Calculate the moments and fores acting on the hinges as well as the sheer forces of the segment of the doorframe that has the hingesdoor weight =200ib
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Question:
Calculate the moments and fores acting on the hinges as well as the sheer forces of the segment of the doorframe that has the hingesdoor weight =200ib
[3 =
wa Kia weight = 55 Ib
z
in
25m
ESL)
28n
25m M
\, A
An Vedct ll.
i
An N
3
*
lo 52 A]
4 4 1 1 1 1 { 1 1 1 1 1 1 1 1 { | { | 1 { 1 1 { | | 1
This is AI generated Response.
Asked by: Jack
Created at: 2025-10-16 14:20:09
Final Answer:
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# Hinge Forces, Shear, and Moments on the Doorframe Segment
Let's break down the problem based on the provided image and data:
---
### **Given Data**
- **Door weight, \( W_d \):** 200 lb
- **Kid weight, \( W_k \):** 55 lb
- **Door width (distance from hinges to door handle), \( d \):** 19 in
- **Vertical distances:**
- **Top hinge B to bottom hinge A:** 25 in apart
- **Bottom hinge A to bottom of the door:** 19 in
- **Top hinge B to top of the door:** 25 in
#### **Assumptions:**
- Door and kid are both exerting vertical forces.
- The kid hangs at the door handle (point K), at the outer edge, 19 in from the hinge line.
- The weight of the door acts at its center of gravity (midway down the door).
---
## **1. Free-Body Diagram (FBD) Overview**
**Forces acting on the door:**
- **At hinges (A and B):** Unknown forces (vertical \(A_y, B_y\), horizontal \(A_z, B_z\))
- **Door weight:** 200 lb, acting at center (G)
- **Kid weight:** 55 lb, acting at handle (K)
---
## **2. Coordinates of Key Points**
Let the origin be at the bottom hinge (A):
- **A:** (, , )
- **B:** (, 25, )
- **G:** (, 19 + (25+25)/2, ) = (, 19+25, ) = (, 44, ) *(since the center is 25 in above A)*
- **K:** (19, 69, ) *(19 in out along x, at the height of the handle / kid)*
---
## **3. Force Balance Equations**
### **ΣF_y = (vertical equilibrium)**
\[
A_y + B_y = W_d + W_k = 200 + 55 = 255 \text{ lb}
\]
---
## **4. Moment Balance (about A, bottom hinge)**
Take moments about A to solve for \( B_y \):
### **Moments in the Z-direction (out of plane):**
- The moment arm for the **door weight** (at G): \( y_G = 44 \) in
- The moment arm for the **kid weight** (at K): \( y_K = 69 \) in
- The moment arm for the **top hinge reaction**: \( y_B = 25 \) in
\[
\sum M_A = = B_y \times 25 - 200 \times 25 - 55 \times 69
\]
\[
B_y \times 25 = 200 \times 25 + 55 \times 69
\]
\[
B_y = \frac{200 \times 25 + 55 \times 69}{25}
\]
\[
B_y = \frac{500 + 3795}{25} = \frac{8795}{25} = 351.8 \text{ lb}
\]
From vertical equilibrium:
\[
A_y = 255 - 351.8 = -96.8 \text{ lb}
\]
(Negative sign means the bottom hinge "pulls up" rather than pushes down.)
---
## **5. Shear Forces on the Doorframe Segment**
The shear force is equal to the sum of the vertical reactions at the hinges.
- **Shear at A (bottom hinge):** \( -96.8 \) lb (upward)
- **Shear at B (top hinge):** \( 351.8 \) lb (downward)
---
## **6. Horizontal Forces (Z-direction) and Moments**
Assume no external horizontal forces (door is not being pushed/pulled sideways), so the horizontal forces at A and B must be equal and opposite to each other.
\[
A_z + B_z =
\]
If there are no other horizontal loads, \( A_z = -B_z \), but their values are indeterminate with the provided information.
---
## **7. Moment About the Hinges**
### **Moment at Each Hinge:**
#### **At bottom hinge (A):**
- Vertical reaction: \( -96.8 \) lb (up)
- No moment about A from A itself.
- The couple formed by the forces at A and B creates a moment that balances the moments from the weights.
#### **At top hinge (B):**
- Vertical reaction: \( 351.8 \) lb (down)
- Again, the couple is what balances the external loads.
---
## **8. Summary Table**
| Hinge | Vertical Reaction (lb) | Shear (lb) |
|-------|------------------------|------------|
| A | -96.8 (up) | -96.8 |
| B | 351.8 (down) | 351.8 |
---
## **9. Answers to Questions**
### **a) Moments on Hinges**
- The moment is shared as a couple by the vertical forces at A and B, separated by 25 in.
### **b) Shear Forces**
- Shear at A: \( -96.8 \) lb (up)
- Shear at B: \( 351.8 \) lb (down)
### **c) Forces on the Doorframe Segment with Hinges**
- These are the reactions found above, transmitted into the frame at hinge locations.
---
### **10. Diagram with Forces**
```
B (top) 351.8 lb (down)
|---------------------|
| |
| | (door)
| |
|---------------------|
A (bottom) -96.8 lb (up)
```
---
## **11. Additional Notes**
- The negative vertical force at A indicates tension (pulling up), common in double-hinge doors with offset loads.
- If you need the horizontal (axial) forces or the moment in the plane of the door (from a child pulling perpendicular to the door), more information is needed.
---
**If you require the full moment vector or further decomposition, let me know!**
Question:
Nice — thanks for laying out the numbers. Your moment equation about A had the right idea, but the moment arms you used for the door and the kid were inconsistent with the distances you listed. I’ll do a step-by-step, digit-by-digit recomputation using the distances in your message so everything is transparent.
---
## Given / interpreted geometry (from your message)
* Door weight (W_d=200\ \text{lb}).
* Kid weight (W_k=55\ \text{lb}).
* Hinge separation (A\leftrightarrow B = 25\ \text{in}).
* Distance from bottom of door to hinge A = 19 in.
* Distance from top hinge B to top of door = 25 in.
→ therefore total door height = (19 + 25 + 25 = 69\ \text{in}.)
* Door centroid (G) is at midheight of door (= \tfrac{69}{2}=34.5\ \text{in}) measured from bottom.
* Distance from hinge A (at 19 in from bottom) to G: (y_G = 34.5 - 19 = 15.5\ \text{in}.)
* You placed the kid at the handle K at the top (y = 69 in). Distance from A to K: (y_K = 69 - 19 = 50\ \text{in}.)
* Horizontal lever arm from hinge line to handle (d = 19\ \text{in}) (for torsion about the hinge axis).
(If the handle was intended to be at midheight instead of at the top, tell me and I’ll recompute; for now I use the positions you listed.)
---
## 1) Vertical equilibrium (sum of vertical forces)
[
A_y + B_y = W_d + W_k = 200 + 55 = 255\ \text{lb}.
]
---
## 2) Moment equilibrium about A (take positive CCW)
Moments about A (using distances from A):
* Moment from door weight: (M_{d} = 200\ \text{lb} \times 15.5\ \text{in} = 200\times15.5 = 3100\ \text{in·lb}.)
* Moment from kid weight: (M_{k} = 55\ \text{lb} \times 50\ \text{in} = 55\times50 = 2750\ \text{in·lb}.)
* These both produce clockwise moments about A (they try to rotate door downwards), so the upward reaction at B must produce the balancing CCW moment:
[
B_y\cdot(25\ \text{in}) - (3100 + 2750) = 0.
]
So
[
B_y = \frac{3100 + 2750}{25} = \frac{5850}{25} = 234.0\ \text{lb}.
]
Then from vertical equilibrium:
[
A_y = 255 - 234.0 = 21.0\ \text{lb}.
]
**Therefore**
* (B_y = 234.0\ \text{lb}) (upward at top hinge).
* (A_y = 21.0\ \text{lb}) (upward at bottom hinge).
(These are positive upward; no negative/tension sign here.)
---
## 3) Shear forces transmitted into the frame at the hinge locations
* Vertical shear at A = (A_y = 21.0\ \text{lb}) (up).
* Vertical shear at B = (B_y = 234.0\ \text{lb}) (up).
(These are the concentrated vertical reactions that the frame must carry at the hinge fastenings.)
---
## 4) Torsional moment (about the hinge / vertical axis) from the kid pulling / hanging at the handle
Horizontal lever arm (d=19\ \text{in}) and horizontal component of kid’s force (if the kid is hanging vertically, the vertical force does not create torsion; **but** if the kid pulls sideways on the knob or the knob force produces a moment about the hinge axis because it is offset horizontally, the torsion is:)
Given your earlier discussion the 55 lb produces torsion by being at the outer edge (19 in offset). The torsional moment magnitude is:
[
M_z = 55\ \text{lb}\times 19\ \text{in} = 1045\ \text{in·lb}.
]
(= (1045/12\approx87.1\ \text{ft·lb}).)
---
## 5) How the two hinges resist that torsion (in-plane horizontal forces)
Two hinges separated by (L_{AB}=25\ \text{in}) resist the torsion by equal & opposite horizontal forces (H) at A and B such that:
[
M_z = H\cdot L_{AB}\quad\Rightarrow\quad H = \frac{M_z}{L_{AB}} = \frac{1045}{25} = 41.8\ \text{lb}.
]
So one hinge pushes the frame in one horizontal direction with ≈ **41.8 lb**, and the other hinge pushes the opposite direction with ≈ **41.8 lb**. (These are the in-plane shear forces applied by the hinge plates to the frame to produce the resisting couple.)
---
## 6) Compact summary (numbers)
* Vertical reactions:
* (A_y = 21.0\ \text{lb}) (up at bottom hinge)
* (B_y = 234.0\ \text{lb}) (up at top hinge)
* Vertical shear transmitted to frame:
* At A: **21.0 lb**
* At B: **234.0 lb**
* Torsional moment about hinge axis from 55 lb at 19 in:
* (M_z = 1045\ \text{in·lb} \approx 87.1\ \text{ft·lb}.)
* In-plane horizontal forces at hinges to resist that torsion (couple):
* (H_A = -41.8\ \text{lb}), (H_B = +41.8\ \text{lb}) (opposite directions; magnitude **41.8 lb** each).
---
## 7) Why your earlier result was different
You used moment arms (25) in and (69) in in the moment equation (you wrote (200\times25 + 55\times69)). Those arms are inconsistent with the geometry you listed (if A is at 19 in from bottom, the door centroid is at 34.5 in from bottom → moment arm to A is 15.5 in, not 25 in; the kid at top (69 in from bottom) has moment arm 50 in from A, not 69 in). Using the correct distances yields the numbers above.
---
## 8) Next useful calculations I can do for you (tell me which you want)
* compute shear / bending / torsion diagrams for the **frame segment** between hinges (and at the base of the frame),
* convert hinge reaction components into fastener loads (e.g., per screw if each hinge has N screws and you want bolt shear/tension),
* recompute if the handle is actually at **midheight** (44 in) instead of the top (69 in), or if the 55 lb is horizontal (pull) rather than vertical (hang).
Which of those would be most helpful?
This is AI generated Response.
Asked by: Jack
Created at: 2025-10-16 14:24:32
Final Answer:
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# Calculation of Forces and Moments on Doorframe Hinges
This document provides a structured approach to calculating the forces and moments acting on the hinges of a doorframe, considering the weights of the door and a hanging child.
## Given Data
- **Door Weight, \( W_d \):** 200 lb
- **Child Weight, \( W_k \):** 55 lb
- **Hinge Separation (A to B):** 25 in
- **Distance from Bottom of Door to Hinge A:** 19 in
- **Total Door Height:** 69 in (calculated from the given dimensions)
## 1. Free-Body Diagram Overview
### **Forces Acting on the Door:**
- **At Hinges (A and B):** Unknown vertical forces \( A_y \) and \( B_y \)
- **Weight of Door:** 200 lb acting at the center (midway down the door)
- **Weight of Child:** 55 lb acting at the handle
### **Key Points:**
- **Hinge A (Bottom):** Position at (0, 0)
- **Hinge B (Top):** Position at (0, 25)
- **Center of Gravity (G):** Position at (0, 34.5) (midway down the door)
- **Child Position (K):** Position at (19, 69)
## 2. Force Balance Equations
### **Vertical Force Equilibrium:**
\[
A_y + B_y = W_d + W_k = 200 + 55 = 255 \text{ lb}
\]
## 3. Moment Balance about Hinge A
Taking moments about hinge A:
### **Moments Calculation:**
1. **Moment from Door Weight:**
\[
M_d = W_d \times \text{(distance from A to G)} = 200 \text{ lb} \times 15.5 \text{ in} = 3100 \text{ lb·in}
\]
2. **Moment from Child Weight:**
\[
M_k = W_k \times \text{(distance from A to K)} = 55 \text{ lb} \times 50 \text{ in} = 2750 \text{ lb·in}
\]
### **Total Moment About A:**
\[
\sum M_A = B_y \times 25 - (3100 + 2750) = 0
\]
\[
B_y \times 25 = 5850
\]
\[
B_y = \frac{5850}{25} = 234.0 \text{ lb}
\]
### **Vertical Reaction at A:**
\[
A_y = 255 - 234.0 = 21.0 \text{ lb}
\]
## 4. Shear Forces on the Doorframe Segment
**Vertical Shear Forces:**
- **At Hinge A:** \( 21.0 \text{ lb} \) (upward)
- **At Hinge B:** \( 234.0 \text{ lb} \) (upward)
## 5. Torsional Moment from Child Weight
### **Torsional Moment Calculation:**
\[
M_z = W_k \times \text{(horizontal distance to hinge)} = 55 \text{ lb} \times 19 \text{ in} = 1045 \text{ lb·in}
\]
## 6. Horizontal Forces to Resist Torsion
Assuming equal and opposite forces at the hinges:
\[
M_z = H \cdot L_{AB} \quad \Rightarrow \quad H = \frac{M_z}{L_{AB}} = \frac{1045}{25} = 41.8 \text{ lb}
\]
## 7. Summary of Results
| Parameter | Value |
|-----------------|--------------|
| \( A_y \) | 21.0 lb (up) |
| \( B_y \) | 234.0 lb (up) |
| Shear at A | 21.0 lb |
| Shear at B | 234.0 lb |
| Torsional Moment | 1045 lb·in |
| Horizontal Force | 41.8 lb (each hinge) |
## Conclusion
This analysis provides the vertical reactions at the hinges and the torsional moment induced by the hanging child. The calculations ensure the doorframe can support the loads effectively while maintaining structural integrity. If further computations or adjustments are needed, please specify.
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door weight =200ib
[3 =
wa Kia weight = 55 Ib
z
in
25m
ESL)
28n
25m M
\, A
An Vedct ll.
i
An N
3
*
lo 52 A]
4 4 1 1 1 1 { 1 1 1 1 1 1 1 1 { | { | 1 { 1 1 { | | 1