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Analysis of Trusses (Joint Method)

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Problem:

For the given truss, determine the forces in all members.

Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram

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Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 1: – Diagram

Problem:

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Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 2: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 2: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 2: – Diagram

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Problem:

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Problem:

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Problem:

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Problem:

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Problem:

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Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 7: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 7: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 7: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 7: – Diagram

Problem:

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Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 8: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 8: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 8: – Diagram

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Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 8: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 8: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 8: – Diagram Analysis of Trusses: Method of Joints | Statics of Rigid Bodies – Problem 8: – Diagram
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Problem (Symmetric truss by joint method):

A triangular truss has supports at A and C, 6m apart, and joint B located 3m horizontally from A and 4m above AC. A 20kN downward load acts at B. Determine the forces in AB and BC.

The geometry of each inclined member is a 3-4-5 triangle. By symmetry, the vertical reactions are 10kN each.

\[ \begin{aligned} \text{Joint B:}\quad 2F\left({4\over5}\right) &= 20 \\ F &= 12.5\ \text{kN} \end{aligned} \] $\boxed{F_{AB}=12.5\ \text{kN compression}},\qquad \boxed{F_{BC}=12.5\ \text{kN compression}}$

Problem (Bottom chord from joint equilibrium):

Using the same triangular truss, determine the force in member AC after finding that AB and BC are each 12.5kN in compression.

Use joint A. Member AB pushes joint A down-left because it is in compression.

\[ \begin{aligned} \text{Joint A:}\quad \sum F_x=0:\quad F_{AC}-12.5\left({3\over5}\right)&=0 \\ F_{AC}&=7.50\ \text{kN} \end{aligned} \] $\boxed{F_{AC}=7.50\ \text{kN tension}}$

Problem (Joint with vertical and horizontal loads):

At a truss joint, member AB is horizontal to the left and member AC is inclined 30° above the horizontal to the right. A 6kN downward load and a 4kN horizontal load to the right act at the joint. Determine the member forces, assuming tension pulls away from the joint.

Analyze the loaded joint directly.

\[ \begin{aligned} \text{Joint:}\quad \sum F_y=0:\quad F_{AC}\sin30^\circ-6&=0 \\ F_{AC}&=12\ \text{kN} \\ \sum F_x=0:\quad -F_{AB}+F_{AC}\cos30^\circ+4&=0 \\ F_{AB}&=14.39\ \text{kN} \end{aligned} \] $\boxed{F_{AC}=12\ \text{kN tension}},\qquad \boxed{F_{AB}=14.39\ \text{kN tension}}$
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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q27

PSAD - Statics / Analysis of Trusses / Engr. Janclyde Espinosa (Clidez)

For the truss shown below, the allowable forces for each member are the ff.
BC=70kN (T)
EC=60kN (C)
DE=25kN (C).

Compute the maximum safe value of P (in kN) if the strength of member EC governs.

  1. 52
  2. 61
  3. 43
  4. 48

Compute the maximum safe value of P (in kN) if the strength of member BC governs.

  1. 61
  2. 43
  3. 52
  4. 48

Compute the maximum safe value of P (in kN) if the strength of member DE governs.

  1. 43
  2. 52
  3. 61
  4. 48
Because the truss is linear elastic, every member force is proportional to the applied load P. Joint equilibrium gives $F_{EC}=1.154P$ (compression), $F_{BC}=1.148P$ (tension), and $F_{DE}=0.581P$ (compression).
Part 1. $1.154P\le60$, so $P=\boxed{52\text{ kN}}$ when EC governs.
Part 2. $1.148P\le70$, so $P=\boxed{61\text{ kN}}$ when BC governs.
Part 3. $0.581P\le25$, so $P=\boxed{43\text{ kN}}$ when DE governs.

Question Bank: q53

PSAD - Statics / Analysis of Trusses / Engr. Janclyde Espinosa (Clidez)

For the truss shown,

q53

If P=3000N and Q=1000N, determine the force in member CD.

  1. 5250 (C)
  2. 2250 (C)
  3. 5250 (T)
  4. 2250 (T)

If P=3000N and Q=1000N, determine the force in member JK.

  1. 2250 (T)
  2. 5250 (C)
  3. 2250 (C)
  4. 5250 (T)

If FCD = 6000N and FGD= 1000N (both under compression), determine the value of P.

  1. 5167.95
  2. 5237.89
  3. 5436.75
  4. 5284.86
Use the method of joints, starting at the supported lower joints. At every joint, apply $\sum F_x=0$ and $\sum F_y=0$; a negative result relative to an assumed tensile force means compression.
Part 1. With $P=3000\text{ N}$ and $Q=1000\text{ N}$, propagating the joint forces upward gives $F_{CD}=\boxed{5250\text{ N (compression)}}$.
Part 2. The same joint solution gives $F_{JK}=\boxed{2250\text{ N (tension)}}$.
Part 3. Use the specified compression forces at CD and GD in the equilibrium equations of joints C and D. Solving for the applied vertical load gives $\boxed{P=5167.95\text{ N}}$.

Question Bank: q324

PSAD - Statics / Analysis of Trusses / Engr. Janclyde Espinosa (Clidez)

For the truss shown, determine the forces in all members by using the method of joints/sections.

q324

Force in AB

  1. 41.408 (T)
  2. 41.408 (C)
  3. 36.592 (T)
  4. 36.592 (C)

Force in AE

  1. 18.262 (C)
  2. 18.262 (T)
  3. 11.408 (C)
  4. 11.408 (T)

Force in BE

  1. 5.74 (C)
  2. 5.74 (T)
  3. 7.493 (T)
  4. 7.493 (C)

Force in EF

  1. 11.408 (C)
  2. 11.408 (T)
  3. 18.262 (C)
  4. 18.262 (T)

Force in BF

  1. 7.493 (T)
  2. 7.493 (C)
  3. 5.74 (C)
  4. 5.74 (T)

Force in BC

  1. 36.592 (T)
  2. 36.592 (C)
  3. 40 (T)
  4. 40 (C)

Force in CD

  1. 36.592 (T)
  2. 36.592 (C)
  3. 40 (T)
  4. 40 (C)

Force in DF

  1. 58.576 (C)
  2. 58.576 (T)
  3. 41.408 (T)
  4. 41.408 (C)

Force in CF

  1. 40 (T)
  2. 40 (C)
  3. 41.408 (T)
  4. 41.408 (C)
### Truss-member forces First obtain the support reactions from the complete truss. Then apply $\sum F_x=0$ and $\sum F_y=0$ joint by joint, starting at joints $A$ and $D$. A positive result denotes tension; a negative result denotes compression. $$AB=\boxed{41.408 \text{kN (T)}},\quad AE=\boxed{18.262 \text{kN (C)}},\quad BE=\boxed{5.74 \text{kN (C)}},$$ $$EF=\boxed{11.408 \text{kN (C)}},\quad BF=\boxed{7.493 \text{kN (T)}},$$ $$BC=CD=\boxed{36.592 \text{kN (T)}},\quad DF=\boxed{58.576 \text{kN (C)}},$$ $$CF=\boxed{40 \text{kN (T)}}.$$

Question Bank: q423

PSAD - Statics / Analysis of Trusses / Engr. Janclyde Espinosa (Clidez)

For the truss shown, if P=20kN

q423

Determine the force of member AB in kN

  1. 38
  2. 46
  3. 28
  4. 21

Determine the force of member AC in kN

  1. 6.84
  2. 5.13
  3. 12.42
  4. 9.44
The $20\text{-kN}$ load is $20^\circ$ below the horizontal. Resolve it at joint $C$:
$$P_x=20\cos20^\circ=18.79\text{ kN},\qquad P_y=20\sin20^\circ=6.84\text{ kN}$$

Member $AC$
At joint $C$, member $AC$ is vertical. Its force balances the vertical component of the load:
$$F_{AC}=P_y=6.84\text{ kN}$$
Thus $AC$ is in tension.

Member $AB$
The horizontal member $BC$ carries $P_x=18.79\text{ kN}$. At joint $B$, the horizontal component of member $AB$, which is at $60^\circ$ to the horizontal, balances this force:
$$F_{AB}\cos60^\circ=18.79$$
$$F_{AB}=\frac{18.79}{0.5}=37.59\text{ kN}\approx38\text{ kN}$$
The direction required at joint $B$ makes $AB$ a compression member.

Therefore, $F_{AB}=\mathbf{38\text{ kN}}$ (compression) and $F_{AC}=\mathbf{6.84\text{ kN}}$ (tension).

Question Bank: q615

PSAD - Statics / Analysis of Trusses / Engr. Janclyde Espinosa (Clidez)

The truss shown is subjected to a lateral load of F=12kN.
a=1.5m
b=4m
c=3.5m

q615

Determine the reaction at A in kN.

  1. 6
  2. 7
  3. 8
  4. 9

Determine the force in member AE, in kN.

  1. 4.19 (T)
  2. 1.91 (T)
  3. 14.40 (C)
  4. 8.98 (C)

Determine the reaction at D, in kN.

  1. 13.4
  2. 6.0
  3. 12.0
  4. 9.1
Resolve the 12-kN lateral load and use $\sum F_x=0$, $\sum F_y=0$, and moments for the truss geometry. The support reactions are $A=6$ kN and $D=13.4$ kN. Joint equilibrium at $A$ gives $F_{AE}=4.19$ kN in tension. Therefore: 6 kN, 4.19 kN (T), and 13.4 kN.

Question Bank: q616

PSAD - Statics / Analysis of Trusses / Mastermatician

A transmission tower is loaded as shown.
P1=10kN
P2=13kN
P3=16kN

q616

Determine the total reaction at H in kN.

  1. 46.55
  2. 25.41
  3. 39.00
  4. 52.11

Determine the force in member CJ, in kN.

  1. 10.7
  2. 9.7
  3. 6.1
  4. 16.1
Solution figure In this solution, we make the two cutting sections as shown to avoid solving the reaction at A. We first obtain the member forces of CD and BC; then, we analyze joint C to determine the force in CJ. In this example, therefore, it is crucial to combine the method of joints and sections. In both cuts, we consider the upper section and we apply the principle of transmissibility to reduce the terms in our equilibrium equations. $$ \sum M_A = 0 $$ $$ -10(5.4)-13(3.6)-16(1.8)-H_y(5.1)=0 $$ $$ H_y=-25.41\,\text{kN}\;(\downarrow) $$ $$ \sum F_x = 0 $$ $$ H_x=10+13+16 $$ $$ H_x=39\,\text{kN} $$ $$ R_H=\sqrt{39^2+25.41^2}=46.55\,\text{kN} $$ $$ \text{@ Section 1-1} $$ $$ \sum M_F=0 $$ $$ -CD\left(\frac{3}{\sqrt{10}}\right)\left(5.1-2(1.2)\right)-10(1.8)=0 $$ $$ CD=-7.03\,\text{kN}\;(\text{C}) $$ $$ \text{@ Section 2-2} $$ $$ \sum M_G=0 $$ $$ -BC\left(\frac{3}{\sqrt{10}}\right)\left(5.1-2(0.6)\right)-10(3.6)-13(1.8)=0 $$ $$ BC=-16.05\,\text{kN}\;(\text{C}) $$ Solution figure $$ \sum F_y = 0 $$ $$ CD\left(\frac{3}{\sqrt{10}}\right) -BC\left(\frac{3}{\sqrt{10}}\right) -CJ\left(\frac{1.8}{2.25}\right)=0 $$ $$ -7.03\left(\frac{3}{\sqrt{10}}\right) -(-16.05)\left(\frac{3}{\sqrt{10}}\right) -CJ\left(\frac{1.8}{2.25}\right)=0 $$ $$ CJ=10.7\,\text{kN}\;(\text{T}) $$ Alternatively, we can make the cutting section 3-3. $$ \text{@ Section 3-3} $$ Solution figure $$ \sum M_F = 0 $$ $$ \left[ -BC\left(\frac{3}{\sqrt{10}}\right) - CJ\left(\frac{1.8}{2.25}\right) \right] \left(5.1 - 2(1.2)\right) - 10(1.8) = 0 $$ $$ \left[ -(-16.05)\left(\frac{3}{\sqrt{10}}\right) - CJ\left(\frac{1.8}{2.25}\right) \right] \left(5.1 - 2(1.2)\right) - 10(1.8) = 0 $$ $$ CJ=10.7\,\text{kN}\;(\text{T}) $$

Question Bank: q624

PSAD - Statics / Analysis of Trusses / Mastermatician

Refer to the figure shown. Diagonals BH, CG, HD, and CI are flexible cables and are, therefore, only capable of carrying tensile forces.
P1=2.5kN
P2=1.2kN
P3=1.2kN

L1=3m
L2=2.25m
L3=3m

q624

Determine the force in member HD.

  1. 0.000
  2. 0.800
  3. 0.784
  4. 0.792

Determine the force in member CI.

  1. 0.800
  2. 0.000
  3. 0.792
  4. 0.784

What is the force in member BH?

  1. 0.000
  2. 0.800
  3. 0.792
  4. 0.784

What is the force in member CG?

  1. 0.792
  2. 0.000
  3. 0.800
  4. 0.784
Because the diagonals are cables, any negative result means the cable is slack and carries zero force. Solve the panel-joint equilibrium equations with $P_1=2.5$, $P_2=P_3=1.2$ kN. The results are $F_{HD}=0$, $F_{CI}=0.800$ kN, $F_{BH}=0$, and $F_{CG}=0.792$ kN. Thus: 0.000 kN, 0.800 kN, 0.000 kN, and 0.792 kN.

Question Bank: q625

PSAD - Statics / Analysis of Trusses / Mastermatician

Refer to the figure shown. Diagonals BH, CG, HD, and CI are flexible cables and are, therefore, only capable of carrying tensile forces.
P1=2.5kN
P2=0.0kN
P3=1.2kN

L1=3m
L2=2.25m
L3=3m

q625

Determine the force in member CG, in kN.

  1. 2.093
  2. 1.850
  3. 0.445
  4. 1.697

Determine the force in member DH, in kN.

  1. 1.850
  2. 0.445
  3. 1.697
  4. 2.093

Determine the force in member DI, in kN.

  1. 1.48
  2. 1.64
  3. 1.66
  4. 0.784
Use the same cable-truss joint equations, setting $P_2=0$. Retain only positive cable forces because flexible diagonals cannot carry compression. The forces are $F_{CG}=2.093 kN$, $F_{DH}=1.850 kN$, and $F_{DI}=1.48 kN$.

Question Bank: q771

PSAD - Statics / Analysis of Trusses / Engr. Clidez

The truss is loaded as shown:

q771

Compute the force in member AF (in kN).

  1. 56.57 (C)
  2. 56.57 (T)
  3. 44.72 (C)
  4. 44.72 (T)

Compute the force in member CF.

  1. 0
  2. 44.72 (T)
  3. 44.72 (C)
  4. 40 (T)

Compute the force in member FG.

  1. 44.72 (C)
  2. 44.72 (T)
  3. 56.57 (C)
  4. 56.57 (C)
Solution figure
The reactions can be solved by taking advantage of the symmetry of the loading. For the reactions $A_y$ and $E_y$, we take the total vertical force and divide it by 2. Solution figure
We analyze joints A and B and determine the forces in the unknown members. We start at joint B and set the vertical forces equal to zero. Since only BF has a vertical internal force, we can directly equate it to the downward joint load at B. However, we cannot find yet the force in member AB and BC since these are two unknown horizontal internal forces.

Hence, we proceed to joint A and solve AF and AB. Since there are only two unknowns on this joint, we can use $\Sigma F_x=0$ and $\Sigma F_y=0$ to solve the unknowns. Solution figure
Here, we use $\Sigma F_x=0$ and $\Sigma F_y=0$ to develop two equations with two unknowns. We then solve this using Mode 5-2 in our calculator. Solution figure
By symmetry, CF and CH are zero-force members. Therefore, $\Sigma F_y=0$ will give us the internal force in CG, which is 40kN (T).