A horizontal beam AB is pinned to a wall at A and propped by the strut CD, pinned to the wall at C, 0.6 m below A. A load of 12 kN hangs at B. Neglect the weights. Find the magnitude of the pin force A on the beam at A.
P = 12 kNAD = 1.5 mAB = 3.6 mAC = 0.6 m
A = 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, 3 m above A. A load of 12 kN hangs at B. Find the tension T in the cable.
W = 12 kNWb = 1 kNAB = 5 mAC = 3 m
T = kN
9.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 4.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 10 mw = 2 kN/m from 0 to 10 mx = 4.5 m
Mx = kN·m
10.Shear force & bending momentHard
The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment Mmax in the beam.
L = 5 mP = 55 kN at 2 mw = 4 kN/m from 0 to 5 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
A 120 mm × 150 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 52.5 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 120 mmh = 150 mmd = 40 mmhole centre: y = 52.5 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A 150 mm × 240 mm plate has a rectangular cut-out 40 mm wide and 40 mm high, centred on the vertical centre line with its centre 144 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 150 mmh = 240 mmcut-out: 40 × 40 mmhole centre: y = 144 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: a·b = 4
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.
Positive: the angle between the vectors is less than 90°.
2.Vectors: dot & cross product
Answer: (a × b)y = −31
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 = 0·5 − (−4)·3 = 12
(a × b)y = azbx − axbz = (−4)·4 − 3·5 = −31
(a × b)z = axby − aybx = 3·3 − 0·4 = 9
Requested component
(a × b)y = −31
Answer
(a × b)y = −31
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 = 30.63 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 = 45·2.5 + 55·3
RB = 277.5/4 = 69.38 kN
ΣFy = 0: RA = 100 − 69.38 = 30.63 kN
Answer
RA = 30.63 kN
4.Beam support reactions
Answer: RA = 29.8 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 = 4·4 = 16 kN, acting at x̄ = 7 m
ΣMA = 0: RB·10 = 50·5 + 4·4·7
RB = 362/10 = 36.2 kN
ΣFy = 0: RA = 66 − 36.2 = 29.8 kN
Answer
RA = 29.8 kN
5.Truss: method of joints
Answer: FBC = −27.95 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 = 30·5 → Cy = 25 kN
ΣFy = 0: Ay = 30 − 25 = 5 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 = 2/2.236 = 0.8944, cos γ = 1/2.236 = 0.4472
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.
Strut force
FCD = 12·3.6/(0.3714·1.5) = 77.55 kN
Components at A
Ax = FCD·cos α = 72 kN (towards the wall)
Ay = 12 − FCD·sin α = −16.8 kN
Magnitude
A = √(72² + −16.8²) = 73.93 kN
Answer
A = 73.93 kN
A negative Ay means the wall pulls the beam down at A.
8.Frames & machines
Answer: T = 24.3 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 = √(5² + 3²) = 5.831 m, sin β = 0.5145
Moments about A
T·0.5145·5 = 12·5 + 1·2.5
T = 24.3 kN
Answer
T = 24.3 kN
9.Shear force & bending moment
Answer: Mx = 24.75 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 = 2·10 = 20 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 2·10·5
RB = 100/10 = 10 kN
ΣFy = 0: RA = 20 − 10 = 10 kN
Cut at x = 4.5 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 = 10·4.5 − 2·4.5·2.25 = 24.75 kN·m
Answer
Mx = 24.75 kN·m
10.Shear force & bending moment
Answer: Mmax = 78 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. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.
Support reactions
Equilibrium of the whole beam.
UDL resultant: W = w·ℓ = 4·5 = 20 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·5 = 55·2 + 4·5·2.5
RB = 160/5 = 32 kN
ΣFy = 0: RA = 75 − 32 = 43 kN
Where is V = 0?
V(2⁻) = RA − w·2 = 35 kN
V(2⁺) = 35 − 55 = −20 kN
The shear changes sign at the load, so the maximum is under the load.
Maximum moment
Mmax = RA·2 − w·2²/2 = 43·2 − 4·2²/2 = 78 kN·m
Answer
Mmax = 78 kN·m
It occurs at x = 2 m from A.
11.Centroid & second moment of area
Answer: Ix = 32.94 × 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 = (18000·75 − 1257·52.5)/(18000 − 1257) = 76.69 mm
Full plate about ȳ
120·150³/12 + 18000·−1.689² = 33.8 × 10⁶ mm⁴
Hole about ȳ
π·40⁴/64 + 1257·−24.19² = 0.8609 × 10⁶ mm⁴
Subtract
Ix = 32.94 × 10⁶ mm⁴
Answer
Ix = 32.94 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: Ix = 171.6 × 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 = (36000·120 − 1600·144)/(36000 − 1600) = 118.9 mm