**Problem Statement:**2. For the beam and loading shown below, a) Draw the shear and bending moment diagrams. b) Determine the maximum absolute values of the shear and bending moment.**Given:**- Uniform distributed load: 3 kips/ft over a length of 6 ft from point A to point B.- Point load: 30 kips at point C.- Reactions: - \( A_x = 0 \) - \( A_y = 6 \) kips downward - \( B_y = 54 \) kips upward**Maximum Values:**- \(|V_{max}| = 30\) kips- \(|M_{max}| = 90\) kip-ft**Diagram Explanation:**The diagram shows a horizontal beam supported at point A by a pin and at point B by a roller. The beam is loaded with a uniform distributed load of 3 kips/ft between points A and B for a total length of 6 ft, followed by a point load of 30 kips applied at point C, which is 3 ft from point B.**Dimensions:**- Distance between A and B = 6 ft.- Distance between B and C = 3 ft.**Loadings:**- Uniform Load: Acts over segment AB, creating a total load of \(3 \text{ kips/ft} \times 6 \text{ ft} = 18 \text{ kips}\).- Point Load: Applied at point C.This setup requires the calculation of shear forces and bending moments along the beam to understand the internal forces and moments at any given section. The problem asks to specifically find the maximum values of these forces/moments, which are provided as 30 kips for shear and 90 kip-ft for the moment.

Free body diagram shown below: Convert uniform distribution load into point load. F=3×6=18 kips.Calculation: Balance the force in horizontal direction. Ax=0 kips Take the moment at point B. Ay×6+30×3=3×6×62Ay=54-906Ay=-6 kips The negative sign shown the assumed direction is wrong, therefore the force is in downward direction. Balance the force in vertical direction. Ax+By=3×6+30-6+By=48By=54 kipsShear force calculation: The shear force at A is VA=-6 kips. The shear force at B is VB=-6-36=-18 kips. The shear force at C is VC=-24+54=30 kips. Bending moment calculation: The bending moment at A is MA=0 kip‐ft. The bending moment at B is MB=-6×6-36×62=-90 kip‐ft. The bending moment at C is MC=-6×9-36×62+3=0 kip‐ft.The forces are Ax=0 kips, Ay=6 kips↓ and By=54 kips↑ and the absolute maximum shear and bending moment is Vmax=30 kips and Mmax=90 kip‐ft.

Engineering

Mechanical Engineering

2. For the beam and loading shown below, a) Draw the shear and bending moment diagrams, b) Determine the maximum absolute values of the shear and bending moment. (Ans: Ax = 0, Ay = 6 kips ↓, By = 54 kips ↑ IVmaxl = 30 kips, Mmaxl = 90 kip ft) A 3 kips/ft. ▬▬▬▬▬▬ 6 ft. B 30 kips 3 ft.

2. For the beam and loading shown below, a) Draw the shear and bending moment diagrams, b) Determine the maximum absolute values of the shear and bending moment. (Ans: Ax = 0, Ay = 6 kips ↓, By = 54 kips ↑ IVmaxl = 30 kips, Mmaxl = 90 kip ft) A 3 kips/ft. ▬▬▬▬▬▬ 6 ft. B 30 kips 3 ft.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial forces, torsion, and so on, then such body is said to be under combined loading. A very practical example of combined loading is the crankshaft of an internal combustion engine (IC engine). The crankshaft of the IC engine is under high rpm, which is caused due to the torsional forces. Due to the presence of piston, counterbalances, and internal imbalances, the crankshaft experiences lateral forces and due to the rise in temperature, the crankshaft undergoes thermal expansion, which pulls the crankshaft along the axial direction and lateral direction. The presence of unbalanced loads along the periphery of the crankshaft also induces a bending moment.

Question

**Problem Statement:**

2. For the beam and loading shown below,
a) Draw the shear and bending moment diagrams.
b) Determine the maximum absolute values of the shear and bending moment.

**Given:**
- Uniform distributed load: 3 kips/ft over a length of 6 ft from point A to point B.
- Point load: 30 kips at point C.
- Reactions:
- \( A_x = 0 \)
- \( A_y = 6 \) kips downward
- \( B_y = 54 \) kips upward

**Maximum Values:**
- \(|V_{max}| = 30\) kips
- \(|M_{max}| = 90\) kip-ft

**Diagram Explanation:**

The diagram shows a horizontal beam supported at point A by a pin and at point B by a roller. The beam is loaded with a uniform distributed load of 3 kips/ft between points A and B for a total length of 6 ft, followed by a point load of 30 kips applied at point C, which is 3 ft from point B.

**Dimensions:**
- Distance between A and B = 6 ft.
- Distance between B and C = 3 ft.

**Loadings:**
- Uniform Load: Acts over segment AB, creating a total load of \(3 \text{ kips/ft} \times 6 \text{ ft} = 18 \text{ kips}\).
- Point Load: Applied at point C.

This setup requires the calculation of shear forces and bending moments along the beam to understand the internal forces and moments at any given section. The problem asks to specifically find the maximum values of these forces/moments, which are provided as 30 kips for shear and 90 kip-ft for the moment.

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