# BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL)

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1 LECTURE Third Edition BES: SHER ND OENT DIGRS (GRPHICL). J. Clark School of Engineering Department of Civil and Environmental Engineering 3 Chapter 5.3 by Dr. Ibrahim. ssakkaf SPRING 003 ENES 0 echanics of aterials Department of Civil and Environmental Engineering University of aryland, College Park LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. Eample 8 The beam is loaded and supported as shown in the figure. Write equations for the shear and bending moment for any section of the beam in the interval B. y 6 kn/m B 9 m

2 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. Eample 8 (cont d) free-body diagram for the beam is shown Fig. 7. The reactions shown on the diagram are determined from equilibrium equations as follows: 8 kn ; 9 kn (9) 0; B B y B R R F R R LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 3 Eample 8 (cont d) y B 6 kn/m R 9 kn R b 8 kn y 9 kn S ( ) 9 for ; 9 for ; 3 < < + + < < + + F S y 6 kn/m 9 m w w kn/m 3 3 Figure 7

3 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 4 Eample 9 timber beam is loaded as shown in Fig. 8a. The beam has the cross section shown in Fig.8.b. On a transverse cross section ft from the left end, determine a) The fleural stress at point of the cross section b) The fleural stress at point B of the cross section. LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 5 Eample 9 (cont d) y 600 lb 5600 ft-lb 300 lb/ft 3500 ft-lb 6 ft 0 ft 5 ft 900 lb 500 lb Figure 8a 8 B 8 Figure 8b 6

4 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 6 Eample 9 (cont d) 8 First, we have to determine the moment of inertia I. From symmetry, the neutral ais is located at a distance y 5 in. either from the bottom or the upper edge. Therefore, 5 6 I 3 3 ( 5) 6( 3) in LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 7 y Eample 9 (cont d) 5600 ft-lb 300 lb/ft 5600 ft-lb ft 900 lb lb 00 lb + F S y 3500 ft-lb 500 lb 0; + (00) 0 for 0 < < ft - lb 0; ()(00)(0.5) 0 at ft

5 5 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No Eample 9 (cont d) 6 a) Stress at r y σ I psi Compression (C) ( )( ) b) Stress at B r y 5500 σ B I psi Tension (T) ( )( 5) psi + 59 psi LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 9 In cases where a beam is subjected to several concentrated forces, couples, and distributed loads, the equilibrium approach discussed previously can be tedious because it would then require several cuts and several free-body diagrams. In this section, a simpler method for constructing shear and moment diagrams are discussed.

6 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 0 a b L P w O w 3 Figure 9 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. The beam shown in the Figure 0 is subjected to an arbitrary distributed loading w w() and a series of concentrated forces and couple moments. We will consider the distributed load w to positive when the loading acts upward as shown in Figure 0.

7 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. P w L L P w R L + C Figure 0a C Figure 0b R + L LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 3 In reference to Fig. 0a at some location, the beam is acted upon by a distributed load w(), a concentrated load force P, and a concentrated couple C. free-body diagram segment of the beam centered at the location is shown in Figure 0b.

8 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 4 The element must be in equilibrium, therefore, ( + ) + Fy 0; L + wavg + P L 0 From which P + w avg (8) LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 5 In Eq. 8, if the concentrated force P and distributed force w are both zero in some region of the beam, then 0 or L R (9) This implies that the shear force is constant in any segment of the beam where there are no loads.

9 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 6 If the concentrated load P is not zero, then in the limit as 0, P or R L + P (30) That is, across any concentrated load P, the shear force graph (shear force versus ) jumps by the amount of the concentrated load. LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 7 Furthermore, moving from left to right along the beam, the shear force graph jumps in the direction of the concentrated load. If the concentrated load is zero, then in the limit as 0, we have wavg 0 (3)

10 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 8 nd the shear force is a continuous function at. Dividing through by in Eq. 3, gives lim0 X d d w (3) That is, the slope of the shear force graph at any section in the beam is equal to the intensity of loading at that section of the beam. LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 9 oving from left to right along the beam, if the distributed force is upward, the slope of the shear force graph (d/d w) is positive and the shear force graph is increasing (moving upward). If the distributed force is zero, then the slope of the shear graph (d/d 0) and the shear force is constant.

11 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 0 In any region of the beam in which Eq. 3 is valid (any region in which there are no concentrated loads), the equation can be integrated between definite limits to obtain d w d LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. Load and Shear Force Relationships Slope of d d Shear Diagram w( ) (33) Distributed Load Intensity

12 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. Load and Shear Force Relationships Change in Shear rea under Loading Curve between d w( ) d and (34) LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 3 Similarly, applying moment equilibrium to the free-body diagram of Fig. 0b we obtain ( + ) ( + ) + a( w ) + c 0; - L L L L avg 0 From which C + L + a wavg ( ) (35)

13 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 4 P w L L P w R L + C Figure 0a C Figure 0b R + L LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 5 In which - / < a < / and in the limit as 0, a 0 and w avg w. Three important relationships are clear from Eq. 35. First, if the concentrated couple is not zero, then in the limit as 0, C or C (36) R L +

14 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 6 That is, across any concentrated couple C, the bending moment graph (bending moment versus ) jumps by the amount of the concentrated couple. Furthermore, moving from left to right along the beam, the bending moment graph jumps upward for a clockwise concentrated couple and jumps downward for a concentrated counterclockwise C. LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 7 Second, if the concentrated couple C and concentrated force P are both zero, then in the limit as 0, we have L + a( wavg ) 0 (37) and dividing by, gives lim d d (38) 0

15 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 8 That is, the slope of the bending moment graph at any location in the beam is equal to the value of the shear force at that section of the beam. oving from left to right along the beam, if is positive, then d/d is positive, and the bending moment graph is increasing LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 9 In the region of the beam I which Eq. 38 is valid (any region in which there are no concentrated loads or couples), the equation can be integrated between definite limits to get d d (39)

16 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 30 Shear Force and Bending oment Relationships Slope of oment Diagram Shear d (40) d LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 3 Shear Force and Bending oment Relationships Change in oment rea under Shear diagram between d d and (4)

17 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 3 Shear and oment Diagrams Procedure for nalysis Support Reactions Determine the support reactions and resolve the forces acting on the beam into components which are perpendicular and parallel to the beam s ais Shear Diagram Establish the and aes and plot the values of the shear at two ends of the beam. Since d/d w, the slope of the shear diagram at any point is equal to the intensity of the distributed loading at the point. LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 33 Shear and oment Diagrams If a numerical value of the shear is to be determined at the point, one can find this value either by using the methods of establishing equations (formulas)for each section under study or by using Eq. 34, which states that the change in the shear force is equal to the area under the distributed loading diagram. Since w() is integrated to obtain, if w() is a curve of degree n, then () will be a curve of degree n +. For eample, if w() is uniform, () will be linear. oment Diagram Establish the and aes and plot the values of the moment at the ends of the beam.

18 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 34 Shear and oment Diagrams Since d/d, the slope of the moment diagram at any point is equal to the intensity of the shear at the point. In particular, note that at the point where the shear is zero, that is d/d 0, and therefore this may be a point of maimum or minimum moment. If the numerical value of the moment is to be determined at a point, one can find this value either by using the method of establishing equations (formulas) for each section under study or by using Eq. 4, which states that the change in the moment is equal to the area under the shear diagram. Since w() is integrated to obtain, if w() is a curve of degree n, then () will be a curve of degree n +. For eample, if w() is uniform, () will be linear. LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 35 Shear and oment Diagrams Loading Shear Diagram, d w d oment Diagram, d d

19 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 36 Shear and oment Diagrams Loading Shear Diagram, d w d oment Diagram, d d LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 37 Shear and oment Diagrams Eample 0 Draw the shear and bending moment diagrams for the beam shown in Figure a. 40 lb/ft 600 lb ft 0 ft 000 lb in Figure a

20 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 38 Shear and oment Diagrams Eample 0 (cont d) Support Reactions The reactions at the fied support can be calculated as follows: + Fy 0; R 40( ) R Figure b + 0; + 40() ( 6) + 600( 0) + 5,880 lb in 40 lb/ft 600 lb 080 lb ft 5,880 lb ft R 080 lb 0 ft 000 lb ft LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 39 Shear and oment Diagrams 5,880 lb ft Eample 0 (cont d) Shear Diagram Using the established sign convention, the shear at the ends of the beam is plotted first. For eample, when 0, 080; and when 0, lb 40 lb/ft ( ) for 0 < < ,880 for 0 < <

21 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 40 Shear and oment Diagrams Eample 0 (cont d) 40 lb/ft 5,880 lb in ft 080 lb ( ) 600 for < < 0 ( )( - 6) for < 0 5, < LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 4 Shear and oment Diagrams Eample 0 (cont d) 40 lb/ft 600 lb Shear Digram ft 0 ft 000 lb in (lb) ( ) for 0 < < (ft)

22 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 4 Shear and oment Diagrams (ft. lb) Eample 0 (cont d) Bending oment Diagram (ft) 40 lb/ft 600 lb ft 0 ft 000 lb in LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 43 Shear and oment Diagrams Eample 0 (cont d) 40 lb/ft 600 lb 5,880 lb ft ft R 080 lb 0 ft 000 lb ft 080 (lb) (+) 600 (ft lb) (-) -000

23 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 44 Shear and oment Diagrams Eample Draw complete shear and bending moment diagrams for the beam shown in Fig. y 8000 lb 000 lb/ft B C D Figure a ft 4 ft 8 ft LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 45 Shear and oment Diagrams Eample (cont d) The support reactions were computed from equilibrium as shown in Fig..b. y 8000 lb 000 lb/ft B C D ft 4 ft 8 ft Figure a R,000 lb R C,000 lb

24 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 46 Shear and oment Diagrams Eample (cont d) 8000 lb 000 lb/ft B C D,000 lb ft 4 ft 8 ft,000 lb,000 8,000 lb (lb) (+) (+) 5.5 ft (-) 3,000 lb 30,50 (ft -lb) (-) (-),000 64,000 LECTURE 3. BES: SHER ND OENT DIGRS (GRPHICL) (5.3) Slide No. 47 Shear and oment Diagrams Eample (cont d),000,000 3, ,000

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