs at B. Find the magnitude of the pin reaction A.
W = 6 kNWb = 1.5 kNAB = 4 mAC = 2.5 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.
45 kN @ 8 m50 kN @ 2 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 = 3.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 4 mP = 25 kN at 1 mw = 10 kN/m from 0 to 4 mx = 3.5 m
Mx = kN·m
11.Centroid & second moment of areaMedium
A symmetric I-section has flanges 200 mm wide and 18 mm thick, an overall depth of 460 mm and a web 6 mm thick. Find its second moment of area Ix about the horizontal centroidal axis x.
B = 200 mmH = 460 mmtf = 18 mmtw = 6 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
The T-section is used as a beam with the flange on top and carries a sagging bending moment of 20 kN·m. Find the size of the bending stress σtop at the top edge.
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.
Negative: the angle between the vectors is more than 90°.
2.Vectors: dot & cross product
Answer: V = 3
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.
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Replace each distributed load by its resultant (its area) acting at the centroid of the load diagram, and take moments about a support to remove that support’s reaction from the equation.
Moments about A
Counter-clockwise positive.
ΣMA = 0: RB·6 = 10·1 + 10·10
RB = 110/6 = 18.33 kN
Vertical equilibrium
RA = 10 + 10 − 18.33 = 1.667 kN
Answer
RA = 1.667 kN
4.Beam support reactions
Answer: RA = 6.75 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 = 2·2 = 4 kN, acting at x̄ = 1 m
ΣMA = 0: RB·4 = 15·3 + 2·2·1
RB = 49/4 = 12.25 kN
ΣFy = 0: RA = 19 − 12.25 = 6.75 kN
Answer
RA = 6.75 kN
5.Truss: method of joints
Answer: member BF
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).
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).
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 = 2.5·1.2/2 = 1.5 kN
Components of the pin force on BC
Cx = T = 1.5 kN, Cy = −By = −2.5 kN
Magnitude
C = √(1.5² + 2.5²) = 2.915 kN
Answer
C = 2.915 kN
8.Frames & machines
Answer: A = 10.83 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 = (6 + 1.5/2)/0.53 = 12.74 kN
Components at A
Ax = T·cos β = 10.8 kN
Ay = 6 + 1.5 − T·sin β = 0.75 kN
Magnitude
A = √(10.8² + 0.75²) = 10.83 kN
Answer
A = 10.83 kN
9.Shear force & bending moment
Answer: E (x = 8 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 = 56.67 kN, R2 = 38.33 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 = 113.3, MC = 126.7, MD = 140, ME = 153.3 kN·m, MF = 76.67, MG = 0
Answer
Mark E: ME = 153.3 kN·m
10.Shear force & bending moment
Answer: Mx = 11.88 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 = 10·4 = 40 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 25·1 + 10·4·2
RB = 105/4 = 26.25 kN
ΣFy = 0: RA = 65 − 26.25 = 38.75 kN
Cut at x = 3.5 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)
Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = Ic + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre.
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.
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 = 220·10 = 2200 mm², yf = 220 + 10/2 = 225 mm