VIPSolutionsGive step-by-step solution with explanation and final answer:11. The thermal stress ditbution in 8 corposie cynical pressure vessel requires the sinulanacus sokton of the adie displacement and the temperate gradient squations. The coupling occurs because the elastic modulus E and the thermal expansion coefficient a sre temperature-dependent. For a steady-state condition where the inner wall is at T and the outer walls at T,, the radial stress a; must satisfy the equilibrium of the differential element. The following multi-physics framework defines the compatibility between the creurerential stain and the integrate thermal expansion cross the vessel tikness do, 0 —0p wt 5 0 : = glos— vo] + aAT 12. A detald structural health moritoing report for a suspension bidge dentfies ass of tension in several secondary hangers. The bridge has a main span of 1200 m and a sag of 120 m. The cables are made of high-strength steel with an ultimate tensile strength of 1860 MPa.
Question:
Give step-by-step solution with explanation and final answer:
11. The thermal stress ditbution in 8 corposie cynical pressure vessel requires the
sinulanacus sokton of the adie displacement and the temperate gradient squations.
The coupling occurs because the elastic modulus E and the thermal expansion coefficient
a sre temperature-dependent. For a steady-state condition where the inner wall is at T
and the outer walls at T,, the radial stress a; must satisfy the equilibrium of the
differential element. The following multi-physics framework defines the compatibility
between the creurerential stain and the integrate thermal expansion cross the vessel
tikness
do, 0 —0p
wt 5 0
:
= glos— vo] + aAT
12. A detald structural health moritoing report for a suspension bidge dentfies ass of
tension in several secondary hangers. The bridge has a main span of 1200 m and a sag of
120 m. The cables are made of high-strength steel with an ultimate tensile strength of
1860 MPa.
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Asked by: Rahul
Created at: 2026-01-24 21:07:11
Asked by: Rahul
Created at: 2026-01-24 21:07:11
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Let's solve ** 12** step by step:
---
## **12. Suspension Bridge Cable Tension**
### **Given Data**
-Main span of the bridge (\(L\))** 120
- **Sag (\(\))** 120 m
-Ultimate tensile strength of cable (\(\_{ult}\**: 186 MPa =1,860,,000\) Pa
are to estimate **maximum tension** in main cable.
### **Step1: Understanding the System**
The cable of a bridge forms aparabolic curve** under uniformly distributed load (the bridge deck). **maximum tension** occurs at the **anchorages**), not at the midspan.
---
### **Step 2: Horizontal and Maximum Tension Formulas**
#### **Horizontal Component of Tension (\(H\)):**
For a uniformly loaded cable:
\[
H = \frac{w L^2}{8 h}
\]
\(w\) = uniformly distributed load per unit horizontalN/m)
- \(L\) = span (m- \(h\) = sag (m)
#### **Maximum Tension (\(T_{}\)):**
\[
T_{max} = \sqrt{H^2 + \left(\fracw L}{2}\right)^2}
\]
most cases, if \(h\) is not too small, \(T_{max} \ H\) near the supports.
---
### **Step 3: Required Cable Cross-Sectional Area**
The **maximum allowable tension** is:
\[
T_{max} = \sigma_{ult} \cdot A
\]
Where:
- \(A\) = cross-sectional area of the cable (m²)
- \(\sigma_{ult}\) = ultimate tensile strength (Pa)
---
### **Step 4: Final Formulae**
If we know the **actual tension** required (from the bridge's weight), we can check if the cable is safe.
But since only ultimate strength is given, **the maximum tension the cable can support** is:
\[
T_{max} = \sigma_{ult} \cdot A
\]
However, **without the area (\(A\)) of the cable or the load (\(w\)), we cannot compute the actual tension**. But if the question is asking for the **maximum possible tension** in the cable based on the tensile strength, it is:
\[
\boxed{T_{max} = \sigma_{ult} \times A}
\]
Where:
- \(\sigma_{ult} = 1.86 \times 10^9\) Pa
- \(A =\) cross-sectional area of the cable (unknown unless given)
If you need to find the **maximum load the cable can support**, use:
\[
w_{max} = \frac{8 h \sigma_{ult} A}{L^2}
\]
---
### **Step 5: If Only Ultimate Strength is Required**
**Final Answer:**
- The **maximum tension** that each cable can safely support (before breaking) is:
\[
\boxed{T_{max} = 186\ \text{MPa} \times A = 1.86 \times 10^9 \ \text{Pa} \times A}
\]
where \(A\) is the cross-sectional area of the cable.
---
### **Note**
If you are given the **cross-sectional area** or the **total load** the cable must support, substitute that value to get the numerical answer.
---
#### **Summary Table**
| Quantity | Value | Unit |
|-------------------|--------------|--------------|
| Span (\(L\)) | 120 | m |
| Sag (\(h\)) | 120 | m |
| Ultimate Strength (\(\sigma_{ult}\)) | 186 | MPa (= \(1.86 \times 10^9\) Pa) |
| Maximum Tension (\(T_{max}\)) | \(1.86 \times 10^9 \times A\) | N (where \(A\) is in m²) |
---
If you have the **cross-sectional area**, plug it in for a numerical value! If you need help with a specific value for \(A\) or \(w\), please provide those details.
Question:
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This is AI generated Response.
Asked by: Rahul
Created at: 2026-01-24 21:14:37
Asked by: Rahul
Created at: 2026-01-24 21:14:37
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# **12. Suspension Bridge Cable Tension**
## **Given Data**
- **Main span of the bridge (\(L\))**: 120 m
- **Sag (\(h\))**: 120 m
- **Ultimate tensile strength of cable (\(\sigma_{ult}\))**: 1860 MPa = \(1,860,000\) Pa
## **Step 1: Understanding the System**
The cable of a suspension bridge forms a parabolic curve under a uniformly distributed load (the bridge deck). The maximum tension occurs at the **anchorages**, not at the midspan.
## **Step 2: Horizontal and Maximum Tension Formulas**
### **Horizontal Component of Tension (\(H\))**
For a uniformly loaded cable, the horizontal component of tension is given by:
\[
H = \frac{w L^2}{8 h}
\]
Where:
- \(w\) = uniformly distributed load per unit length (N/m)
- \(L\) = span (m)
- \(h\) = sag (m)
### **Maximum Tension (\(T_{max}\))**
The maximum tension in the cable is calculated as:
\[
T_{max} = \sqrt{H^2 + \left(\frac{w L}{2}\right)^2}
\]
In many cases, if \(h\) is not too small, \(T_{max} \approx H\) near the supports.
## **Step 3: Required Cable Cross-Sectional Area**
The maximum allowable tension is given by:
\[
T_{max} = \sigma_{ult} \cdot A
\]
Where:
- \(A\) = cross-sectional area of the cable (m²)
## **Step 4: Final Formulas**
To determine the maximum tension the cable can support based on tensile strength:
\[
T_{max} = \sigma_{ult} \cdot A
\]
## **Step 5: If Only Ultimate Strength is Required**
The maximum tension that each cable can safely support (before breaking) is:
\[
\boxed{T_{max} = \sigma_{ult} \times A}
\]
Where:
- \(\sigma_{ult} = 1.86 \times 10^9\) Pa
- \(A =\) cross-sectional area of the cable (unknown unless given)
## **Summary Table**
| Quantity | Value | Unit |
|-------------------|--------------|--------------|
| Span (\(L\)) | 120 m | m |
| Sag (\(h\)) | 120 m | m |
| Ultimate Strength (\(\sigma_{ult}\)) | 1860 MPa (= \(1.86 \times 10^9\) Pa) | Pa |
| Maximum Tension (\(T_{max}\)) | \(1.86 \times 10^9 \times A\) | N (where \(A\) is in m²) |
---
### **Final Note**
If you have the **cross-sectional area** or the total load the cable must support, substitute that value to get a numerical answer. If further assistance is needed with specific values for \(A\) or \(w\), please provide those details.
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