1) The beam ends, and any point at which the load type changes: Points of concentrated force or moment, end points of a distributed loads.

2) Useful points for the M-diagram:

    a) Point of zero shear force within a beam segment having a distributed load.  There is a relative maximum or minimum of the bending moment at such a point.   The point of zero shear can be determined without having to draw the shear force diagram. 

   b) Mid-point of a uniformly loaded beam segment.  This is a helpful point for drawing a parabola.

a) Identify the key points.

b) Compute the internal force of interest at each key point by the method of sections.  If discontinuous, compute the values before and after the point of discontinuity, respectively.  At the beam ends, the internal forces become the external ones acting on the end cross sections.

c) Plot and join the key points by appropriate curves.  The type of curve between two consecutive points depends only on the type of external loading on the beam segment between these two points. 

1.  Curves joining Key Points of M-Diagrams

a) Unloaded beam segment:   Join the end points by a straight line.

b) Uniformly loaded beam segment:   Join the end points and the mid-point by a parabola.  Alternately, joint the end points and a point of relative extremum by a parabola.  A correctly drawn parabola has the following properties:

• The tangent at the midpoint is parallel to the straight line joining the end points.

• If the load acts downward, the parabola will be concave downward, and the mid-point will be above the line joining the ends by wL²/8, where w is the load intensity, and L is the length of the segment. 

• If the beam segment includes a point of zero shear, that point is one of relative maximum or minimum, thus with a horizontal tangent.

c) Segment with a varying load intensity:  For a qualitative diagram, draw M concave downward if the load is downward, with a relative extremum at a point of zero shear.  The slope of this diagram is steeper where the shear force is larger in magnitude.

Discontinuities at Key Points of M-Diagrams

a) At a point of concentrated force, without a concentrated moment, the M-diagram has a kink (slope discontinuity) and a continuous value.  The kink is concave downward if the load acts downward.

b) At a point of concentrated moment, without a concentrated force, the M-diagram has a jump discontinuity, and a continuous slope.

Draw qualitatively the bending moment diagram for the following cases:

2. Curves joining Key Points of V-Diagrams

a) Unloaded beam segment:   V is constant.  Draw a horizontal line using the value at either end point. 

b) Uniformly loaded beam segment:    V is linear.  Join the end points by a straight line.  The line should be decreasing if the load is down.  If the line crosses the x axis, the point of intersection is a key point for the M-diagram at which there is a relative extremum.

c) Segment with a varying load intensity:   The slope of the V-diagram is steeper where the load intensity is larger in magnitude.  At a point of zero load intensity, the V diagram has a horizontal tangent.  A qualitative V-diagram can be drawn based on that property.  For example,  if the load is downward and its intensity increases with x, draw a curve concave downward with a slope increasing in magnitude..

Discontinuities at Key Points of V-Diagrams

a) At a point of concentrated force, the V-diagram has a jump discontinuity equal in magnitude to that force.  The jump from left to right is downward if the force is downward.  The slope of the diagram is continuous unless a discontinuous distributed load is also applied. 

b) At a point of concentrated moment, the V-diagram is not affected. ( A more detailed analysis not to be made here that considers the type of forces that generate the moment,  would result in a better understanding of the shear force behavior, and a possible revision of the V-diagram.)

2. Draw qualitatively the shear force diagram for the following cases:

3.

a) The structure has three vertical force unknown reactions. Why is it statically determinate?  Deduce from a condition at pin c that the reaction at a is upward, and is less than the resultant of the applied load to the left of c.  Deduce also that the reaction at e is upward.

Because the condition of zero internal moment at c determines the reaction at a.

The moment at c of the forces to the left of c must be zero. The reaction at a is thus upward. The resultant of the applied load to the left of c has a shorter moment arm than the reaction at a.  It is thus larger than that reaction.  From the preceding, deduce that the shear force at c acts downward on ceg.  Isolating part ceg, the moment equilibrium equation about point g shows that the reaction at e is upward.

b) At what points is the shear force discontinuous?  At the points of concentrated load: Points e, and f.

c) At what points is the slope of the shear force diagram discontinuous?  At the points of discontinuity of the load intensity:  Points b and d.

d) At what points if any is the bending moment discontinuous?  At no point, since there is no external concentrated moment.

e) At what points is the slope of the bending moment diagram discontinuous?  At the same points as in b).

f) Outline on a sketch of the structure a qualitative V-diagram, assuming the reaction at g is downward.

g) Outline on a sketch of the structure a qualitative M-diagram.  Make sure not to show a kink where the slope is continuous.

s = -My/I

t = VQ/tI

Geometry Review:

The moment of an area about an axis is the product of the area by the distance of its centroid to that axis.

The moment of an area about an axis passing through its centroid is zero.  Equivalently, a centroidal axis divides an area into two parts having equal moments in magnitude about that axis.

4.

a)  In bending, the neutral axis (N.A.) of a cross section is the centroidal axis perpendicular to the plane of bending.   Show that in order that the neutral axis of the T section be located as shown, the flange width must be related to the other dimensions by

b = c²t/[t'(c - t') + t(c/2 - t')²]

Flange: Q_f = bt'(c/2 - t'/2) = bt'(c - t')/2

Web above N.A: Q_wa = t(c/2 - t')(c/2 - t')/2 = t(c/2 - t')²/2

Web under N.A. Q_wu = - ct(t/2) = - ct²/2

Q_f + Q_wa + Q_wu = bt'(c - t')/2  + t(c/2 - t')²/2  - ct²/2 = 0

Solve for b, and obtain given formula.

b) The bending moment diagram of the structure is as shown.  At what point in a cross-section within bc does the maximum tensile stress occur?  What is the ratio of that stress to the maximum compressive stress in the same cross-section?

The bending moment being negative, it causes tension above the neutral axis.  The maximum tensile stress is at the top of the cross-section.  The extreme fibers being at distances c/2 and c, respectively, the maximum tensile stress is one half the maximum compressive stress within bc.

c) Answer the same question as in b) for segment ab.

In segment ab, in the portion where M < 0 the answers are the same as for segment bc.  In the part where M > 0, tension is below the neutral axis, and its maximum is twice the maximum compressive stress.

d) If M_a = 1.75 |M_b|, where do the maximum tensile and compressive stresses in the structure occur? What are their respective values in terms of the given data, and the moment of inertia I of the cross-section?

At a, the bending moment is maximum, and the maximum tensile stress in the cross section occurs at the bottom. Thus,  the maximum tensile stress in the structure occurs at a, and is equal to M_ac/I = 1.75|M_b|c/I

The maximum compressive stress at a, which occurs at the top of the cross-section,  is at distance c/2 from the neutral axis.  At b, the maximum compressive stress occurs at the bottom where the distance to the neutral axis is twice that at a, and the bending moment is larger than M_a/2 in magnitude. Thus, the maximum compressive stress in the structure occurs at the bottom of the cross-section at b, and is equal in magnitude to |M_b|c/I.

a) If M_a = 1.75 |M_b|, what is the maximum shear force (in magnitude)  in the structure in terms of the data shown?

Within ab, V = (M_b - M_a)/L = 2.75M_b/L = - 2.75P

Within bc, V = P

b) Where in a cross section within ab does the maximum vertical shear stress occur, and what is its value?

The maximum shear flow, and thus the maximum vertical shear stress, occur at the neutral axis.

t = VQ/tI

where Q = tc²/2 and V = 2.75P. Thus,

t = 2.75Pc²/2I

c) What is the normal force N (resultant of s) acting on the portion of the cross section below the neutral axis,  if M is the bending moment at that location?

N = ∫sdA = ∫-MydA/I = -(M/I)∫ydA = -MQ/I

Q is the moment of area of the rectangular part below the neutral axis.  Thus

Q = - ct(c/2) = -c²t/2

N = Mc²t/2I, tensile if positive.

d) Consider a free-body consisting of the part of segment ab between the supports, and below the neutral plane.   Using part c), show on the free-body the axial forces N_a and N_b acting on the end cross sections, and, using part b), show the longitudinal shear force acting on the neutral plane.  Are the forces in equilibrium?

F = ttL= 2.75Pc²tL/2I

N_a + N_b = M_ac²t/2I + |M_b|c²t/2I  = 2.75PLc²t/2I

N_a + N_b = F

The forces are in equilibrium.