The A-frame shown stands on a pin at A and a roller at B, 3 m apart. The legs meet at the pin C, 2.4 m high, and the horizontal tie DE is 0.8 m above the ground. A load of 15 kN acts at C. Find the magnitude of the force C that the pin C exerts on leg BC.
P = 15 kNAB = 3 mheight of C = 2.4 mheight of DE = 0.8 m
C = kN
8.Frames & machinesHard
The frame shown has pins at A, B and C; A and C are fixed to the ground 3 m apart, and B is 3 m high and 0.45 m to the right of A. Member AB carries a vertical load of 4 kN at K, 0.27 m (horizontally) from A, and a horizontal load of 6 kN to the right acts at B on member AB. Find the size of the horizontal reaction Ax at A.
P = 4 kNQ = 6 kNL = 3 mH = 3 mhorizontal distance A to B = 0.45 mhorizontal distance A to K = 0.27 m
Ax = kN
9.Shear force & bending momentMedium
The simply supported beam carries a clockwise couple M0 = 50 kN·m at x = 2 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 = 50 kN·m clockwise, at 2 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.
30 kN @ 0 m60 kN @ 6 m20 kN @ 2 m8 kN/m, 1–5 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
Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.
bf = 180 mmtf = 25 mmtw = 12 mmhw = 280 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 30 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 = (−5)·(−4) + (−6)·0 + 4·(−4) = 4
Magnitudes
|a| = √((−5)² + (−6)² + 4²) = 8.775
|b| = √((−4)² + 0² + (−4)²) = 5.657
Angle
cos θ = 4/(8.775·5.657) = 0.08058
θ = arccos(0.08058) = 85.38°
Answer
θ = 85.38°
2.Vectors: dot & cross product
Answer: (a × b)x = −4
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 = (−2)·2 − 0·2 = −4
(a × b)y = azbx − axbz = 0·(−4) − 5·2 = −10
(a × b)z = axby − aybx = 5·2 − (−2)·(−4) = 2
Requested component
(a × b)x = −4
Answer
(a × b)x = −4
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 = −33 kN
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Replace each distributed load by its resultant (its area) acting at the centroid of the load diagram, and take moments about a support to remove that support’s reaction from the equation.
Moments about A
Counter-clockwise positive.
ΣMA = 0: RB·5 = 30·3 + 75·8
RB = 690/5 = 138 kN
Vertical equilibrium
RA = 30 + 75 − 138 = −33 kN
Answer
RA = −33 kN
RA is negative: the pin at A has to hold the beam down.
4.Beam support reactions
Answer: RB = 22.25 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 = 3·5 = 15 kN, acting at x̄ = 6.5 m
ΣMA = 0: RB·10 = 25·5 + 3·5·6.5
RB = 222.5/10 = 22.25 kN
ΣFy = 0: RA = 40 − 22.25 = 17.75 kN
Answer
RB = 22.25 kN
5.Truss: method of joints
Answer: FAB = −39.66 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·5 = 55·3 → Cy = 33 kN
ΣFy = 0: Ay = 55 − 33 = 22 kN
Joint A
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
sin α = 2/3.606 = 0.5547
ΣFy = 0: Ay + FAB·sin α = 0
FAB = −22/0.5547 = −39.66 kN
Answer
FAB = −39.66 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.
Tie force from leg BC
By = 7.5 kN
T = 7.5·1.5/1.6 = 7.031 kN
Components of the pin force on BC
Cx = T = 7.031 kN, Cy = −By = −7.5 kN
Magnitude
C = √(7.031² + 7.5²) = 10.28 kN
Answer
C = 10.28 kN
8.Frames & machines
Answer: Ax = 0.594 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.
Direction of BC
BC = 3.937 m, cos = 0.6476, sin = 0.7619
Whole frame: moments about A
FBC·0.7619·3 = 4·0.27 + 6·3
FBC = 8.347 kN
Horizontal balance
Ax + 6 − 8.347·0.6476 = 0
Ax = −0.594 kN → |Ax| = 0.594 kN
Answer
Ax = 0.594 kN
The minus sign means Ax points to the left.
9.Shear force & bending moment
Answer: Mx = 37.5 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 = 50/8 = 6.25 kN (upward)
RA = −RB = −6.25 kN (downward)
Cut just right of the couple
The left part now contains RA and the couple.
Mx = RA·a + M0 = −6.25·2 + 50 = 37.5 kN·m
Answer
Mx = 37.5 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 = 53.5 kN, R2 = 88.5 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 = −30, MC = −10.5, MD = −19, ME = −35.5, MF = −60 kN·m, MG = 0
Answer
Mark F: MF = −60 kN·m
11.Centroid & second moment of area
Answer: Ix = 66.92 × 10⁶ mm⁴
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 = 180·25 = 4500 mm², yf = 280 + 25/2 = 292.5 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.
Split into two rectangles
Measure every y from the bottom edge.
flange: Af = 180·10 = 1800 mm², yf = 180 + 10/2 = 185 mm