Download presentation

1
**CM 197 Mechanics of Materials Chap 13: Shear Forces and Bending Moments in Beams**

Professor Joe Greene CSU, CHICO Reference: Statics and Strength of Materials, 2nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill, Westerville, OH (1997) CM 197

2
**Chap 13: Shear Forces and Bending Moments**

Topics Introduction Types of Beams Types of Loading Beam Reactions Shear Force and Bending Moment in Beams Shear Force and Bending Moment Diagrams Relationships Among Load, Shear, and Moment Sketching Shear and Moment Diagrams Using Relationships Shear and Moment Formulas

3
Introduction Beams are members that carry transverse loads and are subjected to bending. Any structural member subjected to bending may be referred to as a beam. Shafts, girders, stringers, floor beams, joists, etc. Objective of chapter Determine the internal forces at various sections along a beam. Review Types of beams and supports. Calculation of beam sections. Calculation of internal forces (shear and moments) in the beam. Produce graphical representations of shear and bending moments. Assumptions Beams are straight and of uniform cross sections that possess a vertical plane of symmetry. Fig 13-1 Horizontal positions Subjected to forces applied in the vertical plane of symmetry. Plane of symmetry

4
**Types of Supports for Beams**

Roller supports Roller or link support resists motion of the beam only along the direction perpendicular to the plane of the support (along axis) Reaction at roller support acts along the known direction. Fig 13-2 Have one unknown Reaction force, R Hinge supports Resists motion of the beam at the support in any direction on the plane of loading. Have two unknown Reaction forces- Rx and Ry Fixed supports Beam is built into wall or column. End of beam is fixed and doesn’t move. Have two unknown Reaction forces, Rx and Ry, and one unknown Moment, M R Ry Rx Ry Rx M

5
**Types of Beams Beams Simple Beam (Statically Determinate- 3 unknowns)**

Beam supported at its end with a hinge and a roller Overhanging Beam (Statically Determinate- 3 unknowns) Simple supported beam with an overhang from one or both ends Cantilever Beam (Statically Determinate- 3 unknowns) Beam that is fixed at one end and free at the other Propped cantilever beam (Statically Indeterminate >3 unknowns) Beam that is fixed at one end and simply supported at the other. Fixed Beam (Statically Indeterminate >3 unknowns) Beam with both ends are fixed to supports Continuous Beam (Statically Indeterminate >3 unknowns) Beam is supported on a hinge and two or more roller supports

6
**Types of Loading Loading Concentrated Loads Uniform Loads**

Load is applied at a specific point on the beam and is considered a discrete force acting at the point. Example, weight fastened to a beam by a cable that applies a concentrated load. Uniform Loads Load is distributes over a part or the entire length of the beam. Load is force per unit length of beam (lb/ft or N/m) Replaced with equivalent force = uniform load times length of beam Example, weight of beam. Linearly Varying Loads Load is distributed with a uniform variation of intensity. Occurs on a vertical or inclined wall due to liquid pressure. Load changes along the length of the beam.

7
**Beam Reactions Beam Reactions Review of Chapter 3 Procedure**

Write three independent equilibrium equations to solve for the three unknowns. Three equations could be combinations of force-component or moment equations, as long as they are independent of each other. This chapter assume no horizontal forces. Thus reaction force in horizontal direction is zero. Assume, sum of forces in x direction is 0, thus the horizontal reaction is 0. Example, 13-1 Example, 13-2

8
**Shear Force & Bending Moment in Beams**

Shear Force and Bending Moment Developed in a beam to resist the external forces and to maintain equilibrium. Figure 13-7 Beam is 10 ft long with roller support on the right and hinge support on left. Beam is subjected to two concentrated loads. Results For cross section at 3 feet from the left, the forces and moments need to be balanced. Balance the forces and moments at each section Normal forces: Rx = 0; Ry =150 pounds; Shear force: V = 150 – 100 = 50 lbs; Moment is acting clockwise and = 3*150 – 100*2 = 250lb-ft; Internal resisting moment moment has to be equal and opposite of moment and acting opposite direction. M = 250ft-lb counterclockwise. For results for cross section from the right, the forces and moments are calculated in the same manner. R Ry Rx 100 lb 200 lb 10 ft 1 ft 3 ft

9
**Beam Sign Conventions Positive Shear Positive Moment**

Shear force at a section is positive if the external forces on the beam produce a shear effect that ends to cause the left side of the section to move up relative to the right side. Fig 13-8a Positive Moment Bending moment at a section is positive if external forces on the beam produce a bending effect that causes the beam to concave upward at the section. Fig 13-8b Positive internal shear force, V, at a given section of a beam viewed from both directions is shown in Fig 13-8c. Positive internal moment M at a given section of a beam viewed from both directions is shown in Fig 13-8d. Rule 1 for finding shear forces Internal shear force at any section of a beam is equal to the algebraic sum of the external forces on either segment separated by the section. From the left of beam: upward force as positive From the right of the beam: downward force as positive Rule 2 for finding bending moments Internal bending moment at any section of a beam is equal to the algebraic sum of the moments about a section due to external forces. From either side: moment due to upward force as positive Example, 13-1 and Example, 13-2

10
**Shear Force and Moment Diagrams**

Shear force and bending moment diagrams depict the variation of shear force and bending moment along a beam. Help visualize the shear forces and bending moments along beam Construct diagram Ordinate (y-axis) values are from balance of forces and moments. Compute reaction forces and moment couples. Plot values of the shear just below the free body diagram on the y-axis with positive values above the baseline and negative values below the baseline. Plot values of moments just below the shear diagram on same x-axis scale with positive values above the baseline and negative values below the baseline. Important aspects of diagrams For concentrated loads The shear diagram has a horizontal line between the loads The moment diagram has a line with constant slope For distributed loads The shear diagram has a line with a constant slope. The moment diagram has a curved line between moment points. If the shear diagram crosses the horizontal line (at zero shear) then it has a maximum or minimum moment at that point. Example 13-5

11
**Relationships among Load, Shear, and Moment**

Certain relationships exist among the loading diagram, shear diagram and moment diagram. Assume incremental element of a beam. Relationships between Load and Shear For a distributed load, w, applied to a beam, The slope of the shear diagram (rate of change of the shear force per unit length) at any section is equal to the load intensity at that section. The shear force at a section is equal to the shear force at the previous section plus the total load between the two sections. Shear force at the section immediately to the right of a concentrated load is equal to the shear force immediately to the left plus the load. Shear force has an abrupt change at the concentrated load, upward = + Relationship between shear and moment Moment at a section is equal to the moment at the previous section plus the area under the shear diagram between the two sections.

12
**Using Relationships of Diagrams**

Loading Diagram or free body diagram Show all applied forces and reactions on beam plus dimensions Never replace a distributed load with an equivalent concentrated load Shear diagram procedure Draw shear diagram directly below the loading diagram. Horizontal line is drawn at proper location below loading diagram. Draw vertical lines from controlling sections for supports and loads. Start at left end and compute shear at controlling sections using Eqn VB = VA + LoadB-A Note: Section with concentrated force is applied, shear force diagram has an abrupt change at concentrated load location. Plot points on shear diagram using forces at each load. Connect points Note: Sections with distributed loads have a sloping line between them. Note: Slope of line is the intensity of the load. Concentrated loads have slope = 0

13
**Using Relationships of Diagrams**

Moment diagram procedure Draw moment diagram directly below the shear diagram. Horizontal line is drawn at proper location below loading diagram. Draw vertical lines from controlling sections for supports and loads Calculate areas from shear diagram for each section. Note: moments at free end or ends of a simple beam are always = 0. Start at the left end and compute the moments using Eqn 13-7, MB = MA + AreaB-A Plot points on moment diagram using previous moments and areas at each section Connect points with straight lines (for concentrated loads) or curves (for distributed loads) Note: Maximum or minimum of moment occurs at the location where the shear is = 0 or where the shear changes sign. Example, 13-6, 13-7, 13-8, 13-9

14
**Shear and Moment Formulas**

From previous section, shear force and bending moment diagram can be plotted for some simple but typical loading conditions. Table 13-1 Procedure Match the loading conditions to one or more loading examples in Table 13-1 Get the loading solutions from the example and use it in problem for Reaction forces, Maximum shear, maximum moments, etc. Method of superposition Effect of each load is computed separately and the then combined effect is added algebraically. Example, 13-10, 13-11

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google