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Chapter 5 Stresses in Beam (Basic Topics)
5.1 Introduction Beam : loads acting transversely to the longitudinal axis the loads create shear forces and bending moments, stresses and strains due to V and M are discussed in this chapter lateral loads acting on a beam cause the beam to bend, thereby deforming the axis of the beam into curve line, this is known as the deflection curve of the beam the beams are assumed to be symmetric about x-y plane, i.e. y -axis is an axis of symmetric of the cross section, all loads are assumed to act in the x-y plane, then the bending deflection occurs in the same plane, it is known as the plane of bending the deflection of the beam is the displacement of that point from its original position, measured in y direction
5.2 Pure Bending and Nonuniform Bending pure bending : M = constant V = dM / dx = 0 pure bending in simple beam and cantilever beam are shown
nonuniform bending : M g constant V = dM / dx g 0 simple beam with central region in pure bending and end regions in nonuniform bending is shown
5.3 Curvature of a Beam consider a cantilever beam subjected to a load P choose 2 points m 1 and m 2 on the deflection curve, their normals intersect at point O ', is called the center of curvature, the distance m 1 O ' is called radius of curvature !, and the curvature is defined as
= 1 /! and we have! d = ds if the deflection is small ds j dx , then 1 d d = C = C = C ! ds dx sign convention for curvature
- : beam is bent concave upward (convex downward)
- : beam is bent concave downward (convex upward)
consider the longitudinal line ef , the length L 1 after bending is
y L 1 = (! - y ) d = dx - C dx ! y then < ef = L 1 - dx = - C dx ! and the strain of line ef is
^ < ef^ y x =^ CC^ =^ -^ C^ =^ -^ ^ y dx! x vary linear with y (the distance from N.S.) y > 0 (above N. S.) = - y < 0 (below N. S.) = + the longitudinal strains in a beam are accompanied by transverse strains in the y and z directions because of the effects of Poisson's ratio
Example 5- a simply supported beam AB , L = 4.9 m h = 300 mm bent by M 0 into a circular arc bottom = x = 0. determine !, , and (midpoint deflection) y - 150 ! = - C = - CCCC x 0. = 120 m
= C = 8.33 x 10 -3^ m- ! =! (1 - cos ) ∵! is large, ∴ the deflection curve is very flat L / 2 8 x 12 then sin = CC = CCCC = 0. ! 2 x 2, = 0.02 rad = 1.146 o then = 120 x 10 3 (1 - cos 1.146 o^ ) = 24 mm
5.4 Normal Stress in Beams (Linear Elastic Materials) ∵ x occurs due to bending, ∴ the longitudinal line of the beam is
subjected only to tension or compression, if the material is linear elastic
then " x = E x = - E y " vary linear with distance y from the neutral surface consider a positive bending moment M applied, stresses are positive below N.S. and negative above N.S. ∵ no axial force acts on the cross section, the only resultant is M ,
thus two equations must satisfy for static equilibrium condition
i.e. F (^) x = ∫" dA = - ∫ E y dA = 0 ∵ E and are constants at the cross section, thus we have
" x vary linearly with y " x j M " x j 1 / I the maximum tensile and compressive stresses occur at the points located farthest from the N.A.
"^ M^ c^1 M 1 =^ -^ CC^ =^ -^ C I S 1 "^ M^ c^2 M 2 =^ CC^ =^ C I S 2 I I where S 1 = C , S 2 = C are known as the section moduli c 1 c 2 if the cross section is symmetric w.r.t. z axis (double symmetric cross section), then c 1 = c 2 = c
thus S^ M c^ M 1 =^ S 2 and^ " 1 =^ -^ " 2 =^ -^ CC^ =^ -^ C I S for rectangular cross section
I = CC^ b^ h^3 S = CC b^ h^2 12 6 for circular cross section
I = CC^ ^ d^^4 S = CC ^ d^^3 64 32 the preceding analysis of normal stress in beams concerned pure bending, no shear force in the case of nonuniform bending ( V g 0), shear force produces warping
(out of plane distortion), plane section no longer remain plane after bending, but the normal stress " x calculated from the flexure formula are not significantly altered by the presence of shear force and warping we may justifiably use the theory of pure bending for calculating " x even when we have nonuniform bending the flexure formula gives results in the beam where the stress distribution is not disrupted by irregularities in the shape, or by discontinuous in loading (otherwise, stress concentration occurs)
example 5-
a steel wire of diameter d = 4 mm is bent around a cylindrical drum of radius R 0 = 0.5 m E = 200 GPa " pl = 1200 MPa determine M and " max the radius of curvature of the wire is d ! = R 0 + C 2 EI 2 EI E d^4 M = C = CCCC = CCCCC ! 2 R 0 + d 32(2 R 0 + d ) (200 x 10 3 ) 4 4 = CCCCCCC = 5007 N-mm = 5.007 N-m 32 (2 x 500 + 4) M M M d 2 EI d E d " max = C = CCC = CC = CCCCCC = CCCC S I / ( d /2) 2 I 2 I (2 R 0 + d ) 2 R 0 + d
t = 12 mm b = 300 mm h = 80 mm determine the maximum tensile and compressive stresses in the beam
construct the V-dia. and M-dia. first we can find + Mmax = 2.205 kN-m
- M (^) max = - 3.6 kN-m next, we want to find the N. A. of the section A (mm^2 ) y (mm) A y (mm^3 ) A 1 3,312 6 19, A 2 960 40 38, A 3 960 40 38, total 5,232 96, Ai y (^) i 96, c 1 = CCC = CCC = 18.48 mm A (^) i 5, c 2 = h - c 1 = 61.52 mm moment of inertia of the section is I (^) z 1 = Iz c + A 1 d 12 1 1 I (^) z c = C ( b - 2 t ) t^3 = C 276 x 12 3 = 39744 mm^4 12 12 d 1 = c 1 - t / 2 = 12.48 mm I (^) z 1 = 39,744 + 3,312 x 12.48 2 = 555,600 mm^4 similarly I (^) z 2 = Iz 3 = 956,000 mm^4
then the centroidal moment of inertia I (^) z is I (^) z = Iz 1 + Iz 2 + I (^) z 3 = 2.469 x 10 6 mm^4 I (^) z I (^) z S 1 = C = 133,600 mm^3 S 2 = C = 40,100 mm^3 c 1 c 2 at the section of maximum positive moment
"^ M^ 2.025 x 10^3 x 10^3 t =^ " 2 =^ C^ =^ CCCCCCC^ =^ 50.5 MPa S 2 40, M 2.025 x 10 3 x 10 3 " c = " 1 = - C = - CCCCCCC = - 15.2 MPa S 1 133,
at the section of maximum negative moment
"^ M^ - 3.6 x 10^3 x 10^3 t =^ " 1 =^ -^ C^ =^ -^ CCCCCCC^ =^ 26.9 MPa S 1 133, "^ M^ - 3.6 x 10^3 x 10^3 c =^ " 2 =^ C^ =^ -^ CCCCCCCC^ =^ - 89.8 MPa S 2 40, thus (" t ) max occurs at the section of maximum positive moment (" t ) max = 50.5 MPa and (" c ) max occurs at the section of maximum negative moment (" c ) max = - 89.8 MPa
5.6 Design of Beams for Bending Stresses design a beam : type of construction, materials, loads and environmental conditions
is surfaced consider a rectangular of width b and depth h b h^2 A h S = CC = CC = 0.167 A h 6 6 a rectangular cross section becomes more efficient as h increased, but very narrow section may fail because of lateral bucking
for a circular cross section of diameter d d^3 A d S = CC = CC = 0.125 A d 32 8 comparing the circular section to a square section of same area h^2 = d^2 / 4 => h = √ d / 2 S (^) square 0.167 A h 0.167 √ d / 2 0. CC = CCCCC = CCCCCCC = CCC = 1. S (^) circle 0.125 A d 0.125 d 0. ∴ the square section is more efficient than circular section the most favorable case for a given area A and depth h would have to distribute A / 2 at a distance h / 2 from the neutral axis, then
A h (^) 2 A h^2 I = C (C) x 2 = CC 2 2 4 S = CC^ I^ = CC A h = 0.5 A h h / 2 2 the wide-flange section or an I - section with most material in the flanges would be the most efficient section
for standard wide-flange beams, S is approximately S j 0.35 A h wide-flange section is more efficient than rectangular section of the same area and depth, ∵ much of the material in rectangular beam is located near
the neutral axis where it is unstressed, wide-flange section have most of the material located in the flanges, in addition, wide-flange shape is wider and therefore more stable with respect to sideways bucking
Example 5-
a simply supported wood beam carries uniform load L = 3 m q = 4 kN/m " allow = 12 MPa wood weights 5.4 kN/m^3 select the size of the beam (a) calculate the required S q L^2 (4 kN/m) (3 m)^2 Mmax = CC = CCCCCCC = 4.5 kN-m 8 8 Mmax 4.5 kN-m S = CC = CCCC = 0.375 x 10^6 mm^3 " allow 12 MPa (b) select a trial size for the beam (with lightest weight) choose 75 x 200 beam, S = 0.456 x 10 6 mm^3 and weight 77.11 N/m (c) now the uniform load on the beam is increased to 77.11 N/m
Example 5-
a simple beam AB of length 7 m q = 60 kN/m " allow = 110 MPa select a wide-flange shape firstly, determine the support reactions R (^) A = 188.6 kN R (^) B = 171.4 kN the shear force V for 0 ≦ x ≦ 4 m is V = R (^) A - q x for V = 0, the distance x 1 is
x^ R^ A^ 188.6 kN 1 =^ CC^ =^ CCCC^ =^ 3.143 m q 60 kN/m and the maximum moment at the section is Mmax = 188.6 x 3.143 / 2 = 296.3 kN-m the required section moudlus is Mmax 296.3 x 10 6 N-mm S = CC = CCCCCCCC = 2.694 x 10 6 mm^3 " allow 110 MPa from table E -1, select the HE 450 A section with S = 2,896 cm^3 the weight of the beam is 140 kg/m, now recalculate the reactions, M (^) max , and Srequired , we have
R (^) A = 193.4 kN R (^) B = 176.2 kN V = 0 at x 1 = 3.151 m => Mmax = 304.7 KN-m
M (^) max S (^) required = CC = 2,770 cm^3 < 2,896 cm^3 (O. K.) " allow
Example 5-
the vertical posts B are supported planks A of the dam post B are of square section b x b the spacing of the posts s = 0.8 m water level h = 2.0 m " allow = 8.0 MPa determine b the post B is subjected to the water pressure (triangularly distributed) the maximum intensity q 0 is q 0 = h s the maximum bending moment occurs at the base is
M^ q^0 h^ h^ ^ h^3 s max =^ CC^ (C)^ =^ CCC 2 3 6 M (^) max h^3 s b^3 and S = CC = CCC = C " allow 6 " allow 6 h^3 s 9.81 x 2^3 x 0. b^3 = CC = CCCCCC = 0.007848 m^3 = 7.848 x 10^6 mm^3 " allow 8 x 10 6 b = 199 mm use b = 200 mm
at that section ( x = L /2), the maximum is 128 P L P L " max = CCCC = 4.741 CC 27 dA^3 dA^3 it is 19% greater than the stress at the built-in end
Example 5-
a cantilever beam of length L support a load P at the free end cross section is rectangular with constant width b , the height may vary such that " max = " allow for every cross section (fully stressed beam) determine the height of the beam b hx^2 M = P x S = CC 6 "^ M^ P x^^6 P x allow =^ C^ =^ CCC^ =^ CC S b hx^2 / 6 b hx^2 solving the height for the beam, we have
h^6 P x^ 1/ x =^ (^ CCC^ ) b " allow at the fixed end ( x = L ) 6 P L (^) 1/ h (^) B = ( CCC ) b " allow
then h x^ 1/2^ the idealized beam has the x =^ hB (^ C^ ) L parabolic shape
5-8 Shear Stress in Beam of Rectangular Cross Section
for a beam subjected to M and V with rectangular cross section having width b and height h , the shear stress $ acts parallel to the shear force V assume that $ is uniform across the width of the beam consider a beam section subjected the a shear force V , we isolate a small element mn , the shear stresses $ act vertically and accompanied horizontally as shown ∵ the top and bottom surfaces are free,
then the shear stress must be vanish, i.e. $ = 0 at y =! h / for two equal rectangular beams of height h subjected to a concentrated load P , if no friction between the beams, each beam will be in compression above its N.A., the lower longitudinal line of the upper beam will slide w.r.t. the upper line of the lower beam for a solid beam of height 2 h , shear stress must exist along N.A. to prevent sliding, thus single beam of depth 2 h will much stiffer and stronger than two separate beams each of depth h
