A horizontal beam AB is pinned to a wall at A and propped by the strut CD, pinned to the wall at C, 0.75 m below A. A load of 10 kN hangs at B. Neglect the weights. Find the magnitude of the pin force A on the beam at A.
P = 10 kNAD = 2 mAB = 3.6 mAC = 0.75 m
A = 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 its end B to the point C on the wall, 4 m above A. A load of 5 kN hangs at B. Find the tension T in the cable.
W = 5 kNWb = 2 kNAB = 5 mAC = 4 m
T = kN
9.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 7 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 10 mw = 10 kN/m from 0 to 10 mx = 7 m
Mx = kN·m
10.Shear force & bending momentHard
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 @ 6 m40 kN @ 2 m
1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)
11.Centroid & second moment of areaMedium
A 200 mm × 300 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 210 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 200 mmh = 300 mmd = 40 mmhole centre: y = 210 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 50 kN·m. Find the bending stress σbottom at the bottom 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.
Dot product equals zero
a·b = 2·(−1) + 4·k + (−4)·4 = 0
−18 + 4·k = 0
Solve for k
k = −(−18)/4 = 4.5
Answer
k = 4.5
2.Vectors: dot & cross product
Answer: V = 30
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 b × c
(b × c)x = bycz − bzcy = (−2)·2 − (−1)·(−1) = −5
(b × c)y = bzcx − bxcz = (−1)·1 − 2·2 = −5
(b × c)z = bxcy − bycx = 2·(−1) − (−2)·1 = 0
Dot product with a
a·(b × c) = 1·(−5) + 5·(−5) + (−1)·0 = −30
V = |−30| = 30
Answer
V = 30
3.Beam support reactions
Answer: RB = 17.92 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·6 = 40·1 + 15·4.5
RB = 107.5/6 = 17.92 kN
ΣFy = 0: RA = 55 − 17.92 = 37.08 kN
Answer
RB = 17.92 kN
4.Beam support reactions
Answer: RB = 22.5 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 = 10·1 = 10 kN, acting at x̄ = 3.5 m
ΣMA = 0: RB·6 = 40·2.5 + 10·1·3.5
RB = 135/6 = 22.5 kN
ΣFy = 0: RA = 50 − 22.5 = 27.5 kN
Answer
RB = 22.5 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 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).
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 = 10·3.6/(0.3511·2) = 51.26 kN
Components at A
Ax = FCD·cos α = 48 kN (towards the wall)
Ay = 10 − FCD·sin α = −8 kN
Magnitude
A = √(48² + −8²) = 48.66 kN
Answer
A = 48.66 kN
A negative Ay means the wall pulls the beam down at A.
8.Frames & machines
Answer: T = 9.605 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 + 4²) = 6.403 m, sin β = 0.6247
Moments about A
T·0.6247·5 = 5·5 + 2·2.5
T = 9.605 kN
Answer
T = 9.605 kN
9.Shear force & bending moment
Answer: Mx = 105 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 = 10·10 = 100 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 10·10·5
RB = 500/10 = 50 kN
ΣFy = 0: RA = 100 − 50 = 50 kN
Cut at x = 7 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 = 50·7 − 10·7·3.5 = 105 kN·m
Answer
Mx = 105 kN·m
10.Shear force & bending moment
Answer: F (x = 5 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 = 15 kN, R2 = 70 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 = 15, MC = 30, MD = 5, ME = −20, MF = −45 kN·m, MG = 0
Answer
Mark F: MF = −45 kN·m
11.Centroid & second moment of area
Answer: Ix = 445.3 × 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 = (60000·150 − 1257·210)/(60000 − 1257) = 148.7 mm
Full plate about ȳ
200·300³/12 + 60000·1.284² = 450.1 × 10⁶ mm⁴
Hole about ȳ
π·40⁴/64 + 1257·61.28² = 4.845 × 10⁶ mm⁴
Subtract
Ix = 445.3 × 10⁶ mm⁴
Answer
Ix = 445.3 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: σbottom = 209 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 = 280·15 = 4200 mm², yf = 200 + 15/2 = 207.5 mm