A horizontal beam AB is pinned to a wall at A and propped by the strut CD, pinned to the wall at C, 1 m below A. A load of 15 kN hangs at B. Neglect the weights. Find the magnitude of the pin force A on the beam at A.
P = 15 kNAD = 1.2 mAB = 3.6 mAC = 1 m
A = kN
8.Frames & machinesHard
The beam AB of the wall bracket shown is pinned at A and propped by the strut CD (C is 0.6 m below A). The beam carries a uniform load of 4 kN/m over its full length and a point load of 15 kN at its free end B. Find the force FCD in the strut.
w = 4 kN/mP = 15 kNAD = 1.6 mAB = 3.6 mAC = 0.6 m
FCD = kN
9.Shear force & bending momentMedium
The simply supported beam carries a clockwise couple M0 = 30 kN·m at x = 1 m from A, and no other load. Find the bending moment Mx just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.
L = 8 mM0 = 30 kN·m clockwise, at 1 m
Mx = kN·m
10.Shear force & bending momentHard
The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment Mmax in the beam.
L = 10 mw = 6 kN/m from 0 to 8 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
A channel section lies with its open side up: a base plate 120 mm wide with two upright legs, 120 mm high overall. Base and legs are all 8 mm thick. Find its second moment of area Ix about the horizontal axis x through its centroid.
b = 120 mmh = 120 mmt = 8 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A symmetric I-section has flanges 140 mm wide and 22 mm thick, an overall depth of 220 mm and a web 6 mm thick. It carries a bending moment of 150 kN·m about the axis x. Find the size of the bending stress σj at the junction of the web and the top flange.
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 = (−6)·(−4) + 3·(−2) + (−2)·k = 0
18 + (−2)·k = 0
Solve for k
k = −18/(−2) = 9
Answer
k = 9
2.Vectors: dot & cross product
Answer: n̂y = 0.9631
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 = 1·(−3) − (−5)·1 = 2
(a × b)y = azbx − axbz = (−5)·(−2) − 2·(−3) = 16
(a × b)z = axby − aybx = 2·1 − 1·(−2) = 4
Normalise
The z-component of a × b is already positive.
|a × b| = √(2² + 16² + 4²) = 16.61
n̂ = (2, 16, 4)/16.61 = (0.1204, 0.9631, 0.2408)
n̂y = 0.9631
Answer
n̂y = 0.9631
3.Beam support reactions
Answer: RA = 57.5 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.
ΣMA = 0: RB·6 = 50·1.5 + 30·2
RB = 135/6 = 22.5 kN
ΣFy = 0: RA = 80 − 22.5 = 57.5 kN
Answer
RA = 57.5 kN
4.Beam support reactions
Answer: RA = 44.06 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 = 5·1 = 5 kN, acting at x̄ = 3.5 m
ΣMA = 0: RB·8 = 55·2 + 5·1·3.5
RB = 127.5/8 = 15.94 kN
ΣFy = 0: RA = 60 − 15.94 = 44.06 kN
Answer
RA = 44.06 kN
5.Truss: method of joints
Answer: FAB = −46.58 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 = 50·1 → Cy = 8.333 kN
ΣFy = 0: Ay = 50 − 8.333 = 41.67 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 2/2.236 = 0.8944
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −41.67/0.8944 = −46.58 kN
Answer
FAB = −46.58 kN (compression)
6.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).
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 = 15·3.6/(0.6402·1.2) = 70.29 kN
Components at A
Ax = FCD·cos α = 54 kN (towards the wall)
Ay = 15 − FCD·sin α = −30 kN
Magnitude
A = √(54² + −30²) = 61.77 kN
Answer
A = 61.77 kN
A negative Ay means the wall pulls the beam down at A.
8.Frames & machines
Answer: FCD = 142.3 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.
Resultant of the uniform load
w·L = 4·3.6 = 14.4 kN at 1.8 m
Moments about A for the beam
FCD·0.3511·1.6 = 14.4·1.8 + 15·3.6 = 79.92 kN·m
FCD = 142.3 kN
Answer
FCD = 142.3 kN
9.Shear force & bending moment
Answer: Mx = 26.25 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. An applied couple makes the bending moment diagram jump by the size of the couple.
Support reactions
Moments about A, counter-clockwise positive: RB·L − M0 = 0. Vertical balance: RA + RB = 0.
RB = M0/L = 30/8 = 3.75 kN (upward)
RA = −RB = −3.75 kN (downward)
Cut just right of the couple
The left part now contains RA and the couple.
Mx = RA·a + M0 = −3.75·1 + 30 = 26.25 kN·m
Answer
Mx = 26.25 kN·m
10.Shear force & bending moment
Answer: Mmax = 69.12 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 = 6·8 = 48 kN, acting at x̄ = 4 m
ΣMA = 0: RB·10 = 6·8·4
RB = 192/10 = 19.2 kN
ΣFy = 0: RA = 48 − 19.2 = 28.8 kN
Zero shear
The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.
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 rectangles
Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.
base: A1 = 120·8 = 960 mm², y1 = 4 mm
two legs: A2 = 2·8·112 = 1792 mm², y2 = 8 + 112/2 = 64 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.