CHAPTER 4. Stresses in Beams

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1 CHAPTER 4 Stresses in Beams Problem 1. A rolled steel joint (RSJ) of -section has top and bottom flanges 150 mm 5 mm and web of size 00 mm 1 mm. t is used as a simply supported beam over a span of 4 m to carry an uniformly distributed load of 0 kn/m over its entire span. Draw bending and shearing stresses across a section at 1/4th span. ig. 1 Solution: Due to symmetry Y NA 175 mm from top as welll as bottom most fibres. At 1 th 4 Bending stress: mm 4. WL 0 4 R A R B kn span i.e., at 1 m from support, 0 0 kn 0 N Wx 0 1 M R A kn-m 10 N-mm f t f b M Ymax N/mm.5475 (c) ig. 1 Shearing stress: ay b Shearing stress at bottom of top flange N/mm

2 Shearing stress at the same level but in web N/mm or shearing stress at NA N/mm. Therefore, bending stress and shearing stress diagrams are as shown in ig. 1(a) and 1(b). Problem. Unsymmetric -section shown in ig. is subjected to a bending moment of 15 kn-m, the top flange being in compression. Draw the stress variation diagram across the section marking the salient points and compute the total moment resisted by the top flange. ig. Solution: Here, A mm y 1 15 mm A mm A 10 mm y mm y 5 mm y Distance of NA from bottom-most fibre mm ( ) mm 4 Y top Bending stress: Now f top ( 17.91) ( ) f M y M 15 kn-m 15 N-mm N/mm 4

3 f top web f b web f b flange N/mm Moment Resisted by top flange alone N/mm N/mm f bdy y y 40 dy y [ ] 594 N-mm 5.94 kn-m. Ans. Problem. f the section shown in ig. is subjected to a shear force of 50 kn, draw the variation of shear stress across the depth and find the total shear resisted by web N/mm 1.94 N/mm N/mm Solution: q bottom of top flange or shear stress at N-A, q top of web 0.7 N/mm 1. N/mm ig. 50 kn 50 N q ( ay ) b N/mm N/mm [40 ( )] 40 ( ) ay 40 ( ) + (17.0) mm 5

4 q NA ay b Shear stress at bottom of web: or this ay is calculated from bottom side N/mm ay 10 ( ) 04. q bottom web N/mm q in bttom flange at junction with web N/mm SD is shown in the figure y ay (147.0 y) (147.0 y ) q ay b Shear resisted by web 1 web ( (147.0 y ) (147.0 y ) 0y kn 4.5 kn Problem 4. A cantilever beam of 1 m span in a machine part is having cross-section as shown in ig. 4. The permissible stress in tension and compression are 0 N/mm and 0 N/mm respectively. ind the maximum load W it can carry at free end. [Note: Cantilever is having tension at top] Solution: A mm y mm 0 A 50 0 mm y mm y Distance of N-A from bottom fibre y c mm y t ( ) mm mm ( ) ig. 4

5 (0 1.05) 1.90 mm 4 n cantilever maximum moment occurs at support and is WL W 1 kn-m, if W is in kn Tensile stress occurs at top and compressive stress occurs at bottom side of cantilever beam. Hence (1) from consideration of tensile stress, M f y t E 0 W 1.90 W 11.0 kn. () from consideration of compressive stress, 0 W 1.90 W 0.9 kn (50 1.0) 1.0 Load carrying capacity W 0.9 kn. Ans. Problem 5. A timber beam is to be designed to carry a load of 5 kn/m over a simply supported span of m. Permissible stress is N/mm. Take depth as twice the breadth. Design the beam. f the permissible stress in shear is 1 N/mm, check for shear. Solution: R A R B 5 Max. S 15 kn 15 kn ig. 5 WL 5 Maximum moment M.5 kn-m.5 N-mm Let b be the breadth of the beam and d be the depth d b bd b( b) b 1 1 Moment of resistance of beam 1 1 bd fper b ( b) 0 b Equating moment of resistance to applied moment, we get 0 b.5 b 150 mm 7

6 d 00 mm. Check for shear: q max N/mm < 1 N/mm. bd Hence safe. Use mm section. Problem. A cantilever beam of span m has linearly varying cross-section of size 00 mm 00 mm at fixed end and 0 mm 0 mm at the free end. f it carries a concentrated load of 4 kn at free end, find the maximum stress developed in the beam. Solution: Consider a section at distance y from top. Let the size of the square be a. Then, (00 0) a 0 + y y mm At this section, 1 a a a 1 1 a Section modulus Z 1 a a. Bending moment at this section M 4y kn-m Equating moment of resistance to bending moment, we get f Z M 4 y 4 y 4 y f a a ( y) f or maximum stress 0 y i.e. 4 [(0 + 50y) + y ( ) (0 + 50y) 4 50] y 4 ( y) ( y) y 150y 4 y 1 m f max 4 y (0 + 50) N/mm. Ans. Problem 7. A wooden beam is strengthened by rigidly connecting two steel plates as shown in ig. 7(a). ind how much concentrated load at quarter span it can carry over a span of mm. Take: E w 500 N/mm f w 5 N/mm ig. Ans.

7 Solution: Modular ratio E s 5 N/mm f s 140 N/mm E E s w [Ans. W.94 kn] The composite section may be replaced by an equivalent steel section. The 0 mm 00 mm wooden component is replaced by a steel component of 00 mm deep but having a width of mm.59 y distance of N A from bottom most fibre (50 + (1 4) (0 + ) + 0 4) mm ( ) Stress in steel ( 7.707) ( ) mm 4. E E s t stress in timber > 140 N/mm 500 Hence stress of 140 N/mm in steel governs the load carrying capacity. MR y max f N-mm ( ) kn-m... (a) f W is the load in kn, M W ab W kn-m t... (b) Equating (a) and (b), we get W 14.5 kn. Ans. Problem. A composite beam is made up of cast iron and steel and has the cross-section as shown in ig.. ind the Cast ron moment carrying capacity of the section if the cast iron is in compression. Given E c 1 5 N/mm f c 0 N/mm and E s 5 N/mm f s 140 N/mm Steel ig. 7(a) mm 00 mm 0 ig. 7(b) mm mm 0 mm 0 mm 90 mm 9 0 mm ig.

8 Solution: Modular ratio m E E s c 5 1 Equivalent width of cast iron plate 0 0 mm A mm y mm A mm y 40 mm y ( ) ( ).91 mm 4 Considering the compressive strength of cast iron, moment carrying capacity. Equivalent steel stress m 0 0 M f c.91 y ( ) N-mm kn-m f steel stress in tension governs the moment carrying capacity MR f N-mm y kn-m Moment carrying capacity M kn-m. Ans. Problem 9. The cross-section of a plate girder consists of web plate 000 mm mm, two flanges of size 400 mm 0 mm and 4 angles 150 mm 150 mm mm as shown in ig. (a). Draw variation of bending and shearing stresses, if the moment and shearing force acting at the section are 4000 kn-m and 00 kn-m respectively. A 1 A 0 mm 0 ig. 9 0 mm 0 mm 0

9 N/mm q 1 q q 000 q 4 q 5 Q N/mm (a) (b) (c) ig. Solution: Due to symmetry N-A is at mid depth y b y t 0 mm (0 ) mm 4 M 4000 kn-m 4000 N-mm 00 kn 00 f q 1 q q q (0 5) M 4000 y min N/mm.7 ay b 0.59 N/mm N/mm (400 0 () N/mm ( ) ( ). N/mm 1

10 q 5 q q N/mm N/mm 00.7 [ ] [ ] [ ].1 N/mm Problem. igure 11 shows the cross-section of a welded plate girder. ind the ratio of maximum shear stress to average shear stress calculated by neglecting overhanging portion of flanges. ind also shear resisted by web All Dimensions in mm ig. 11 Solution: Due to symmetry neutral axis is at mid depth mm Maximum shear occurs at N-A q max ( ay) ( ) b 1.9 q q q av max av Area of rectangle Ans.

11 Shear Resisted by web: Consider an element at y from N-A in the web. ts width mm Depth be dy. Depth of web above the level of y 0 y 0 y 0 + y C.G of this portion y y ay (0 ) + (0 y) (0 y ) ( ay) q [ (0 y )] b 1.9 Shear resisted by the elemental strip 1.9 Shear resisted by web [ (0 y )] δy. 0 [ (0 y ) dy y 40000y 5 0 y Ans. Problem 11. A simply supported beam of 4 m span carries concentrated loads of kn at distances 1 m, m and m from the support. The cross-section of the beam is 0 mm 40 mm. At the section 0.75 m from the left support, calculate bending and shearing stresses at a distance of 0, 0 and 10 mm above the neutral axis. Solution: kn kn kn mm A 1 m 1 m 1 m 1 m (a) B ig. 1 Due to symmetry, R A R B + + kn At 0.75 m from support kn M kn-m mm 4 (b) 40 mm

12 f f 0 0 f 0 f 10 M.5 y y Shear stress q ( ay ) b q 0 q 0 Ans N/mm. Ans N/mm. Ans. 0 (10 y) 10 + y y N/mm. Ans N/mm. Ans q Ans Problem 1. The cross-section of a beam consists of two identical channels and two cover plates as shown in ig. 1. Draw the shear stress variation if the section has to resist a shear force of 00 kn q 1 q q q q 5 (a) 0 1 ig. 1 Solution: Due to symmetry. N-A is at mid depth ŷ 0 mm [ (00 ) ] mm 4 00 kn 00 N (b) 4

13 q 1 q ( ay ) 0. N/mm b N/mm q q 4 00 ( ) ( ) N/mm.0 N/mm q 5.7 N/mm Variation is as shown in the figure. 00 [ ]

2012 MECHANICS OF SOLIDS

R10 SET - 1 II B.Tech II Semester, Regular Examinations, April 2012 MECHANICS OF SOLIDS (Com. to ME, AME, MM) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~