Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
span = 5 mheight = 2 mB is 2 m from Avertical load = 45 kNhorizontal load = 15 kN →
FBC = kN
7.Frames & machinesMedium
The A-frame shown stands on a pin at A and a roller at B, 2 m apart, with its apex pin C 3.6 m high and the horizontal tie DE 1.2 m above the ground. A vertical load of 8 kN acts on leg AC at K, 0.75 m (horizontally) from A. Find the tension T in the tie.
P = 8 kNAB = 2 mheight of C = 3.6 mheight of DE = 1.2 mhorizontal distance A to K = 0.75 m
T = kN
8.Frames & machinesHard
The beam AB of the wall bracket shown is pinned at A and propped by the strut CD (C is 1.5 m below A). The beam carries a uniform load of 5 kN/m over its full length and a point load of 12 kN at its free end B. Find the force FCD in the strut.
w = 5 kN/mP = 12 kNAD = 1.2 mAB = 3 mAC = 1.5 m
FCD = kN
9.Shear force & bending momentMedium
Find the bending moment Mx at the section x = 5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 10 mw = 12 kN/m from 0 to 10 mx = 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 = 6 mP = 30 kN at 1 mw = 10 kN/m from 0 to 6 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
A symmetric I-section has flanges 200 mm wide and 16 mm thick, an overall depth of 480 mm and a web 10 mm thick. Find its second moment of area Ix about the horizontal centroidal axis x.
B = 200 mmH = 480 mmtf = 16 mmtw = 10 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 10 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.
Dot product
a·b = 2·1 + 7·3 + 2·(−4) = 15
Length of b
|b| = √(1² + 3² + (−4)²) = 5.099
Scalar projection
compb a = a·b/|b| = 15/5.099 = 2.942
Answer
compb a = 2.942
2.Vectors: dot & cross product
Answer: n̂y = −0.5657
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
(a × b)x = aybz − azby = 0·(−3) − 3·(−3) = 9
(a × b)y = azbx − axbz = 3·(−1) − 5·(−3) = 12
(a × b)z = axby − aybx = 5·(−3) − 0·(−1) = −15
Normalise
The z-component of a × b is negative, so use −(a × b).
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·5 = 10·1 + 55·2
RB = 120/5 = 24 kN
ΣFy = 0: RA = 65 − 24 = 41 kN
Answer
RA = 41 kN
4.Beam support reactions
Answer: RB = 31.13 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 = 6·1 = 6 kN, acting at x̄ = 1.5 m
ΣMA = 0: RB·8 = 40·6 + 6·1·1.5
RB = 249/8 = 31.13 kN
ΣFy = 0: RA = 46 − 31.13 = 14.88 kN
Answer
RB = 31.13 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).
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·5 = 45·2 + 15·2 → Cy = 24 kN
ΣFy = 0: Ay = 45 − 24 = 21 kN
ΣFx = 0: Ax = −15 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 = 2/3.606 = 0.5547, cos γ = 3/3.606 = 0.8321
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 = 8·0.75/2 = 3 kN
Leg BC: moments about C
3·1 = T·2.4
T = 1.25 kN
Answer
T = 1.25 kN
8.Frames & machines
Answer: FCD = 62.43 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.
Resultant of the uniform load
w·L = 5·3 = 15 kN at 1.5 m
Moments about A for the beam
FCD·0.7809·1.2 = 15·1.5 + 12·3 = 58.5 kN·m
FCD = 62.43 kN
Answer
FCD = 62.43 kN
9.Shear force & bending moment
Answer: Mx = 150 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 = 12·10 = 120 kN, acting at x̄ = 5 m
ΣMA = 0: RB·10 = 12·10·5
RB = 600/10 = 60 kN
ΣFy = 0: RA = 120 − 60 = 60 kN
Cut at x = 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 = 60·5 − 12·5·2.5 = 150 kN·m
Answer
Mx = 150 kN·m
10.Shear force & bending moment
Answer: Mmax = 61.25 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 = 10·6 = 60 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 30·1 + 10·6·3
RB = 210/6 = 35 kN
ΣFy = 0: RA = 90 − 35 = 55 kN
Where is V = 0?
RA − P − w·x0 = 0 → x0 = (55 − 30)/10 = 2.5 m
Maximum moment
Mmax = RA·x0 − P·(x0 − 1) − w·x0²/2 = 61.25 kN·m
Answer
Mmax = 61.25 kN·m
It occurs at x = 2.5 m from A.
11.Centroid & second moment of area
Answer: Ix = 419.5 × 10⁶ mm⁴
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 = 180·10 = 1800 mm², yf = 220 + 10/2 = 225 mm