A horizontal beam AB is pinned to a wall at A and propped by the strut CD, pinned to the wall at C, 0.5 m below A. A load of 4 kN hangs at B. Neglect the weights. Find the magnitude of the pin force A on the beam at A.
P = 4 kNAD = 1.2 mAB = 3.6 mAC = 0.5 m
A = kN
8.Frames & machinesHard
The A-frame shown stands on a pin at A and a roller at B, 4 m apart. The legs meet at the pin C, 4 m high, and are tied by the horizontal bar DE at 1 m. A horizontal force of 2 kN to the right acts at C. Find the tension T in the tie.
P = 2 kNAB = 4 mheight of C = 4 mheight of DE = 1 m
T = kN
9.Shear force & bending momentMedium
The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 2 m from A. Find the bending moment Mx at the section x = 1.5 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.
L = 4 mP = 65 kN at 2 mw = 5 kN/m from 0 to 4 mx = 1.5 m
Mx = kN·m
10.Shear force & bending momentHard
Find the bending moment Mx at the section x = 0.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 4 mP = 45 kN at 3 mw = 6 kN/m from 0 to 4 mx = 0.5 m
Mx = kN·m
11.Centroid & second moment of areaMedium
A 120 mm × 300 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 195 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 120 mmh = 300 mmd = 40 mmhole centre: y = 195 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.
Multiply matching components and add
a·b = (−4)·(−2) + 6·3 + 4·4 = 8 + 18 + 16 = 42
Answer
a·b = 42
Positive: the angle between the vectors is less than 90°.
2.Vectors: dot & cross product
Answer: (a × b)z = −14
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, component by component
(a × b)x = aybz − azby = 3·6 − 0·(−1) = 18
(a × b)y = azbx − axbz = 0·4 − 2·6 = −12
(a × b)z = axby − aybx = 2·(−1) − 3·4 = −14
Requested component
(a × b)z = −14
Answer
(a × b)z = −14
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
3.Beam support reactions
Answer: RA = 25.63 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.
ΣMA = 0: RB·4 = 20·1.5 + 35·2.5
RB = 117.5/4 = 29.38 kN
ΣFy = 0: RA = 55 − 29.38 = 25.63 kN
Answer
RA = 25.63 kN
4.Beam support reactions
Answer: RB = 30.38 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 = 2·7 = 14 kN, acting at x̄ = 4.5 m
ΣMA = 0: RB·8 = 30·6 + 2·7·4.5
RB = 243/8 = 30.38 kN
ΣFy = 0: RA = 44 − 30.38 = 13.63 kN
Answer
RB = 30.38 kN
5.Truss: method of joints
Answer: FBC = −8.004 kN
Method of joints: every pin of a truss is in equilibrium, so ΣFx = 0 and ΣFy = 0 at each joint. Start at a joint with at most two unknown member forces. A negative result means the member is in compression.
Support reactions
Equilibrium of the whole truss. Moments about A, counter-clockwise positive.
ΣMA = 0: Cy·6 = 30·1 → Cy = 5 kN
ΣFy = 0: Ay = 30 − 5 = 25 kN
Joint C
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
sin γ = 4/6.403 = 0.6247, cos γ = 5/6.403 = 0.7809
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.
Strut force
FCD = 4·3.6/(0.3846·1.2) = 31.2 kN
Components at A
Ax = FCD·cos α = 28.8 kN (towards the wall)
Ay = 4 − FCD·sin α = −8 kN
Magnitude
A = √(28.8² + −8²) = 29.89 kN
Answer
A = 29.89 kN
A negative Ay means the wall pulls the beam down at A.
8.Frames & machines
Answer: T = 1.333 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.
Whole frame: moments about A
By·4 = 2·4
By = 2 kN
Leg BC: moments about C
2·2 = T·3
T = 1.333 kN
Answer
T = 1.333 kN
A positive T means the tie is in tension, as assumed.
9.Shear force & bending moment
Answer: Mx = −48.13 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.
Keep the free-end part
P: −65·0.5 = −32.5 kN·m
w: −5·2.5²/2 = −15.63 kN·m
Mx = −32.5 − 15.63 = −48.13 kN·m
Answer
Mx = −48.13 kN·m
Negative means hogging: tension on the top face.
10.Shear force & bending moment
Answer: Mx = 10.88 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 = 6·4 = 24 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 45·3 + 6·4·2
RB = 183/4 = 45.75 kN
ΣFy = 0: RA = 69 − 45.75 = 23.25 kN
Cut at x = 0.5 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 = 23.25·0.5 − 6·0.5·0.25 = 10.88 kN·m
Answer
Mx = 10.88 kN·m
11.Centroid & second moment of area
Answer: Ix = 267.2 × 10⁶ mm⁴
Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has Ic = bh³/12 and a circle Ic = πd⁴/64.
Centroid
ȳ = (36000·150 − 1257·195)/(36000 − 1257) = 148.4 mm
Full plate about ȳ
120·300³/12 + 36000·1.628² = 270.1 × 10⁶ mm⁴
Hole about ȳ
π·40⁴/64 + 1257·46.63² = 2.858 × 10⁶ mm⁴
Subtract
Ix = 267.2 × 10⁶ mm⁴
Answer
Ix = 267.2 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: σtop = 61 MPa
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 = 140·20 = 2800 mm², yf = 160 + 20/2 = 170 mm