The A-frame shown stands on a pin at A and a roller at B, 4 m apart, with its apex pin C 4.5 m high and the horizontal tie DE 1.8 m above the ground. A vertical load of 15 kN acts on leg AC at K, 1.2 m (horizontally) from A. Find the tension T in the tie.
P = 15 kNAB = 4 mheight of C = 4.5 mheight of DE = 1.8 mhorizontal distance A to K = 1.2 m
T = kN
8.Frames & machinesHard
A uniform horizontal boom AB of weight 1 kN is pinned to a wall at A and held by a cable from its end B to the point C on the wall, 1.5 m above A. A load of 10 kN hangs at B. Find the tension T in the cable.
W = 10 kNWb = 1 kNAB = 2 mAC = 1.5 m
T = kN
9.Shear force & bending momentMedium
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 = 10 mw = 5 kN/m from 0 to 10 mx = 3.5 m
Mx = kN·m
10.Shear force & bending momentHard
Find the bending moment Mx at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 5 mP = 45 kN at 2 mw = 8 kN/m from 0 to 5 mx = 1 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 = 280 mmtf = 10 mmtw = 12 mmhw = 100 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A symmetric I-section has flanges 160 mm wide and 14 mm thick, an overall depth of 280 mm and a web 14 mm thick. It carries a bending moment of 120 kN·m about the axis x. Find the size of the maximum bending stress σmax in the section.
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
a·b = 6·(−4) + (−1)·5 + (−5)·(−1) = −24
Length of b
|b| = √((−4)² + 5² + (−1)²) = 6.481
Scalar projection
compb a = a·b/|b| = −24/6.481 = −3.703
Answer
compb a = −3.703
Negative: a has a component pointing against b.
2.Vectors: dot & cross product
Answer: A = 9.95 m²
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·5 = 45·3 + 65·8
RB = 655/5 = 131 kN
Vertical equilibrium
RA = 45 + 65 − 131 = −21 kN
Answer
RB = 131 kN
RA is negative: the pin at A has to hold the beam down.
4.Beam support reactions
Answer: RA = 34.13 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 = 4·2 = 8 kN, acting at x̄ = 1 m
ΣMA = 0: RB·4 = 45·1.5 + 4·2·1
RB = 75.5/4 = 18.88 kN
ΣFy = 0: RA = 53 − 18.88 = 34.13 kN
Answer
RA = 34.13 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 G
FG and GH 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.
Whole frame: moments about A
By = 15·1.2/4 = 4.5 kN
Leg BC: moments about C
4.5·2 = T·2.7
T = 3.333 kN
Answer
T = 3.333 kN
8.Frames & machines
Answer: T = 17.5 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 angle
BC = √(2² + 1.5²) = 2.5 m, sin β = 0.6
Moments about A
T·0.6·2 = 10·2 + 1·1
T = 17.5 kN
Answer
T = 17.5 kN
9.Shear force & bending moment
Answer: Mx = 56.88 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 = 5·10 = 50 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 5·10·5
RB = 250/10 = 25 kN
ΣFy = 0: RA = 50 − 25 = 25 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)
Mx = 25·3.5 − 5·3.5·1.75 = 56.88 kN·m
Answer
Mx = 56.88 kN·m
10.Shear force & bending moment
Answer: Mx = 43 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 = 8·5 = 40 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 45·2 + 8·5·2.5
RB = 190/5 = 38 kN
ΣFy = 0: RA = 85 − 38 = 47 kN
Cut at x = 1 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 = 47·1 − 8·1·0.5 = 43 kN·m
Answer
Mx = 43 kN·m
11.Centroid & second moment of area
Answer: Ix = 3.564 × 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 = 280·10 = 2800 mm², yf = 100 + 10/2 = 105 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.