Centroid or center of gravity is the point within an object from which the force Centroid of gravity appears to act. Centroid of 3D objects often (but not always) lies somewhere along the lines of symmetry.

Centroidal axis or Neutral axis

Centroid of Area (Center of Gravity of an area): Point () , yx that defines the geometric center of the area δAy ∫∫ δAx y = A x = A A A Q Q δ ⋅== AyAy First Moment of an Area with x ∫ VQ respect to the x-axis A τ = First Moment of an Area with Qy δ ⋅== AxAx Ib respect to the y-axis ∫ A Calculate the center of gravity of the rectangle: (a)Without a hole (b)With a hole of dimensions c and d a) Without a hole δAy b ∫ δyay ab2 b The centroid of any area can be found by taking y A ∫0 ==== moments of identifiable areas (such as A ab ab 22 rectangles or triangles) about any axis. δAx a The moment of an area about any axis is equal ∫ δxax 2ba a x A ∫0 ==== to the algebraic sum of the moments of its A ab ab 22 component areas. b) With a hole n ⎛ b ⎞ d δAy Ay ⎜ ⎟ ( fba ) ⋅⋅+−⋅⋅ dc ∑∫ ii 2 2 y = A y = 1 y = ⎝ ⎠ A A ⋅−⋅ dcba n ⎛ a ⎞ c δAx Ax ⎜ ⎟ (eba ) ⋅⋅−−⋅⋅ dc ∑∫ ii 2 2 x = A x = 1 x = ⎝ ⎠ A A ⋅−⋅ dcba Moment of Inertia also known as the Second Moment of the Area is a term used to describe the capacity of a cross-section to resist bending. It is a mathematical property of a section concerned with a surface area and how that area is distributed about the reference axis. The reference axis is usually a centroidal

Second Moment or Moment of Inertia of an Area

= 2δ = 2δAxIAyI x y ∫∫ A A Rectangular Moments of Inertia: Moments of inertia with respect to an axis Polar moments of inertia: Moment of inertia with respect to a point 2δρ +== IIAJ O ∫ yx A Simple rectangular shape = 2 = bdydAdAyI x ∫ A

h h 2 3 2 ⎡ 3 ⎛ hhby 3 ⎞ ⎤ 2 ⎜ ⎟ x = = bbdyyI ⎢ ⎜ −−= ⎟ ⎥ ∫ h 8833 − h − ⎣ ⎝ ⎠ ⎦ 2 2 bh3 hb3 I = I y = x 12 12

= 2 = hdxdAdAxI y ∫ A

b b 2 3 2 ⎡ 3 ⎛ bbhx 3 ⎞ ⎤ 2 ⎜ ⎟ y = = hhdxxI ⎢ ⎜ −−= ⎟ ⎥ ∫ b 8833 − b − ⎣ ⎝ ⎠ ⎦ 2 2 Polar Moment of Inertia The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs. 2 0 = ∫ dArJ The polar moment of inertia is related to the rectangular moments of inertia,

2 ( 22 ) 2 +=+== 2dAydAxdAyxdArJ 0 ∫ ∫ ∫ ∫ 0 = + IIJ xy

Simple rectangular shape

bh3 hbbh 33 I x = III +=+= 12 Polar yx 1212 bh hb3 22 I = I P ()+= bh y 12 12 Moment of Inertia: Parallel Axis Theorem for an Area If the moment of inertia of an area is known about its neutral axis (centroid axis), we can determine the moment of inertia of area about a corresponding parallel axis using the parallel axis theorem Consider moment of inertia of the shaded area

()+= 2 dAydI x ∫ 1 dA

2 += 2 + 2 dAdydAddAyI x 1 1 ∫∫∫ dA dA dA

First integral represents the moment of inertia of the area about the centroidal axis ⋅+= dAII 2 Second integral is equal to zero, since xc passes xcx 1 through the area’s centroid C Third integral represents the total area A 2 Similarly: ycy ⋅+= dAII 2 2 Simple rectangular shape ⎛ ⎞ bhbhh 33 xBB += AII ⎜ ⎟ += bh3 ⎝ ⎠ 4122 I x = bh3 12 I = BB 3

Note:

= 2 = 2 ArIArI Radius of Gyration xx yy 2 222 = OO += rrrArJ yxO Moment of Inertia of Composite Areas

Determine the moment of inertia of the area shown with respect to the x-axis

The moment of inertia of the area shown can be obtained by sustracting the circle from the rectangle

Circle 2 X-axis xx ' += AdII y 1 = ()25 4 + ππ ()()22 = ()104.117525 mm 46 4 Rectangle 2 xx ' += AdII y 1 = ()()()()()3 + 2 = ()105.11275150100150100 mm46 12 6 6 46 I x ×=×−×= 10101104.11105.112 mm Determine the moment of inertia of the area shown with respect to the centroids.

Rectangle A

2 xx ' += AdII y 1 = ()()()()()300100 3 + 2 ×= 10425.1200300100 mm 49 12 2 yy ' += AdII x 1 = ()()()()()100300 3 + 2 ×= 1090.1250300100 mm 49 12

Rectangle B

2 xx ' += AdII y 1 = ()()3 ×= 1005.0100600 mm 49 12 2 yy ' += AdII x 1 = ()()3 = 8010.1600100 mm 49 12 Rectangle D

2 xx ' += AdII y 1 = ()()()()()300100 3 + 2 ×= 10425.1200300100 mm 49 12 2 yy ' += AdII x 1 = ()()()()()100300 3 + 2 ×= 1090.1250300100 mm 49 12

Adding each segment

9 I x ( )×++= 10425.105.0425.1 49 I x ×= 1090.2 mm 9 I y ()×++= 1090.180.190.1 49 I y ×= 1060.5 mm Determine the moment of inertia of the area shown with respect to the x-axis centroid.

-

2 Bodies Ai yi yi*Ai Ii di=yi-ybar di Ai

1 21600 90 1944000 58320000 0 0 2 -9600 90 -864000 -11520000 0 0

12000 1080000 46800000 0 ybar 90 mm I 46800000 mm4

46 I x ×108.46 mm Radius of Gyration r == x A 12000mm2

rx = 45.62 mm Determine the moment of inertia of the area shown with respect to the x- axis centroid. Moments of Inertia of Composite Areas Moments of Inertia of Composite Areas The strength of a W14x38 rolled steel beam is increased by attaching a plate to its upper flange. Determine the moment of inertia and radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the section. Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section. Section A, in 2 , Ayy ,in. in 3 Plate 6.75 12.50425.7 Section Beam Section 11.20 00 ∑ A = 95.17 ∑ Ay = 12.50

∑ Ay 12.50 in 3 = ∑∑ YAyAY == = 792.2 in. ∑ A 17.95 in 2 Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis. 2 2 Ix′ beam, section x YAI +=+= ( )( 792.220.11385 ) = in3.472 4 3 AdII 2 =+= 1 ()()3 + ()(− 792.2425.775.69 2 ) x′ plate, x 12 4 = 2.145 in 4

I = Ixx ′′ beam, section + Ix′ plate, = + 2.1453.472

4 Ix′ = 618 in Calculate the radius of gyration from the moment of inertia of the composite section.

I 5.617 in 4 r x′ == = 87.5 in x′ A 17.95 in 2 Relationship between Relationship between Bending Moments and Bending Moments and Curvatures Normal Stresses 1 M κ == ρ EI My σ −= x I

Maximum stresses

Mc1 M σ1 −=−= I S1

Mc2 M σ 2 == I S2

S1 and S2 are known as the section moduli. Stresses in a simple beam.

∑ y RF A ()( ) 125.1220 +−−== RB

∑ M @ B RA ()()()()()(−== =−− 09221211225.1220 )

RA = 59.23 Kips RB = 41.21 Kips Rectangular Cross Section ()(2775.8 )3 I = = 14352in4 z 12 I 14352 S === 1063in3 c 27 2 dV =−= 0 =−= CqdxVq dx x ∫

≤ ≤ 90 x = − 5.1 ⇒+ ()01 = A ⇒ 1 = RCRVCxVx A

x xV +−= 59.235.1 2 M x 59.2375.0 2 MCxx ()0 C2 =⇒=⇒++−= 00 2 M x +−= 59.2375.0 xx

V()9− = 09.10 VKips ()9+ −= 91.1 MKips ()9 = 6.151 − ftKips ≤≤ 229 x = − 5.1 + ⇒ VCxVx ()93 + = 91.1 ⇒− C3 = 59.11

x xV +−= 59.115.1 2 M x 59.1175.0 4 MCxx ()22 0 C4 =⇒=⇒++−= 02.108 2 M x xx ++−= 02.10859.1175.0

V()22 = 41.21 MKips ()9 = 6.151 − ftKips

M Max = 6.151 − ftkips M × ×1210006.151 σ Max == = 1710 psi Max S 1063 A cast-iron machine part is acted upon by 3 kN-m Knowing E = 165GPa and neglecting the effects of fillets, determine (a) the maximum tensile and compressive stresses, (b) the radius of curvature.

(a) Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.

(b) Apply the elastic flexural formula to find the maximum tensile and compressive stresses.

(c) Calculate the curvature (a) Based on the cross section geometry, calculate the location of the section centroid and moment of inertia. (b) Apply the elastic flexural formula to find the maximum tensile and compressive stresses.

− 49 I z ×10868 m − 36 A = 022.0 Smc A == ×= 1045.39 m cA 022.0 m − 49 I z ×10868 m − 36 B = 038.0 Smc B == ×= 108.22 m cB 038.0 m M − 3000 ⋅ mN σ A −=−= − 36 = 0.76 MPa S A ⋅105.394 m M 3000 ⋅− mN σ B == − 36 −= 3.131 MPa SB ⋅108.22 m

(c) Calculate the curvature 1 M 3000 ⋅ mN == = 0209.0 m−1 ρ EI ()()⋅ 9 Pa ⋅1086810165 − m49 ρ = 7.47 m Stresses in a beam with an overhang. Cross Section

Reactions − ( )( 5.42.3 )( 5.4 ) R = 2 = 8.10 kN B 3

RA = ()()=− 6.38.105.42.3 kN

≤ ≤ 30 x = − 2.3 + ⇒ ()01 = 6.3 ⇒ 1 = 6.3 kNCkNVCxVx 6.3 06.32.3 xVxV ==⇒=⇒+−= 125.1 m x x 2.3 2 x 6.36.1 2 MCxxM ()0 C2 =⇒=⇒++−= 00 2 x +−= 6.36.1 xxM

V()3− −= 0.6 VkN ()3+ = 8.4 kN

M()125.1 025.2 −= MmkN ()0.3 6.3 −−= mkN 2.3 5.43 x −=≤≤ 2.3 + ⇒ ()33 = 8.4 ⇒ 3 = 4.14 kNCkNVCxVx x xV +−= 4.142.3 2 x 4.146.1 4 MCxxM ()3 6.3 C4 −=⇒−=⇒++−= 4.32 2 x xxM −+−= 4.324.146.1

Maximum Bending M()125.1 = 025.2 − MmkN ()0.3 = − 6.3 − mkN Moments 4 I z 2468761mm 3 c1 = 48.18 Smm 1 == = 133591mm c1 48.18 mm 4 I z 2468761mm 3 c2 = 52.61 Smm 2 == = 40129mm c2 52.61 mm x = 125.1 m x = 0.3 m M 2025 − mN M 3600 −− mN σ1 −=−= − 33 −= 2.15 MPa σ1 −=−= − 33 = 9.26 MPa S1 ×10336.1 m S1 ×10336.1 m M 2025 − mN M 3600 −− mN σ 2 == − 33 = 5.50 MPa σ 2 == − 33 −= 8.89 MPa S2 ×10401.0 m S2 ×10401.0 m

Maximum Tensile Stress = 50.5MPa at x=1.125m Maximum Compressive Stress = 89.8MPa at x=3.0m Which type of cross section is the most efficient in resisting the bending stresses??

Consider beams made of the same material, subjected to M the same moment and with similar cross section areas. S = σ Allowed πd 2 Circular cross section A = 4 πd 4 I πd 3 I = S === 098175.0 d 3 64 circle d 322 2 Square cross section πd hA 2 =⇒== 886.0 dh 4 4 4 h 3 h h I = S 12 === 6955.0 d 3 12 square h 6 2 h2 πd 2 nπ b=h/n Rectangular cross section bhA h =⇒=== d n 4 2 4 4 h 3 h h I = S 12n === 116.0 dn 3 h 12n tan glerec h 6n 2 116.0 1 == 116.0 dSn 3 164.0 2 == 164.0 dSn 3 = 10 = 367.0 dSn 3

Design of a post using solid wood or aluminum tube.

Max = = ( )( )= 305.212 − mkNmkNPhM

Solid wood:

3 π Md Max 36 SWood ⋅=== 102 mm 32 σ Allowed

1 = 273mmd

Aluminum tube

π 4 4 4 Solid wood: σ =15MPa []2 ()2 tddI =−−= 03356.0 d2 Allowed 64 Aluminum tube: σAllowed=50MPa I M Max 33 S2 ⋅=== 10600 mm d2 σ 2 Allowed

2 = 208mmd Bending Members Made of Several Materials

Consider a composite beam formed

from two different materials E1 and E2. The normal strain y varies linearly. ε −= x ρ

Piecewise linear normal stress variation 1 yE Elemental forces on the E11 εσx −== Neutral axis does not pass through ρ section are yE section centroid of composite section. yE 1 E εσ−== 2 σ11 dAdF −== dA 22 x ρ ρ

2 yE Define a transformed σ 22 dAdF −== dA section such that ρ yE ynE yE dF 2 dA 1 dA −=−=−= 1 ()ndA 2 ρ ρ ρ E n = 2 E1 A bar is made from bonded pieces of steel (ESteel=29000ksi) and brass (Ebrass=15000ksi). Determine the maximum stress in the steel and brass when a moment of 40kips-in is aplied. (a) Transform the bar to an equivalent cross section made entirely of brass E 29000ksi n Steel == = 933.1 EBrass 15000ksi

T inb += ()()+ = 25.24.075.0933.14.0 ininin Evaluated the transformed cross sectional properties hb 3 ( )(325.2 3 ) I T == = 063.5 in4 12 12

Calculated the maximum stresses Mc ()⋅ ( 5.140 ininkips ) ()σ =−= = 85.11 ksi Brass Max I 063.5 in4

(σ Steel )Max = n(σ Brass )Max = × = 9.2285.11933.1 ksi Shear Stresses in Beams of Rectangular Cross Section Note that at the top or at the bottom of the beam the horizontal stresses must vanish.

Glued beam: When it is loaded horizontal shear stresses must develop along the glued surface in order to prevent the sliding.

dM x = V The change of moment with the distance x dx x generates a shear force. My σ −= 1 I ()+ ydMM σ −= 2 I

Isolate a subelement mm1p1p and find the forces acting (assuming equilibrium). My σ dAF == dA ∫∫11 I ()+ ydMM σ dAF == dA ∫∫22 I ()+ ydMM My FFF =−= dA − dA 123 ∫ I ∫ I dMy dM F = dA = ydA 3 I I ∫∫ = τbdxF 3 VQ dM τ = τbdx = ydA Ib I ∫ 1 dM V τ = ydA = ydA Ib dx Ib ∫∫ This equation is known as the shear formula.V, I (for the entire section) and b (width at y1) are constants with y, while Q varies with y. Distribution of Shear Stresses acting on a Beam The first moment Q of the shaded part of the cross-sectional area is obtained by multiplying the area by the distance from its centroid to the neutral axis. ⎡ ⎛ h ⎞⎤⎡ 1 ⎛ h ⎞⎤ ⎛ hb 2 ⎞ byAQ ⎜ −== ⎟ yy ⎜ −+ y ⎟ ⎜ −= y2 ⎟ Shaded 1 ⎢ ⎥⎢ 11 1 ⎥ ⎜ 1 ⎟ ⎣ ⎝ 2 ⎠⎦⎣ ⎝ ⎠⎦ ⎝ 4222 ⎠ For y1=h/2 then Q=0 For y1=0 then Q is maximum VQ V ⎛ hb 2 ⎞ τ == ⎜ − y2 ⎟ ⎜ 1 ⎟ Ib Ib ⎝ 42 ⎠ Parabolic distribution V ⎛ h2 ⎞ τ ⎜ −= y2 ⎟ ⎜ 1 ⎟ I ⎝ 42 ⎠ h y Q 0 τ ⎛ h ⎞ =⇒=⇒= 0 2 ⎜ ⎟ ⎝ 2 ⎠ 2 2 V (bh ) 3V bh 8 3 V 0 Qy Max τ Max =⇒=⇒= 3 = τ Max = 8 (bh )b 2 A 2A 12 h y Q 0 τ ⎛ h ⎞ =⇒=⇒−= 0 2 ⎜ − ⎟ ⎝ 2 ⎠ Shear Flow The shear flow represents the force over a unit length of the beam that would be required to hold the beam together. If the beam started out as two pieces separated along a horizontal line, and the two pieces were welded together, the strength of the weld would have to be at least equal to the shear flow. It would represent the required strength of a unit length of the weld. It can also be used to determine how many nails are required to hold together two pieces of a fabricated wooden beam. From the equation:

3 = τbdxF

dM τbdx = ydA I ∫ 1 dM V VQ τ = ydA ydA == Ib dx Ib ∫∫ Ib VQ V Shear _ qFlow τb ==== Horizontal I Δx

A beam is made of three planks, nailed together. Knowing that the spacing between nails is 25 mm and that the vertical shear in the beam is V = 500 N, determine the shear force in each nail. Calculating Q: To determine Q, cut the cross section horizontally at the NA and choose either the top or the bottom portion, it doesn't matter which because the same answer will result using either portion.

− 36 z == []()()mmmyAQ ×= 1012006.01.002.0 m Calculation of the moment of inertia about the centroid 3 3 ⎡()()1.002.0 2 ⎤ ⎡( )( 02.01.0 ) ⎤ I z = ⎢ + ()()()⎥ + 206.01.002.0 ⎢ ⎥ ⎣ 12 ⎦ ⎣ 12 ⎦ − 46 I z ×= 1020.16 m

Calculating the shear flow (q): Calculating the shear force per nail for a nail spacing of 25 mm. VQ × − 36 )m10120)(N500( q == 46- F = q = 3704)(m025.0()m025.0( mN I × m1016.20 q = 3704 N F = N6.92 m Determination of the Shearing Stress in a Beam

The average shearing stress on the horizontal face of the element is obtained by dividing the shearing force on the element by the area of the face. ΔH Δxq VQ Δx τ = = = ave ΔA ΔA I Δxt VQ τ = ave It

On the upper and lower surfaces of the beam, τyx= 0. It follows that τxy= 0 on the upper and lower edges of the transverse sections. Shearing Stresses in Common Beam Types

For a narrow rectangular beam,

VQ 3V ⎛ y2 ⎞ τ ⎜1−== ⎟ xy ⎜ 2 ⎟ Ib 2 A⎝ c ⎠ 3V τ = max 2 A

For American Standard (S-beam) and wide-flange (W-beam) beams VQ τ = ave It V τmax = Aweb The diagram shows a simply supported 20 ft. beam with a load of 10,000 lb. acting downward at the center of the beam. The beam used is a rectangular 2" by 4" steel beam. We would like to determine the maximum bending (axial) stress and the maximum shear stress Reactions:

y = 5000lbA

y = 5000lbC

M Max 50000 −= ftlb

= 2iny σ Max = 112500 psi ()()42 3 I == 67.10 in4 12

= 5000lbV 3V τ == 5.937 psi A ()()== 842 in2 Max 2A Determine the normal and shear stresses at point C.

Rectangular section

A = 2880 B = 2880 lbRlbR

VC ()()−=−−= 16008361602880 lb bh3 ( )( 0.40.1 )3 ()− 836 I == = 333.5 in4 M = ()()()()−−− 8361608362880 17920 ⋅= inlb 12 12 C 2 yM ()17920 ⋅ ( 0.1 ininlb ) σ C −=−= −= 3360 psi C I 333.5 in4

3 CC C == ()()( )= 5.15.10.10.1 ininininyAQ QV ()( 5.11600 inlb ) τ CC == = 450 psi C Ib ()4 ()0.1333.5 inin Determine the maximum permissible value PMax if the allowable stresses for bending and shear are 11MPa and 1.2MPa respectively. Data : h=150mm ; b=100mm and a=0.5m

Max = Max = PaMPV 3 bh 2 I bh S 12 === = bhA c h 6 2 M 6Pa σ bh2 σ Max PBending =⇒== Allowed Max S bh2 Max 6a 3V 3P 2τ bh τ Max PShear =⇒== Allowed Max 2A 2bh Max 3

Bending σ Allowed 11 PMPa Max =⇒= 25.8 kN Shear τ Allowed = 2.1 PMPa Max =⇒ 0.12 kN

The bending stress governs the design. A timber beam is to support the three concentrated loads shown. Knowing that for the grade of timber used, σ all = psi1800 τall = psi120 determine the minimum required depth d of the beam. Determine the maximum shear force and moment. Vmax = kips3 Mmax ⋅=⋅= inkip90ftkip5.7

1 3 Determine the = 12 dbI section moduli I S 1 db 2 === 1 ()in.5.3 d 2 c 6 6 S = ()in.5833.0 d 2 Determine the beam depth based on allowable normal stress. M 3 ⋅× in.lb1090 σ max == 1800 psi = all S ()in.5833.0 d 2 d = in.26.9

3V Determine the beam depth based on allowable τ = max shear stress. all 2 A 3 lb3000 psi120 = 2 ()in.3.5 d d = in.71.10

Required beam depth is equal to the larger of the two. d = in.71.10 Shear Stresses in Beam of Circular Section = ⋅ φ = ⋅CosrxSinry φ = 2 ⋅ ⋅ φ ⋅δyCosrdA δ = ⋅Cosry φ ⋅δφ

π 4 4 2 2 2 π ⋅r x 2 CosSinrI δφφφ=⋅⋅⋅⋅= ∫−π π 2 4 2 4 4 Q ()(2 ⋅⋅⋅⋅== .dCosrCosrSinrydA φφφφ) π ⋅r π ⋅r Max ∫∫ I = IIJ =+= 0 y 4 yxz 2 π 2 = 2 3 ()()2 φφφ φ −=⇒=⇒ dSinduCosudCosSinrQ φφ Max ∫ Hollow Circular Cross Section 0 3 3 π 2(2 − rr 1 ) 2 2r3 π 2r3 QMax = 2()−=⇒ rrb 12 −= 2 3 ()2 duurQ −= ()u3 2 = 3 Max ∫ 0 0 3 3 2 2 V (2r ) π (2 −= rrA 1 ) VQ 3 4V 4V τ Max == 4 2 == Ib (πr )()2r 3πr 3A ⎛ 2()3 − rr 3 ⎞ 4 V ⎜ 2 1 ⎟ ⎜ 3 ⎟ 4V ⎛ 2 ++ rrrr 2 ⎞ τ = ⎝ ⎠ = ⎜ 2 112 ⎟ Max 4 4 ⎜ 2 2 ⎟ ⎛π ()2 − rr 1 ⎞ 3A ⎝ 2 + rr 1 ⎠ ⎜ ⎟()2()− rr 12 ⎝ 4 ⎠ Design of a post using solid wood or aluminum tube.

Solid wood: 4V 4V τ == = τ Max 3A ⎛πd 2 ⎞ Allowed 3⎜ 1 ⎟ ⎝ 4 ⎠

2 16V ()1216 kN − 23 d1 = = ⋅= 1072.2 m 3πτ Allowed π ()5.73 MPa

1 = 1.52 mmd

Aluminum tube 7π π ( 2 2 )=−= rrrA 2 2 1 16 2

⎛ ⎞ Solid wood: τ =7.5MPa 2 2 ⎜ ⎟⎛ 37 2 ⎞ Allowed 4V ⎛ ++ rrrr ⎞ 4V 1 ()r2 τ = ⎜ 2 112 ⎟ = ⎜ ⎟⎜ 16 ⎟ Max ⎜ 2 2 ⎟ 2 Aluminum tube: τAllowed=25MPa 7π 2 ⎜ 25 ⎟ 3A ⎝ 2 + rr 1 ⎠ 3 ⎜ ⎟ ()r2 ⎜ r2 ⎟⎝ 16 ⎠ ⎝ 16 ⎠

2 V ()12436.1 kN −3 r2 = 436.1 = ⋅10689.0 m τ Allowed 25MPa

d2 = 5.52 mm

Shearing Stresses in Thin-Walled Members Consider a segment of a wide-flange beam subjected to the vertical shear V. The longitudinal shear force on the element is VQ H Δ=Δ x I The corresponding shear stress is ΔH VQ τ τ ≈= = xzzx Δxt It Previously found a similar expression for the shearing stress in the web VQ τ = xy It

τ ≈ 0 in the flanges NOTE: xy τ xz ≈ 0 in the web Shear Stresses in the Webs of Beams with Flanges Shearing Stresses in Thin-Walled Members

The variation of shear flow across VQ the section depends only on the τtq == variation of the first moment. I For a box beam, q grows smoothly from zero at A to a maximum at C and C’ and then decreases back to zero at E. The sense of q in the horizontal portions of the section may be deduced from the sense in the vertical portions or the sense of the shear V.

For a wide-flange beam, the shear flow increases symmetrically from zero at A and A’, reaches a maximum at C and the decreases to zero at E and E’.

The continuity of the variation in q and the merging of q from section branches suggests an analogy to fluid flow. Knowing that the vertical shear is 50 kips in a W10x68 rolled-steel beam, determine the horizontal shearing stress in the top flange at the point a.

For the shaded area, Q = ( )( )( in815.4in770.0in31.4 ) Q = in98.15 3

VQ ( )( in98.15kips50 3) The shear stress at a, τ == It ( 4 )()in770.0in394 τ = ksi63.2 Shear Stresses in the Webs of Beams with Flanges

h1 V _ yFor 1 τWeb []()()()1 1 +−+=⇒= thhhhb w 0 2 8Itw V V τWeb []()1 ()2thhb f =+= []()2 f ()+ hhbt 1 8Itw 8Itw

V ⎡ ⎛ + hh 1 ⎞⎤ τWeb = ⎢()Af ()22 ⎜ ⎟⎥ 8Itw ⎣ ⎝ 2 ⎠⎦

⎡⎛ h h1 ⎞ ⎤ V ⎜ − ⎟ τ = ⎢ 22 ()A ⎥ Web ⎢⎜ ⎟ f ⎥ Itw ⎜ 2 ⎟ ⎣⎢⎝ ⎠ ⎦⎥ V 2 yFor 1 0_ τWeb [()()1 1 +−+=⇒= w ()hthhhhb 1 ] 8Itw V τWeb []()1 ()2 ++= ()()wf hhtthhb 11 8Itw

V ⎡ ⎛ hh 1 ⎞ ⎛ h1 ⎞⎤ = ⎢()Af 22 ⎜ + ⎟ + ()Aw 22 ⎜ ⎟⎥ 8Itw ⎣ ⎝ 22 ⎠ ⎝ 2 ⎠⎦

V ⎡ ⎛ 1 ⎞⎛ + hh 1 ⎞ ⎛ 1 ⎞⎛ h1 ⎞⎤ τWeb = ⎢()Af ⎜ ⎟⎜ ⎟ + ()Aw ⎜ ⎟⎜ ⎟⎥ Itw ⎣ ⎝ ⎠⎝ 22 ⎠ ⎝ ⎠⎝ 22 ⎠⎦ A cantilever beam with T cross section is loaded at the tip by a vertical force. Determine at the section n-n: (a) Maximum compressive stress. (b) Maximum tensile stress. (c) Maximum shear stress.

Calculation of the centroid using bb as reference

Ay ()( )(+− ( ))( )(15.0225.05.04 ) y ∑ == = 1667.0 in ∑ A ()()()()+ 5.025.04

Calculation of the moment of inertia about the centroid 3 3 ⎡()(5.04 ) 2 ⎤ ⎡( )(25.0 ) 2 ⎤ I z = ⎢ + ()( )(+ 1667.025.05.04 ⎥ + ⎢ ) + ()()(− 1667.0125.0 ⎥ ) ⎣ 12 ⎦ ⎣ 12 ⎦ 4 z = 417.1 inI Calculating the bending stresses z = PxM = ( )= 18000121500 − inlb

M z 18000 Compressive Stress σ ,bottomx −= ybottom −= ()833.1 −= 23300 psi I z 417.1

M z 18000 Tensile Stress σ ,topx ytop ()=−−=−= 8.8472667.0 psi I z 417.1 To determine Q, cut the cross section horizontally at the NA and choose either Calculating Q: the top or the bottom portion, it doesn't matter which because the same answer will result using either portion.

⎛ 833.1 ⎞ 3 z yAQ == []()5.0833.1 ⎜ ⎟ = 84.0 in ⎝ 2 ⎠

Calculating the maximum shear stress

QV zy ()()84.01500 τ ,Maxxy == = 1778psi ztI ()()5.0417.1 A beam shown is loaded by a 3 kN force. For points A and B located at section n-n, (0.5m from the support) shown in the figure below, determine the average shear stress.

Calculation of the centroid using the bottom as reference

Ay ()()()+ ( )( )( )+ ( )( )(10206060208011020100 ) y ∑ == ∑ A ( )()()()()()++ 2060208020100 Ay 328000 y ∑ === 33.68 mm ∑ A 4800 Calculation of the moment of inertia about the centroid 3 3 3 ⎡()()20100 2 ⎤ ⎡( )(8020 ) 2 ⎤ ⎡( )(2060 ) 2 ⎤ I z = ⎢ + ()()()667.4120100 ⎥ + ⎢ ()()(−+ 33.88020 ⎥ + )⎢ ()()(−+ 33.582060 ⎥ ) ⎣ 12 ⎦ ⎣ 12 ⎦ ⎣ 12 ⎦ − 46 I z ×= 1063.8 m

Calculating the bending stresses z = PxM = ()= 7505.01500 − mN

M z 750 σ ,bottomx −= ybottom −= −6 ()=− 93.50683.0 MPa I z ×1063.8

M z 750 σ ,topx ytop −=−= −6 ()−=− 49.40517.0 MPa I z ×1063.8 yAQ == [( )(]( ) 0683.002.006.0 02.0 ) ×=− 100.7 − m37 Calculating Q for point A and B: z 2 − 35 z yAQ == []()()()×=− 1067.4005.00517.001.01.0 m

−7 −5 ( )( ×1071500 ) VQ ()()×1067.41500 τ Average = −6 = 608.0 MPa τ Average == = 081.0 MPa ()× ()02.01063.8 It ()× −6 ()1.01063.8 A beam 12in long is to support a load of 488lb as shown. Basing the design only on a bending stress, the designer has selected a 3in column channel as shown. The direct shear has been neglected.

Mc ( )( 5.11098 ) σ ±=±= Max I 66.1

σ Max ±= 992 psi

However, due to the combined bending and direct shear the stress should be maximum just before 3in and at the point where the web joints the flange

y1=1.5-0.273=1.227 Mc ( )( 227.11098 ) σ −=−= −= 812 psi Max I 66.1

1 3 VQ ( )( 525.0366 ) AyQ '' ()()()+== = 525.0273.0410.1227.15.1 in τ =−= − 681psi 2 xy It ()()170.066.1 ⎡ ⎛ ⎞ ⎛ hh t f ⎞ ⎛ 1 ⎞⎛ h t f ⎞⎤ = ⎢ f btQ ⎜ ⎟ ⎜ −−+ ⎟ty w ⎜ ⎟⎜ −− y⎟⎥ ⎜ 222 ⎟ ⎜ 222 ⎟ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎦ y(in) Q(in3) σ(psi) τ(psi) ⎡ 2 ⎤ 1.5 0 992 0 ⎛ ⎞ ⎛ 1 ⎞⎛ hh t f ⎞ = ⎢ btQ ⎜ ⎟ + ⎜ ⎟⎜ −− ⎟ ty ⎥ f 2 ⎜ 222 ⎟ w 1.227 0.525 812 681 ⎣⎢ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎦⎥ 1.0 0.568 661 737 0.75 0.605 496 785 0.5 0.631 331 818 0.25 0.648 165 840 0 0.653 0 847 For the simply supported W 10x45 beam (ABCD) determine the maximum bending stress at 6 ft from the left of the beam and the maximum horizontal shear stress at a point 4 in above the bottom of the beam.

D=10.10 F=8.02 Tf=0.62 Tw=0.35

Reactions

∑ 0 yy 5000 −−+== ( )(4/2000 ftftlblbDBF )

∑ 0 DBF yy =+⇒= 13000lb

∑ M @ D −== ( )()− ( )()()+ By 82420001250000 ()

y 9500 y =⇒= 3500lbDlbB = 4500lbV At 6 ft from the left: 6 M 6 20000 +−−= ()()lblbft 24500

M 6 11000 −−= lbft

M 11000 −−= ftlb σ −= 2688psi = 1.49 inS 3 Max - - - Flange Flange Web Cross Section Info. Cross Section Info.

Designation Area Depth Width thick thick x-x axis x-x axis x-x axis y-y axis y-y axis y-y axis

- A d bf tf tw I S r I S r - in2 in in in in in4 in3 in in4 in3 in

W 10x45 13.20 10.12 8.022 0.618 0.350 249.0 49.1 4.33 53.20 13.30 2.00 = 4500lbV At 4 in above the bottom of 6 ⎤ 4 ⎥ = 249inI VQ ()()889.264500 the beam ⎥ τ ==⇒ = 1388psi = 35.0 inb ⎥ Hor It ()()35.0249 D=10.10 2 ⎥ = ()()37.4153.6 ininQ F=8.02 ⎦ Tf=0.62 6 = 4500lbV ⎤ Tw=0.35 4 ⎥ = 249inI VQ ()()003.274500 ⎥ τ ==⇒ = 1.1394 psi = 35.0 inb ⎥ Max It ()()35.0249 At the centroid of the beam 3 ⎥ Q = 003.27 in ⎦ V 4500 lb τ Max == = 9.1447 psi Aweb ()− ()()35.062.0212.10 VQ V τ [( 2 2 ) ( 2 −+−== 4yhthhb 2 )] web It 8It 1 1 1 ()4500lb τ = []()()2 ()2 +− ()()()2 − 486.835.086.81.1002.8 y2 web ()in4 ()35.02498 1 h y 1 τ =⇒= 8.1216 psi 1 2 web min,

y1 0 τ web max, =⇒= 3.1394 psi For a WT 8 x 25 T-beam determine the maximum bending and shear stress in the beam. We will also determine the bending stress at 4 ft from the left end of the beam. Reactions

∑ 0 CBF yy −+== ( )( )− ( )(4/15004/1000 ftftlbftftlb )

∑ 0 CBF yy =+⇒= 10000lb

∑M @ C −== ( )()()+ ( )()()+ By 62415008410000 ()

y 3330 y =⇒= 6670lbClbB M 12000 −= ftlb σ = 21270 psi = 77.6 inS 3 Max

At 4ft and 2in above the bottom of the beam 8000 −= ftlbM

y =−−= 24.4289.113.8 in σ ()= 24.6 iny = 9450 psi = 2.42 inI 4 Designation Area of T Width thick thick - x-x axis x-x axis x-x axis x-x axis

- A d bf tf tw d/tw I S r y - in2 in in in in - in4 in3 in in WT8x25 7.36 8.13 7.073 0.628 0.380 21.40 42.20 6.770 2.400 1.890

D=8.13 F=7.073 Tf=0.628 Tw=0.38

Calculation of the centroid using bb as reference Ay ( )( )( )+− ( )( )( 751.338.0502.7314.0628.0073.7 ) At the centroid of the beam y ∑ == = 275.1 in ∑ A ()()()()+ 38.0502.7628.0073.7 Calculating Q: To determine Q, cut the cross section horizontally at the NA and choose either the top or the bottom portion, it doesn't matter which because the same answer will result using either portion.

⎛ 227.6 ⎞ 3 VQ ( lb)( 367.76000 ) z yAQ == []()38.0227.6 ⎜ ⎟ = 367.7 in τ == = 2756 psi ⎝ 2 ⎠ web It ()in4 ()38.02.42 Maximum Stresses in Beams My We can obtain the normal and shear stresses from flexure σ −= and shear formulas: I The normal stresses obtained from the flexure formula have VQ their maximum values at the farthest distance from the τ = neutral axis. Ib The normal stresses are calculated at the cross section of maximum bending moment. Normal stresses in a beam of linearly elastic material: (a) side view of beam showing distribution of normal stresses, and (b) cross section of beam showing the z axis as the neutral axis of the cross section.

The shear stress obtained from the shear formula usually have their highest value at the neutral axis. The shear stresses are calculated at the cross section of maximum shear force. In most circumstances, these are the only stresses that are needed for design purposes. However to obtain a more complete picture of the stresses, we will need to determine the principal stresses and maximum shear stresses at various points in the beam. Beams of Rectangular Cross Section Points A and E are at the top and bottom of the beam. Point C is in the midheight of the beam and points B and D are in between. If Hooke’s law applies, the normal and shear stresses at each of these five points can be readily calculated from the flexure and shear formulas. All the elements of vertical and horizontal faces, are in plane stress, because there is no stresses acting perpendicular to the plane of the figure. Points A and E elements are in uniaxial compressive and tensile stresses respectively. Point C (neutral axis) element is in pure shear. Points B and D elements have both normal and shear stresses.

Stresses in a beam of rectangular cross section: (a)simple beam with points A, B, C, D, and E on the side of the beam; (b)normal and shear stresses acting on stress elements at points A, B, C, D, and E; (c)principal stress; and (d)maximum shear stresses. Stress trajectory: Gives the directions of the principal stresses. Stress Contours: Curves connecting points of equal principal stress.

Principal-stress trajectories for beams of rectangular cross section: (a) cantilever beam, and (b) simple beam. (Solid lines represent tensile principal stresses and dashed lines represent compressive principal stresses.) Stress contours for a cantilever beam (tensile principal stresses only). Wide-Flange Beams As for a rectangular beam we identified the point A to E, where A and E are located at the top and bottom of the beam, point C at the neutral axis and points B and D are in the web where it meets the flange. The stresses at these points can be determined using the flexure and shear formulas. They have the same general appearance as in the rectangular beam section but the stresses are different. The largest principal stresses usually occur at the top or bottom of the beam (points A and E) where the stresses obtained from the flexure formula have their largest value. However, depending upon the relative magnitudes of the bending moment and shear force, the largest stresses sometimes occur in points B and D (in the web where it meets the flange). The maximum shear acting directly on a cross section of a wide flange beam stresses always occur at the neutral axis (point C). However, the maximum shear stresses acting on inclined planes usually occur either at the top and bottom of the beam (points A and E) or in the points B and D because of the presence of normal stresses.

A beam AB with a span length L = 6ft supports a concentrated load P = 10800lb acting a distance c = 2ft from the right-hand support (see figure below). The beam is made of steel and has a rectangular cross section (width b=2in and height h = 6in).

Determine the principal stresses and maximum shear stresses at cross section mn located at a distance x = 9in from the end A of the beam. (Consider only the in- plane stresses)

Solution ≤ x ≤ 40 = 9inx ∑ 0 RRF BA =−+⇒= 010800 V()9 = 3600lb V x = 3600 lb ∑M @ B RA ()−⇒= = ()021080060 M ()3240093600 −== inlb M x = 3600 x ()9 A = 3600 B = 7200lbRlbR My My (3240012 − )yinlb σ X −=−= −= −= 900 y Bending stresses: I bh 3 ()()62 inin 3 ( 12 )

Where y has units in inches and σx has units in psi. The stresses calculated are positive when in tension. Note that a positive value of y (upper half of the beam) gives a negative stress, as expected. ⎛ h − y ⎞ 2 VQ ⎛ h ⎞⎜ 2 ⎟ ⎛ hb 2 ⎞ Shear stresses: τ = bQ ⎜ −= ⎟ yy + ⎜ −= y ⎟ ⎝ 2 ⎠⎜ ⎟ 422 Ib ⎝ ⎠ ⎝ ⎠ VQ 12V ⎛ ⎞⎛ hb 2 ⎞ 6V ⎛ h2 ⎞ τ == ⎜ − y2 ⎟ ⎜ −= y2 ⎟ 3 ⎜ ⎟⎜ ⎟ 3 ⎜ ⎟ Ib ()()bbh ⎝ ⎠⎝ 42 ⎠ bh ⎝ 4 ⎠

The figure shows a stress element cut from the side of the beam

at cross section mn. The normal stress σx and the shear stress τxy are shown acting in their positive directions.

The shear stresses τxy acting on the x face of the stress element are positive upwards, whereas the actual shear stresses τ act downward. Therefore ()36006 lb ⎛ (6in)2 ⎞ τ = ⎜ − y 2 ⎟ ()950 −−= y 2 XY 3 ⎜ ⎟ ()()62 inin ⎝ 4 ⎠

In which y has units of inches and τxy has units of psi Calculation of stresses on cross section mn We divide the height of the beam into six equal intervals and label the corresponding points from A to G. Point y (in) σx (psi) τxy (psi) A 3 -2700 0 B 2 -1800 -250 C 1 -900 -400 D 0 0 -450 E -1 900 -400 F -2 1800 -250 G-327000

The normal stresses vary linearly from a compressive stress of -2700psi at the top of the beam (point A) to a tensile stress of 2700psi at the bottom of the beam (point G). The shear stresses have a parabolic distribution with a maximum stress at the neutral axis (point D). 2 Principal Stresses and Maximum ⎛ + ⎞ ⎛ −σσσσ⎞ Shear Stresses ⎜ yx ⎟ ⎜ yx ⎟ ()2 σ 2,1 = ⎜ ⎟ ± ⎜ ⎟ + τ xy ⎝ 2 ⎠ ⎝ 2 ⎠ The principal stresses and maximum shear stresses at each of the seven 2 ⎛ − σσyx ⎞ 2 points A through G may be determined τ MAX = ⎜ ⎟ + ()τ xy ⎜ 2 ⎟ from the following equations: ⎝ ⎠

Point y (in) σx (psi) τxy (psi) σ1 (psi) σ2 (psi) τmax (psi) A 3 -2700 0 0 -2700 1350 B 2 -1800 -250 34 -1834 934 C 1 -900 -400 152 -1052 602 D 0 0 -450 450 -450 450 E -1 900 -400 1052 -152 602 F -2 1800 -250 1834 -34 934 G-3270002700 0 1350