I need help REALLY URGENTLY. I have an assignment due in less than 3 hours and I don't dnerstand what the fuck I'm doing. does anyone know how to calclate forces in a truss using method of joints please
Okay — breathe. We’ve got <3 hours, so I’ll give you a
quick, exact, walk-through of the
Method of Joints and a fully worked example you can copy-style for your assignment. I’ll be explicit about every arithmetic step.
Quick rules / plan
- Draw the truss and label every joint and member.
- Replace supports with reactions (pin → two reactions, roller → one vertical reaction).
- Find support reactions first from equilibrium of the whole truss (∑Fx=0, ∑Fy=0, ∑M=0).
- Start at a joint that has at most two unknown member forces (so you can solve two equilibrium equations: ∑Fx=0 and ∑Fy=0).
- For each joint: assume each member force is tension (pulling away from the joint). If your solved force is negative → it’s actually compression.
- Use geometry to get direction cosines (cosθ, sinθ) for inclined members. Solve algebraically.
- Repeat joint-by-joint until all member forces are found.
Worked example (simple triangular truss)
Truss geometry: triangle with base AB, top C.
Supports: A = pin (Ax, Ay). B = roller (By only vertical).
Members: AB (base), AC (left incline), BC (right incline).
Dimensions: AB = 4.0 m (A at x=0, B at x=4.0), C at midpoint x=2.0 and height 3.0 m → coordinates:
- A = (0,0)
- B = (4.0,0)
- C = (2.0,3.0)
External load: at C, a downward vertical load P = 10.0 kN.
1) Support reactions (whole-truss equilibrium)
Sum of vertical forces: Ay + By − 10.0 = 0. …(1)
Take moments about A to eliminate Ay and get By. Moment sign: counterclockwise positive.
Moment from By about A: By * 4.0 (By upward produce CCW).
Moment from 10.0 kN at C about A: 10.0 * (horizontal distance from A to C) = 10.0 * 2.0 = 20.0 (produces clockwise so negative when balancing, but I'll set equation By
4.0 = 10.02.0).
So:
By * 4.0 = 10.0 * 2.0
By * 4.0 = 20.0
By = 20.0 / 4.0 = 5.0 kN.
Then from (1):
Ay + 5.0 − 10.0 = 0 → Ay = 10.0 − 5.0 = 5.0 kN.
So reactions:
Ay = 5.0 kN up,
By = 5.0 kN up.
2) Geometry for member AC (and BC)
Vector A→C = (2.0, 3.0).
Length AC = √(2.0² + 3.0²) = √(4.0 + 9.0) = √13.
Compute √13 exactly: √13 ≈ 3.605551275463989. (I’ll use 3.605551 for intermediate display.)
Direction cosines (from A toward C):
- cosθ = Δx / length = 2.0 / √13 = 2.0 / 3.605551275463989 ≈ 0.5547001962252291
- sinθ = Δy / length = 3.0 / √13 = 3.0 / 3.605551275463989 ≈ 0.8320502943378437
(Those are the exact fractions cosθ = 2/√13, sinθ = 3/√13.)
3) Joint A (solve two unknowns: FAB and FAC)
At joint A, unknown member forces:
- (F_{AB}) (horizontal along AB, positive to the right)
- (F_{AC}) (along AC, assumed tension pointing away from A toward C)
Equilibrium at A:
Sum of vertical forces (∑Fy = 0):
Ay + (F_{AC}) * sinθ = 0
5.0 + (F_{AC})*(3/√13) = 0
Solve for (F_{AC}):
(F_{AC}) = −5.0 * (√13 / 3)
Compute step-by-step:
- √13 ≈ 3.605551275463989
- √13 / 3 ≈ 3.605551275463989 / 3 = 1.201850425154663
- −5.0 * 1.201850425154663 = −6.009252125773315
So
(F_{AC} = −6.009252125773315) kN (negative → the member is
in compression, magnitude 6.009 kN).
Now sum of horizontal forces (∑Fx = 0):
(F_{AB}) + (F_{AC}) * cosθ = 0
So (F_{AB}) = −(F_{AC}) * cosθ.
Compute cosθ = 2/√13 ≈ 0.5547001962252291.
Plug in:
(F_{AB}) = −(−6.009252125773315) * 0.5547001962252291
= 6.009252125773315 * 0.5547001962252291
Multiply:
6.009252125773315 × 0.5547001962252291 ≈ 3.333333333333333
So
(F_{AB} = +3.333333333333333) kN (positive →
tension, magnitude 3.333 kN).
4) Joint B (symmetry — same magnitudes)
Because the truss and loading are symmetric, joint B will give identical magnitudes:
- (F_{BC} = −6.009252125773315) kN (compression, same as AC)
- (F_{AB} = 3.333333333333333) kN (same AB we already found)
(If you prefer to solve it explicitly, do the joint B equilibrium the same way as A.)
5) Joint C (check equilibrium)
At C the vertical components of AC and BC must balance the 10.0 kN downwards. Each inclined member contributes vertical component:
Vertical component of one member (magnitude 6.009252... kN) inwards toward the joint: for AC at C the force sign flips relative to joint A, but total vertical contribution = 2 * (|F_AC| * sinθ). Compute:
|F_AC| * sinθ = 6.009252125773315 * (3/√13)
But from earlier, we know 6.009252... * (3/√13) = 5.0 (because that was how we computed FAC). So each gives 5.0 kN vertical, two of them give 10.0 kN up, balancing the 10.0 kN down. Good — equilibrium check passes.
Final answers (rounded sensibly)
- Member AB (bottom): 3.333 kN — TENSION
- Member AC (left incline): 6.009 kN — COMPRESSION
- Member BC (right incline): 6.009 kN — COMPRESSION
- Reactions: Ay = 5.0 kN up, By = 5.0 kN up, Ax = 0 (no horizontal external loads)
(If your assignment expects 3 s.f.: AB = 3.33 kN T; AC = 6.01 kN C; BC = 6.01 kN C.)
Quick checklist you can hand in
- Free-body diagram of full truss with loads and support reactions labelled.
- Equations for support reactions (show sum M about A → By = 5.0 kN; then sum Fy → Ay = 5.0 kN).
- Joint A equilibrium: ∑Fx and ∑Fy shown, geometry used (cosθ = 2/√13, sinθ = 3/√13), solved for FAC and FAB and sign interpreted.
- State symmetry (or solve joint B similarly).
- Check joint C equilibrium as verification.
- State final forces and whether each is T or C and the units (kN).
If your truss in the assignment is
different (different geometry, different load positions, or more members), paste the exact geometry and loads here
right now and I’ll set up the equations and solve them the same way. I won’t ask you to wait — I’ll compute it right away.
