SHEAR AND MOMENT DIAGRAMS Beams with transverse loads is subjected to shear and moment at any of its section . The shear and moment at any section can graphically shown by what are called the shear and moment diagrams. The relationship of load, shear and moment ca be determined from the following discussion.

SHEAR AND MOMENT DIAGRAMS for BEAMS

Beams with transverse loads is subjected to shear and moment at any of their section . The shear and moment at any section can be graphically shown by what are called the shear and moment diagrams.

The relationships of load, shear and moment can be determined from the following discussion.

w M O M + dM dx

V V + dV Consider a differential element of the beam with FBD shown at the left. dx Fy = 0 Mo = 0 wdx + V = V + dV M + vdx + w(dx)(dx/2) = M + dM dV = wdx (1) dM = vdx (3) dV/dx = w (2) dM/dx = v (4)

Relationships of load and shear

1. The change in shear between two points in the shear diagram is equal to the area between the same two points in the load diagram. A concentrated load is considered an area in the load diagram. 2. The slope at a point in the shear diagram is equal to the intensity of the load at the same point in the load diagram. 3. In simply supported and cantilever beams, maximum shear occurs at one of the supports.

Relationships of shear and moment

1. The change in moment between two points in the moment diagram is equal to the area between the same two points in the shear diagram. A concentrated moment is considered an area in the shear diagram. 2. The slope at a point in the moment diagram is equal to the intensity of shear at the same point in the shear diagram. 3. The maximum moment occurs at the point of zero shear. - the degree of the curve increases by one degree from load diagram to shear diagram, and from shear diagram to moment diagram.

sheaSign Convention:

load, w : +ve -ve

shear ,V : +ve -ve

moment, M:

+ve -ve

slope: +ve , uniform -ve, uniform

+ve, decreasing -ve, decreasing

+ve, increasing -ve, increasing Sign Convention:

load, w : +ve -ve

shear ,V : +ve -ve

Problem: Draw the shear and moment diagrams for the beam loaded as shown below.

60 kN 30 kN

A B C D Solution:

MA= 0 2m 4m 1m 60(2) + 30(7) = R2 (6) R2 = 55 kN.

Fy = 0 R1 + R2 = 60 + 30 R1 + 55 = 60 + 30 R1 = 35 kN.

The shear V at any point is equal to shear at a previous point plus the area in the load diagram between the two points. The moment M at any point is equal to moment at a previous point plus the area in the shear diagram between the two points.Shear diagram: 1. Before point A, VA = 0 Just after point A, VA = 0 + 35 = +35 kN. 2. Just before point B, VB = +35 + 0 = +35 kN. Just after point B, VB = +35 - 60 = -25 kN. 3. Just before point C, VC = -25 + 0 = -25 kN. Just after point C, VC = -25 +55 = +30 kN. 4. Just before point D, VD = 30 + 0 = + 30 kN. Just after point D, VD = +30 - 30 = 0 kN.

Moment diagram:

1. At A, MA = 0 2. At B, MB = 0 + 35(2) = +70 kN. m. 3. At C, MC = 70 - 25(4) = -30 kN m. 4. At D, MD = -30 + 30(1) = 0

Note: the line connecting moments at A and B is straight (uniform slope) because the intensity of shear between the same points is also uniform.

Problem: Draw the shear and moment diagrams for the beam loaded as shown below. 100 lbs. 120 lbs/ft 120 lbs/ft Solution:

MA = 0 A B C D R2(4) = 120(2)(1) + 100(2) + 120(2)(5) R2 = 410 lbs. 2 ft. 2 ft. 2 ft. R1 R2 Fy = 0 R1 + R2 = 120(2) + 100 + 120(2) R1 + 410 = 240 + 100 + 240 R1 = 170 lbs. The shear V at any point is equal to shear at a previous point plus the area in the load diagram between the two points. The moment M at any point is equal to moment at a previous point plus the area in the shear diagram between the two points. Shear: 1. Just before A: VA = 0 Just after A: VA = 0 + 170 = 170 lbs. 2. Just before B : VB =170 120(2) = - 70 lbs. Just after B: VB = -70 100 = -170 lbs. 3. Just before C: VC = -170 + 0 = - 170 lbs. Just after C: VC = -170 + 410 = +240 lbs. 4. At point D : VD = 240 120(2) = 0

- line A to B (and also C to D) is straight(uniform slope) ,because load between the two points is uniform.

- the point of zero shear is obtained by similar s:

170/x = 240/2 x = 1.42ft.

Moment: 1. At MA = 0 2. At point of zero shear: M = 0 + (170)(1.42) M = 120.42 lb.ft. 3. At B: MB = 120.42 + (-70)(.58) = 100 lb.ft 4. At C: MC = 100 170(2) = -240 5. At D: MD = -240 + (240)(2) = 0

Notes:

-- The moment diagram from point A to the point of maximum shear, is concave downward with slopes that are positive but decreasing, because the intensity of shear between the same point is also positive and decreasing. The curve is already second degree because the the corresponding curve in the shear diagram is first degree.

-- The moment diagram from the point of maximum shear to point B is still concave downward with slopes that are negative and increasing, because the intensity of shear between the same point is also negative and increasing. The curve is also second degree because the the corresponding curve in the shear diagram is first degree.

-- The moment diagram from the point B to C is a straight line with uniform slope because the intensity of shear between the same point is also uniform.

-- The moment diagram from C to D is still concave downward with slopes that are positive but decreasing, because the intensity of shear between the same point is also positive but decreasing. The curve is also second degree because the corresponding curve in the shear diagram is first degree.

Problem: Draw the shear and moment diagrams for the beam loaded as shown below. 60 kN

Solution: 20 kN/m MB = 0 M = 120 kN-m. R2(6) = 20(4)(0) + 60(2) + 120 R2 = 40 kN. A B C D E Fy = 0 2m 2m 2m 2m R1 + R2 = 20(4) + 60 R1 R2 R1 + 40 = 80 + 60 R1 = 100 kN.

The shear V at any point is equal to shear at a previous point plus the area in the load diagram between the two points. The moment M at any point is equal to moment at a previous point plus the area in the shear diagram between the two points. Shear: 1. At A: VA = 0 2. Just before B : VB = 0 20(2) = -40 kN. Just after B : VB = -40 + 100 = +60 kN. 3. Just before C : VC = +60 20(2) = +20 kN. Just after C : VC = +20 60 = - 40 kN. 4. Just before E : VE = -40 + 0 = -40 kN. After E : VE = -40 + 40 = 0

Moment: 1. At A : MA = 0 2. At B : MB = 0 + (-40)(2) = -40 kN.m. 3. At C : MC = - 40 + (60+20)(2) = +40 kN.m. 4. Just before D: MD = +40 40(2) = -40 kN .m. Just after D: MD = - 40 + 120= +80 kN .m. 5. At E : ME = +80 40(2) = 0

Problem: Draw the shear and moment diagrams for the beam loaded as shown in below. 1000 lbs. 400 lbs/ft. Solution: Fy = 0 A B C D MD RD + 500 = 1000 + 400(4) 2ft. 2ft. 4ft. RD = 2100 lbs. 500 lb. RD MD = 0 1000(8) + 400(4)(2) = 500(6) + MD MD = 8200 lbs. ft. -500 lbs. -1000 lbs.

shear diagram -2100 lbs.

-2000 lbs.ft -3000 lbs. ft

moment diagram -8200 lbs. ft.

MOVING LOADS on BEAMS

In the case of moving loads, the usual concerns are the maximum shear and moment caused in the supporting beam.

In simply supported beams with moving loads, the maximum shear will occur at either support and is obtained by placing, successively, one of the loads over either support, and compare the maximum shear at the support for each case.

In simply supported beams with moving loads, the maximum moment will occur under any of the loads. The problem therefore is to find the position of one moving load that will produce maximum moment under it. The absolute maximum moment in the beam is then obtained by comparing the maximum moments under each of the moving loads. Consider a beam with moving loads as shown in the figure. We determine what should be the position of a moving load, say load P2, which will produce maximum moment under. Assume that resultant of the moving loads R, is located at a distance e from P2. The resultantis used to replace the moving loads for the purpose of obtaining reactions, R1, and the moment under load P2. P1 P2 R P3 P4 Mright support = 0 a b c R1 = R/L ( L- e x)

cutting the beam under P2, x e the moment under P2 is: R1 L R2 M = R1 x - P1 a = R/L ( L- e x)x - P1 a

differentiating M with respect to x, and equating to zero: dM/dx = R/L(L - e x) + x (-R/L) = 0 2x = L e L/2 = x + e/2 , thus, to get the position of the load for maximum moment under it, midway between the load and the resultant is made to coincide with midspan of the beam.Problem: Three wheel loads roll as a unit across a 44-ft span. The loads are P1 = 4 kips and P2 = 8 kips separated by 9 ft, and P3 = 6 kipsat 18 ft from P2. Determine the maximum moment in the simply supported span. RSolution: 4 kips 8 kips 6 kips 9ft 18ft To locate the resultant R: 1 2 3 R = 4 + 8 + 6 = 18 kips. R(e1) = 8(9) + 6(27) e2 e1 = [8(9) + 6(27)]/18 e1 = 13 ft. e1 e3 e2 = 13 9 = 4 ft. e3 = 27 13 = 14 ft.

Note: the magnitude of the resultant is the sum of the moving loads, and its moment at any point is equal to the sum of the moments of the moving loads at the same point.

To get maximum moment under load 1, 18kips position the load as shown in the left: 4kips 8kips 6kips Mright support = 0 9ft 18ft R1 (44) = 18 (22- 6.5) 1 2 3 R1 = 6.341 kips. cut beam under load 1 and obtain 6.5ft 6.5ft moment under load 1: 22 ft 22ft M1 = R1 (22- 6.5) R1 R2 = 98.285 kip.ft 18 kipsTo get maximum moment under load 2, 4 kips 8 kips 6kips position the load as shown in the left: 9 ft 18 ft Mright support = 0 R1 (44) = 18 (22- 2) 1 2 3 R1 = 8.18 kips cut the beam under load 2 and obtain 2ft 2ft moment under load 2: 22 ft 22ft M 2 = R1 (22-2) 4(9) R1 18 kips R2 = 127.6 kip ft. 4 kips 8 kips 6 kipsSimilarly under load 3: 9 ft 18 ft M left support = 0 R2 (44) = 18 (22-7) 1 2 3 R2 = 6.136 kips. cut under load 3 and obtain M3 7ft 7ft M3 = R2 ( 22-7) = 92.04 kip ft. R1 22 ft 22 ft R2 Maximum moment = 127.6 kip ft.