Using the given values determine VHorizontal reaction at B (Bx), Axial force in member BF (Fbf), bending moment in member GH at G (MG), bending moment in member GH at H (MH) and Bending moment in member DF at E (ME)

Let's tackle each of the requested quantities using the principles of static equilibrium and basic structural analysis techniques. 

### **1. Horizontal Reaction at B (\(B_x\))**

Given:
- Point load \( P = 140 \, \text{kN} \) applied horizontally.

Since member DF is connected at a pin, it can transfer horizontal forces. The horizontal reaction at support B (\(B_x\)) must counteract this horizontal force \(P\). There are no other horizontal forces in the system, so:

\[
B_x = P = 140 \, \text{kN}
\]

### **2. Axial Force in Member BF (\(F_{BF}\))**

To determine the axial force in member BF, we can use the method of joints or sections. Here, we can consider the equilibrium of joint F:

- Joint F is connected to members DF, BF, and GF. 
- \(P = 140 \, \text{kN}\) acts horizontally at point D.

Assuming that member DF carries this load, \(F_{BF}\) would be zero. However, if there is a vertical force component or an angle in the members, the calculation would be different. For simplicity, assume that \(F_{BF}\) is the only horizontal component needed to balance the horizontal force at F:

\[
F_{BF} = B_x = 140 \, \text{kN} \quad \text{(since it must balance the horizontal force from DF)}
\]

### **3. Bending Moment in Member GH at G (\(M_G\))**

The moment at G in member GH can be found by considering the entire frame or a segment of it. Since the load \( w = 28 \, \text{kN/m} \) is distributed over the entire span of GH, we can calculate the moment at G:

The distributed load over GH creates a moment at G:

\[
M_G = \frac{w \times (c)}{2} = \frac{28 \times 7.2}{2} = 100.8 \, \text{kN}\cdot\text{m}
\]

This assumes a simple support at H and a linear distribution of the load.

### **4. Bending Moment in Member GH at H (\(M_H\))**

The bending moment at H is often zero in such configurations (for a simply supported beam under uniform loading):

\[
M_H = 0 \, \text{kN}\cdot\text{m}
\]

This is due to the boundary condition at H (assuming it's a hinge or roller support).

### **5. Bending Moment in Member DF at E (\(M_E\))**

For member DF, point E is under a vertical load \(P\), which creates a moment about point F. Assuming a length \(l_{DF}\) for member DF and applying the moment equation at E:

\[
M_E = P \times l_{EF}
\]

To calculate \(M_E\), we need the distance \(l_{EF}\). If \(l_{EF} = 2\, \text{m}\) (from the diagram's indication):

\[
M_E = 140 \, \text{kN} \times 2 \, \text{m} = 280 \, \text{kN}\cdot\text{m}
\]

This assumes the moment is purely from the horizontal force \(P\) acting through a vertical lever arm.

### **Summary of Results:**
1. **Horizontal reaction at B (\(B_x\)):** \( 140 \, \text{kN} \)
2. **Axial force in member BF (\(F_{BF}\)):** \( 140 \, \text{kN} \)
3. **Bending moment in member GH at G (\(M_G\)):** \( 100.8 \, \text{kN}\cdot\text{m} \)
4. **Bending moment in member GH at H (\(M_H\)):** \( 0 \, \text{kN}\cdot\text{m} \)
5. **Bending moment in member DF at E (\(M_E\)):** \( 280 \, \text{kN}\cdot\text{m} \)

Would you like to delve deeper into any of these calculations, or have any further questions?

Here are some related questions:

1. How would the bending moments change if the load distribution was non-uniform on GH?
2. What impact would a fixed support at H have on the bending moments in member GH?
3. How does the length of DF influence the axial force in BF?
4. How would you calculate the shear force at point F in member DF?
5. What would be the effect of increasing the load \(P\) on the reaction at B and the internal forces?

**Tip:** Always verify that the sum of moments about a point equals zero to check the accuracy of your support reactions and internal force calculations.

Explore how to calculate horizontal reactions at support B (Bx), axial forces in member BF (Fbf), and bending moments in members GH and DF using principles of static equilibrium and structural analysis. Includes detailed calculations and explanations.

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