Wize University Statics Textbook (Master) > Structural Analysis - Trusses, Frames, Machines

The Frames & Machines section of the course is nothing new, but an application of what you’ve learned so far. The idea of this section is that rather than analyzing structures as a whole, we analyze them one piece at a time, and all sub-sections must be in equilibrium for the overall structure to remain in equilibrium. 

Frames and Machines are composed of pin-connected multi-force members (members subjected to two or more forces.)  Remember that TRUSS members are all TWO-FORCE members.

Frames - Stationary, Support Loads 

Machines - Contain Moving parts that are designed to transfer/transmit or alter the effect of forces

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Steps for solving Frames & Machines Questions: 

     1) Solve for External Equilibrium Reactions where possible (sometimes we will not find all of the reactions.) 
     2) Identify 2 Force Members (Members that are pin connected with no forces acting on the member. )
     3) Break Apart / Disassemble the structure & Draw the Free Body Diagram of Each Piece Except the 2 Force Members. 
     4) Solve using the three Equations of Equilibrium (→+ΣFx=0\stackrel{+}{\rightarrow} \Sigma F_{x}=0→+ΣFx​=0, +↑ΣFy=0+\uparrow \Sigma F_{y}=0+↑ΣFy​=0, ΣMo=0\Sigma M_o=0ΣMo​=0). 

There are some key points to keep in mind:
Do not break down the structure more than you need to.
Everywhere you take things apart, the internal force are equal and opposite. For example.......
It’s always beneficial to consider if you have 2 or 3 force members, and whether they simplify the setup.
Remember that every FBD in 2D allows you to solve for a maximum of 3 unknowns.
You may see pulleys in this section. Remember that any continuous rope around a pulley has the same tension, and that the pulley is typically pin connected at its center unless told otherwise.

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Example
Neglecting the weight of the members in the frame shown, determine the magnitude of the reaction force components at A, B, and C. 

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1) Solve for Global Equilibrium Reactions where possible.

CCW(+)ΣMA=0:−462N(1.155m)+800N(2m)+By(2m+1.15m)=0By=−338N(↓)\begin{array}{l}{C C W(+) \Sigma M_{\mathrm{A}}=0:-462 N(1.155 m)+800 N(2 m)+B_{y}(2 m+1.15 m)=0} \\ {B_{y}=-338 N(\downarrow)}\end{array}CCW(+)ΣMA​=0:−462N(1.155m)+800N(2m)+By​(2m+1.15m)=0By​=−338N(↓)​
  ** By = -338 (as drawn, 'up') which is the same as 338 'down'.

+↑ΣFy=0:Ay−462Ncos⁡(60)+By=0Ay−462Ncos⁡(60)+(−338N)=0Ay=−569N\begin{array}{l}{+\uparrow \Sigma F_{y}=0: A_{y}-462 N \cos (60)+B_{y}=0} \\ {A_{y}-462 N \cos (60)+(-338 N)=0} \\ {A_{y}=-569 N}\end{array}+↑ΣFy​=0:Ay​−462Ncos(60)+By​=0Ay​−462Ncos(60)+(−338N)=0Ay​=−569N​
**  Ay = +569 'up'

2) Identify 2 Force Members. 

-There is none in this case!

3) Break Apart the Assembly and Draw the Free Body Diagram of Each Piece Except the 2 Force Members. 

4) Solve using the Three Equations of Equilibrium. 

From FBD II (assembly on the right),

+↑ΣFy=0:By−Cy=0−338N−Cy=0Cy=−338N(↑)\begin{array}{l}{+\uparrow \Sigma F_{y}=0: B_{y}-C_{y}=0} \\ {-338 N-C_{y}=0} \\ {C_{y}=-338 N(\uparrow)}\end{array}+↑ΣFy​=0:By​−Cy​=0−338N−Cy​=0Cy​=−338N(↑)​
  ** Cy = -338  means it is actually 338 'up', opposite to how it was drawn in the FBD.

CCW(+)ΣMB=0:−Cx(2m)+Cy(2m)+800N(2m)=0−Cx(2m)+−338(2m)+800N(2m)=0Cx=462N\begin{array}{l}{C C W(+) \Sigma M_{B}=0:-C_{x}(2 m)+C_{y}(2 m)+800 N(2 m)=0} \\ {-C_{x}(2 m)+-338(2 m)+800 N(2 m)=0} \\ {C_{x}=462 N}\end{array}CCW(+)ΣMB​=0:−Cx​(2m)+Cy​(2m)+800N(2m)=0−Cx​(2m)+−338(2m)+800N(2m)=0Cx​=462N​

→+ΣFx=0:Cx−800N+Bx=0462N−800N+Bx=0Bx=338N\begin{array}{l}{\stackrel{+}{\rightarrow} \Sigma F_{x}=0: C_{x}-800 N+B_{x}=0} \\ {462 N-800 N+B_{x}=0} \\ {B_{x}=338 N}\end{array}→+ΣFx​=0:Cx​−800N+Bx​=0462N−800N+Bx​=0Bx​=338N​
  ** the direction is 'to the right' as drawn in the FBD

From FBD I, 

→+ΣFx=0:Ax+462Nsin⁡(60)−Cx=0Ax+462Nsin⁡(60)−462N=0Ax=61.9N\begin{array}{l}{\stackrel{+}{\rightarrow} \Sigma F_{x}=0: A_{x}+462 N \sin (60)-C_{x}=0} \\ {A_{x}+462 N \sin (60)-462 N=0} \\ {A_{x}=61.9 \mathrm{N}}\end{array}→+ΣFx​=0:Ax​+462Nsin(60)−Cx​=0Ax​+462Nsin(60)−462N=0Ax​=61.9N​

Magnitude Reactions at A,

RA=(A)x2+(A)y2=(61.9N)2+(−569N)2=572N( ans )R_{A}=\sqrt{(A)_{x}^{2}+(A)_{y}^{2}}=\sqrt{(61.9 N)^{2}+(-569 N)^{2}}=572 N(\text { ans })RA​=(A)x2​+(A)y2​​=(61.9N)2+(−569N)2​=572N( ans )

Magnitude Reactions at B,

RB=(B)x2+(B)y2=(338N)2+(−338N)2=478N( ans )R_{B}=\sqrt{(B)_{x}^{2}+(B)_{y}^{2}}=\sqrt{(338 N)^{2}+(-338 N)^{2}}=478 N(\text { ans })RB​=(B)x2​+(B)y2​​=(338N)2+(−338N)2​=478N( ans )

Magnitude Force at C, 

FC=(C)x2+(C)y2=(462N)2+(−338N)2=572N( ans )F_{C}=\sqrt{(C)_{x}^{2}+(C)_{y}^{2}}=\sqrt{(462 N)^{2}+(-338 N)^{2}}=572 N(\text { ans })FC​=(C)x2​+(C)y2​​=(462N)2+(−338N)2​=572N( ans )

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Example
Neglecting the weight of the members in the frame shown, determine the magnitude of the reaction force components at A, B, and C. 

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1) Solve for Global Equilibrium Reactions where possible.

CCW(+)ΣMA=0:−462N(1.155m)+800N(2m)+By(2m+1.15m)=0By=−338N(↓)\begin{array}{l}{C C W(+) \Sigma M_{\mathrm{A}}=0:-462 N(1.155 m)+800 N(2 m)+B_{y}(2 m+1.15 m)=0} \\ {B_{y}=-338 N(\downarrow)}\end{array}CCW(+)ΣMA​=0:−462N(1.155m)+800N(2m)+By​(2m+1.15m)=0By​=−338N(↓)​

+↑ΣFy=0:Ay−462Ncos⁡(60)+By=0Ay−462Ncos⁡(60)+(−338N)=0Ay=−569N\begin{array}{l}{+\uparrow \Sigma F_{y}=0: A_{y}-462 N \cos (60)+B_{y}=0} \\ {A_{y}-462 N \cos (60)+(-338 N)=0} \\ {A_{y}=-569 N}\end{array}+↑ΣFy​=0:Ay​−462Ncos(60)+By​=0Ay​−462Ncos(60)+(−338N)=0Ay​=−569N​

2) Identify 2 Force Members. 

-There is none in this case!

3) Break Apart the Assembly and Draw the Free Body Diagram of Each Piece Except the 2 Force Members. 

4) Solve using the Three Equations of Equilibrium. 

From FBD II (assembly on the right),

+↑ΣFy=0:By−Cy=0−338N−Cy=0Cy=−338N(↑)\begin{array}{l}{+\uparrow \Sigma F_{y}=0: B_{y}-C_{y}=0} \\ {-338 N-C_{y}=0} \\ {C_{y}=-338 N(\uparrow)}\end{array}+↑ΣFy​=0:By​−Cy​=0−338N−Cy​=0Cy​=−338N(↑)​

CCW(+)ΣMB=0:−Cx(2m)+Cy(2m)+800N(2m)=0−Cx(2m)+−338(2m)+800N(2m)=0Cx=462N\begin{array}{l}{C C W(+) \Sigma M_{B}=0:-C_{x}(2 m)+C_{y}(2 m)+800 N(2 m)=0} \\ {-C_{x}(2 m)+-338(2 m)+800 N(2 m)=0} \\ {C_{x}=462 N}\end{array}CCW(+)ΣMB​=0:−Cx​(2m)+Cy​(2m)+800N(2m)=0−Cx​(2m)+−338(2m)+800N(2m)=0Cx​=462N​

→+ΣFx=0:Cx−800N+Bx=0462N−800N+Bx=0Bx=338N\begin{array}{l}{\stackrel{+}{\rightarrow} \Sigma F_{x}=0: C_{x}-800 N+B_{x}=0} \\ {462 N-800 N+B_{x}=0} \\ {B_{x}=338 N}\end{array}→+ΣFx​=0:Cx​−800N+Bx​=0462N−800N+Bx​=0Bx​=338N​

From FBD I, 

→+ΣFx=0:Ax+462Nsin⁡(60)−Cx=0Ax+462Nsin⁡(60)−462N=0Ax=61.9N\begin{array}{l}{\stackrel{+}{\rightarrow} \Sigma F_{x}=0: A_{x}+462 N \sin (60)-C_{x}=0} \\ {A_{x}+462 N \sin (60)-462 N=0} \\ {A_{x}=61.9 \mathrm{N}}\end{array}→+ΣFx​=0:Ax​+462Nsin(60)−Cx​=0Ax​+462Nsin(60)−462N=0Ax​=61.9N​

Magnitude Reactions at A,

RA=(A)x2+(A)y2=(61.9N)2+(−569N)2=572N( ans )R_{A}=\sqrt{(A)_{x}^{2}+(A)_{y}^{2}}=\sqrt{(61.9 N)^{2}+(-569 N)^{2}}=572 N(\text { ans })RA​=(A)x2​+(A)y2​​=(61.9N)2+(−569N)2​=572N( ans )

Magnitude Reactions at B,

RB=(B)x2+(B)y2=(338N)2+(−338N)2=478N( ans )R_{B}=\sqrt{(B)_{x}^{2}+(B)_{y}^{2}}=\sqrt{(338 N)^{2}+(-338 N)^{2}}=478 N(\text { ans })RB​=(B)x2​+(B)y2​​=(338N)2+(−338N)2​=478N( ans )

Magnitude Force at C, 

FC=(C)x2+(C)y2=(462N)2+(−338N)2=572N( ans )F_{C}=\sqrt{(C)_{x}^{2}+(C)_{y}^{2}}=\sqrt{(462 N)^{2}+(-338 N)^{2}}=572 N(\text { ans })FC​=(C)x2​+(C)y2​​=(462N)2+(−338N)2​=572N( ans )

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Example
Determine the force in cable CD. Neglect the weight of the members in the frame. 

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1) Solve for Global Equilibrium Reactions where possible.

- Not possible in this case. 

2) Identify 2 Force Members. 

-There is none in this case!

3) Break Apart the Assembly and Draw the Free Body Diagram of Each Piece Except the 2 Force Members. 

4) Solve using the Three Equations of Equilibrium. 
  
From FBD II,

CCW(+)ΣMA=0:−1080N(5.4m2)+By(5.4m)=0By=540N\begin{array}{l}{C C W(+) \Sigma M_{A}=0:-1080 N\left(\frac{5.4 m}{2}\right)+B_{y}(5.4 m)=0} \\ {B_{y}=540 N}\end{array}CCW(+)ΣMA​=0:−1080N(25.4m​)+By​(5.4m)=0By​=540N​

→ΣFx=0:Ax+Bx=0Ax=−Bx\begin{array}{l}{\rightarrow \Sigma F_{x}=0: A_{x}+B_{x}=0} \\ {A_{x}=-B_{x}}\end{array}→ΣFx​=0:Ax​+Bx​=0Ax​=−Bx​​

From FBD III, 

CCW(+)ΣMF=0:TC(0.9m)+By(3.6m)+Bx(3.6m)=0TC(0.9m)+540N(3.6m)+Bx(3.6m)=0\begin{array}{l}{C C W(+) \Sigma M_{F}=0: T_{C}(0.9 m)+B_{y}(3.6 m)+B_{x}(3.6 m)=0} \\ {T_{C}(0.9 m)+540 N(3.6 m)+B_{x}(3.6 m)=0}\end{array}CCW(+)ΣMF​=0:TC​(0.9m)+By​(3.6m)+Bx​(3.6m)=0TC​(0.9m)+540N(3.6m)+Bx​(3.6m)=0​

0.9TC+3.6Bx=−1944  (1)0.9 T_{C}+3.6 B_{x}=-1944 \ \ (1)0.9TC​+3.6Bx​=−1944  (1)

From FBD II, 

CCW(+)ΣME=0:Ax(5.4m)−TC(2.7m)=0C C W(+) \Sigma M_{E}=0: A_{x}(5.4 m)-T_{C}(2.7 m)=0CCW(+)ΣME​=0:Ax​(5.4m)−TC​(2.7m)=0
−2.7Tc+5.4Ax=0  -2.7 T_{c}+5.4 A_{x}=0 \ \ −2.7Tc​+5.4Ax​=0  

Since Ax = -Bx from above, 

−2.7TC−5.4Bx=0  (2)-2.7 T_{C}-5.4 B_{x}=0 \ \ (2)−2.7TC​−5.4Bx​=0  (2)

Solve for Tc from Eqs (1) and (2) by using substitution or elimination,

Using Elimination Method, 

(0.9TC+3.6Bx=−1944)×1.5−2.7TC−5.4Bx=0\begin{array}{l}{\left(0.9 T_{C}+3.6 B_{x}=-1944\right) \times 1.5} \\ {-2.7 T_{C}-5.4 B_{x}=0}\end{array}(0.9TC​+3.6Bx​=−1944)×1.5−2.7TC​−5.4Bx​=0​

1.35TC+5.4Bx=−2916−2.7TC−5.4Bx=0\begin{array}{l}{1.35 T_{C}+5.4 B_{x}=-2916} \\ {-2.7 T_{C}-5.4 B_{x}=0}\end{array}1.35TC​+5.4Bx​=−2916−2.7TC​−5.4Bx​=0​

TC=2160NT_{C}=2160 NTC​=2160N

Determine the force P requires to carry the 100N load.

Looking at pulley D:

T1=33.3NT_1=33.3NT1​=33.3N
Looking @ pulley E:

The compound beam is fixed at A and supported by a rock at B and C. There are hinges (pins) at D and E.
Determine the reactions at the supports (A & C). 

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SOLUTION:

THIS IS A COMPOUND BEAM!!! You MUST separate it at the pins, and apply the equations of equilibrium to EACH section (AD, DE, EC).

1) draw FBD of each section.

2) apply equations of equilibrium to EACH member.

ANSWERS:

Practice
Determine the tension in the cable at B when the 180 N force is applied to the pedal at E. Also, calculate the reaction force components at D. If the cable at B has a diameter of 7 .0 mm and the yield stress of the cable material is 70 MPa, determine the actual load (safety) factor for the cable. In your opinion, is this a 'safe' design? Neglect friction and mass of the components. 

Tension \ \ at \ \ B

Enter Answer to 3 Sig Figs. Do not Enter Units.

D_{x}

Enter Answer to 3 Sig Figs. Do not Enter Units.

D_{y}

Enter Answer to 3 Sig Figs. Do not Enter Units.

I don't know

Practice
For the frame shown below joints B, E, and C are pinned connections. Determine the components of the reactions at the fixed support at D and find the components of the forces at joints B, C, and E. 

M_{D}=

Enter Answer to 3 Sig Figs - Do Not enter units.

D_{x}=

Enter Answer to 3 Sig Figs - Do Not enter units.

D_{y}=

Enter Answer to 3 Sig Figs - Do Not enter units.

C_{x}=E_{x}

Enter Answer to 3 Sig Figs - Do Not enter units.

C_{y}=E_{y}

Enter Answer to 3 Sig Figs - Do Not enter units.

B_{x}=

Enter Answer to 3 Sig Figs - Do Not enter units.

B_{y}=

Enter Answer to 3 Sig Figs - Do Not enter units.

I don't know

The mechanism below has pin joints at A and B is driven by a couple moment M = 200 N-m at A. Joint C is a frictionless slider which is attached to member AC by a frictionless pin.
 a) Calculate the force P required to maintain the system in equilibrium. P acts at a right angle to BD, and AC is vertical.
 b) Determine the reactions at A and B.

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Force \ \ P

Enter Answer to 3 Sig Figs. Do not Enter Units.

A_{x}

Enter Answer to 3 Sig Figs. Do not Enter Units.

A_{y}

Enter Answer to 3 Sig Figs. Do not Enter Units.

B_{x}

Enter Answer to 3 Sig Figs. Do not Enter Units.

B_{y}

Enter Answer to 3 Sig Figs. Do not Enter Units.

I don't know

Practice

Determine the forces in each of the hydraulic pistons connected at C and E.

BC =

Enter Answer to 3 Sig Figs.

EC =

Enter Answer to 3 Sig Figs.

I don't know

Quiz: Frames and Machines Quiz

Determine the force P needed to lift the 800N load. Determine where the hook must be place (x) for the bar AB to remain horizontal.

P (in N)

x (in mm)

I don't know

Note: This is a frames and machines question
Determine the support forces at pins A, B and C when the load is 200lbs.

Enter only the magnitudes in lb

I don't know

Determine the support reactions at D (which’s a fixed connection).

Moment

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Determine the tension that must be applied at the winch (w) to support the 700lb force. Determine the force through member BD.

Tension at winch

Force in BD

I don't know

Determine the clamping force at E, if a force F = 50 N is applied.

Answer

I don't know

Determine the force in the piston AB when a 2 kN engine is carried.

Answer

I don't know