structural and stress analysis second edition pdf

9.2 kN 3.5 kN B The normal force at any section of the beam between A and C is constant and given by NAC= 9.2 kN the vertical 3 kN load has no effect on the normal force.. The complete d

Structural and Stress Analysis

Second Edition

by

Dr T.H.G Megson

S o l u t i o n s t o C h a p t e r 2 P r o b l e m s

S.2.1

(a) Vectors representing the 10 and 15 kN forces are drawn to a suitable scale as shown

in Fig S.2.1 Parallel vectors AC and BC are then drawn to intersect at C Theresultant is the vector OC which is 21.8 kN at an angle of 23.4◦to the 15 kN force

(b) Resolving forces in the positive x direction

F x= 10 + 8 cos 60◦− 12 cos 30◦− 20 cos 55◦= −7.9 kN Then, resolving forces in the positive y direction

F y= 8 cos 30◦+ 12 cos 60◦− 20 cos 35◦= −3.5 kN The resultant R is given by

S.2.3 Initially the forces are resolved into vertical and horizontal components as shown

40 kN 80 kN 69.3 kN

35.4 kN

40 kN 34.6 kN(
1, 1.25)

(0, 0.5)

(1.25, 0.25) (1.0, 1.6)

R y ¯x = 40.0 × 1.0 + 60.0 × 1.25 − 34.6 × 1.0

so that

¯x = 0.81 m The resultant R is then given by

RA= 65 kN

MA= 400 kN mAgain the signs of the reactions are positive so that they are in the directionsshown.

(c) In Fig S.2.4(c) there are horizontal and vertical reactions at A and a verticalreaction at B

RB× 8 + 20 × 5 − 5 × 2 × 9 − 15 × 6 − 10 × 2 = 0which gives

RB= 12.5 kN

Finally, resolving vertically

RA,V+ RB− 10 − 15 − 5 × 2 = 0

(a) The loading on the truss shown in Fig P.2.5(a) produces only vertical reactions at

the support points A and B; suppose these reactions are RAand RBrespectivelyand that they act vertically upwards Then, taking moments about B

RA× 10 − 5 × 16 − 10 × 14 − 15 × 12 − 15 × 10 − 5 × 8 + 5 × 4 = 0which gives

RA= 57 kN (upwards)Now resolving vertically

RB+ RA− 5 − 10 − 15 − 15 − 5 − 5 = 0from which

RB= −2 kN (downwards).

(b) The angle of the truss is tan−1(4/10) = 21.8◦ The loads on the rafters are metrically arranged and may be replaced by single loads as shown in Fig S.2.5.These, in turn, may be resolved into horizontal and vertical components and willproduce vertical reactions at A and B and a horizontal reaction at A.

S o l u t i o n s t o C h a p t e r 3 P r o b l e m sS.3.1 Fig S.3.1(a) shows the mast with two of each set of cables; the other two cables

in each set are in a plane perpendicular to the plane of the paper

F IGURE S.3.1 (a)

A B

C D

314.9 kN

247.4 kN 166.4 kN 74.6 kN

Then,

θB= tan−1

2035



= 29.7◦, θC= tan−1

2025



= 38.7◦, θD= tan−1

2015



= 53.1◦.

The normal force at any section of the mast will be compressive and is the sum of theself-weight and the vertical component of the tension in the cables Furthermore theself-weight will vary linearly with distance from the top of the mast Therefore, at

a section immediately above B,

S.3.2 The beam support reactions have been calculated in S.2.4(a) and are as shown

in Fig S.3.2(a); the bays of the beam have been relettered as shown in Fig P.3.2

9.2 kN

3.5 kN B

The normal force at any section of the beam between A and C is constant and given

by NAC= 9.2 kN (the vertical 3 kN load has no effect on the normal force).

Then

NCD= 9.2 − 3.5 = 5.7 kN

and

NDE= 9.2 − 3.5 − 5.7 = 0 Note that NDEcould have been found directly by considering forces to the right ofany section between D and E The complete distribution of normal force in shown inFig S.3.2(b)

Shear force

The shear force in each bay of the beam is constant since only concentrated loads areinvolved

At any section between A and B,

Since only concentrated loads are present it is only necessary to calculate values of

bending moment at the load points Note that MA= ME= 0

Alternatively, MD= 7.9 × 5 = 39.5 kN m; the difference in the two values is due to

rounding off errors The complete distribution is shown in Fig S.3.2(d)

S.3.3 There will be vertical and horizontal reactions at E and a vertical reaction at

B as shown in Fig S.3.3(a) The inclined 10 kN load will have vertical and tal components of 8 and 6 kN respectively, the latter acting to the right Resolving

horizon-horizontally, RE,H= 6 kN Now taking moments about E

RB× 10 − 2 × 8 × 11 − 8 × 3 = 0which gives

RB= 20 kNResolving vertically

RE,V+ RB− 2 × 8 − 8 = 0

F IGURE S.3.3

Normal force

The normal force at all sections of the beam between A and D is zero since there

is no horizontal reaction at B and no horizontal forces between A and D In CD,

NCD= RE,H= 6 kN (compressive and therefore negative); the distribution is shown

Therefore, when x = 0, SAB= 0 and when x = 5 m (i.e at section just to the left of B),

SAB= 10 kN Also, at any section between B and C a distance x from A

SBC= +2x − RB= 2x − 20 Therefore, when x = 5 m (i.e at a section just to the right of B), SBC= −10 kN and

when x = 8 m (i.e at a section just to the left of C), SBC= −4 kN

Between C and D the shear force is constant and SCD= 2 × 8 − 20 = −4 kN Finally,

between D and E, SDE= 2 × 8 − 20 + 8 = +4 kN (or, alternatively, SDE= +RE,V=+4 kN) The complete distribution is shown in Fig S.3.3(c)

distribution is parabolic and that when x = 0, (dMAB/dx) = 0.

At any section between B and C

MBC= −x2+ RB(x − 5) = −x2+ 20(x − 5)

When x = 5 m, MBC= −25 kN m and when x = 8 m, MBC= −4 kN m The distribution

of bending moment between B and C is parabolic but has no turning value between

F IGURE S.3.4(a)

From symmetry, MA= MD= 0 and MB= MC= WL/4 giving the distribution shown in

Fig S.3.4(c)

S.3.5 The support reactions for the beam have been calculated in S.2.4(b) However,

in this case, if forces and moments to the right of any section are considered, thecalculation of the support reactions is unnecessary

65 kN

15 kN

ve A

Shear force

At any section a distance x, say, from B the shear force is given by

SAB= −15 − 5x

Then, when x = 0, SAB= −15 kN and when x = 10 m, SAB= −65 kN; the distribution

is linear as shown in Fig S.3.5(b)

When x = 0, MAB= 0 and when x = 10 m, MAB= −400 kN m The distribution, shown

in Fig S.3.5(c), is parabolic and does not have a turning value between A and B

S.3.6 Only vertical reactions are present at the support points Referring to

Fig S.3.6(a) and taking moments about C

(b)

5.5 kN m

ve 3.75 m

12.5 kN m

25 kN m (c)

F IGURE S.3.6

At A, where x = 0, SAB= 0 and at B where x = 5 m, SAB= +5 kN.

In BC the shear force at any section a distance x from A is given by

Then, when x = 5 m, SBC= −3.75 kN and when x = 15 m, SBC= 6.25 kN.

In CD the shear force is constant and given by

SCD= −5 kN (considering forces to the right of any section)

The complete distribution is shown in Fig S.3.6(b)

When x = 5 m, MBC= −12.5 kN m and when x = 15 m, MBC= −25 kN m The

distri-bution is parabolic and has a turning value when SBC= 0 (see Eq (3.4)) and, from

Eq (i), this occurs at x = 8.75 m Alternatively, but lengthier, Eq (ii) could be ferentiated with respect to x and the result equated to zero When x = 8.75 m, from

dif-Eq (ii), MBC= −5.5 kN m.

In CD the bending moment distribution is linear, is zero at D and −25 kN m at C Thecomplete distribution is shown in Fig S.3.6(c)

S.3.7 Referring to Fig S.3.7(a) and taking moments about C

In AB the shear force is constant and equal to −RA= −5.6 kN.

At any section in BC a distance x from A

Therefore, when x = 3 m, SBC= 4.4 kN and when x = 6 m, SBC= 7.4 kN.

In CD the shear force varies linearly from zero at D to −1 × 1.5 = −1.5 kN at C The

complete distribution is shown in Fig S.3.7(b)

so that when x = 3, MBC= 16.8 kN m and when x = 6 m, MBC= −0.9 kN m Note that

the bending moment at C, by considering the overhang CD, should be equal to

−1 × 1.52/2 = −1.125 kN m; the discrepancy is due to rounding off errors We see

from Eq (i) that there is no turning value of bending moment in BC and that the

slope of the bending moment diagram at D is zero Also, the value of x at which

MBC= 0 is obtained by setting Eq (ii) equal to zero and solving This gives x = 5.9 m

so that MBC= 0 at a distance of 2.9 m from B The complete distribution is shown inFig S.3.7(c)

S.3.8 The vertical reaction at A is given by (taking moments about B)

RA× 10 − w × 10 × 5 + 10 × 2 = 0

which gives

RA= 5w − 2

The position of the maximum sagging bending moment in AB is most easily found

by determining the shear force distribution in AB Then, at any section a distance x

from A

From Eq (i), SAB= 0 when x = (5w − 2)/w This corresponds to a turning value, i.e.

a maximum value, of bending moment (see Eq (3.4)) Therefore, for x = 10/3 m,

w = 1.2 kN/m The maximum value of bending moment in AB is then

MAB(max) = (5 × 1.2 − 2) ×10

3 − 1.2 ×



10 3

The shear force at any section of the beam between A and B a distance x from A is

occurs at L/3 from the right hand support, i.e when x = 2L/3 Then, from Eq (ii)

n = 43

A point of contraflexure occurs at a section where the bending moment changes sign,

i.e where the bending moment is zero At a section of the beam a distance x from A

the bending moment is given by

Substituting in Eq (iii) for RAfrom Eq (i) and the calculated value of n and equating

to zero gives a quadratic equation in x whose factors are L/3 and −L Clearly the required value is L/3.

S.3.10 Referring to Fig S.3.10(a) the support reaction at A is given by, taking

moments about B

RA× 20 − 5 × 20 × 10 + 20 × 5 = 0from which

RA= 45 kNResolving forces vertically

RB+ RA− 5 × 20 − 20 = 0which gives

When x = 0, SAB= −45 kN and when x = 20 m, SAB= 55 kN

In BC, considering forces to the right of any section, SBC= −20 kN and the completedistribution is shown in Fig S.3.10(b)

S.3.11 The beam and its loading are shown in Fig S.3.11.

Suppose that the maximum bending moment occurs in the bay CD The shear force

in CD, at any section a distance x from A is given by

S.3.12 The beam and its loading are shown in Fig S.3.12(a)

Taking moments about B

RA× 6 −1

2 × 6 × 2 × 2 = 0from which

RA= 2 kN

The shear force at any section a distance x from A is given by

SAB = −2 +

12



xw where w = (1/3)x from similar triangles Then

When x = 0, SAB= −2 kN and when x = 6 m, SAB= 4 kN Examination of Eq (i) shows

that SAB= 0 when x = 3.46 m and that (dSAB/dx) = 0 when x = 0 The distribution is

From Eq (ii) MAB= 0 when x = 0 and x = 6 m Also, from Eq (i), MABis a maximum

when x = 3.46 m Then, from Eq (ii)

MAB(max) = 4.62 kN m.

and the distribution is as shown in Fig S.3.12(c)

S.3.13 The arrangement is shown in Fig S.3.13.

The bending moment at B due to the cantilever overhang is −wa2/2 which is

numer-ically equal to the maximum value of MAB Therefore substituting for RAin Eq (ii)and adding the two values of bending moment (their sum is zero) we obtain

2a2− 4aL + L2= 0

Solving gives

a = 0.29L

Alternatively, as in Ex 3.11, the relationship of Eq (3.7) could have been used Then,the area of the shear force diagram between A and the point of maximum saggingbending moment is equal to minus half the area of the shear force diagram betweenthis point and B

S.3.14 The truss of Fig P.3.14 is treated in exactly the same way as though it were a

simply supported beam The support reactions are calculated by taking moments andresolving forces and are as shown in Fig S.3.14(a)

Bending moment

The bending moment at the supports is zero At B the bending moment is equal

to +60 × 8 = 480 kN m while at C the bending moment is +140 × 4 = 560 kN m; thedistribution is shown in Fig S.3.14(c)

S.3.15 The support reactions have been previously calculated in S.2.5(a) and are as

shown in Fig S.3.15(a)

F IGURE S.3.15

10 kN

30 kN

5 kN A

In AB the shear force is equal to +5 kN

In BC the shear force is equal to +5 + 10 = +15 kN

In CD the shear force is equal to +5 + 10 + 15 = +30 kN.

In DE the shear force is equal to +5 + 10 + 15 + 15 − 57 = −12 kN

In EF the shear force is equal to +5 + 10 + 15 + 15 − 57 + 5 = −7 kN or, more simply,considering forces to the right of any section, −5 − 2 = −7 kN

In FG the shear force is equal to −5 kN while in GH the shear force is zero Thecomplete distribution is shown in Fig S.3.15(b)

S.3.16 In this problem it is unnecessary to calculate the support reactions Also, the

portion CB of the cantilever is subjected to shear and bending while the portion BA

is subjected to shear, bending and torsion as shown in Fig S.3.16

Consider CB

The shear force in CB is constant and equal to +3 kN (if viewed in the direction BA).The bending moment is zero at C and varies linearly to −3 × 2 = −6 kN m at B Thetorque in CB is everywhere zero

Consider AB

The shear force in AB is constant and equal to −3 kN The bending moment varieslinearly from zero at B to −3 × 3 = −9 kN m at A The torque is constant and equal to

6 kN m

3 m

3 kN

6 kN m Equivalent

loading on BA

S.3.17 From Fig P.3.17 the torque in CB is constant and equal to −300 N m In BA

the torque is also constant and equal to −300 − 100 = −400 N m

S.3.18 Referring to Fig S.3.18(a)

1500 Nm

1000 Nm

ve (a)

In BA the torque is constant and equal to 1 × 1000 + 500 = 1500 N m

The complete distribution is shown in Fig S.3.18(b)

S.3.19 From symmetry the support reactions are equal and are each 2400 N m Then,

referring to Fig S.3.19(a)

In CB the torque is given by

TCB= 400 + 2x − 2400 = 2x − 2000 and when x = 0.5 m, TCB= −1000 N m Note that TCB= 0 when x = 1.0 m.

The remaining distribution follows from antisymmetry and is shown in Fig S.3.19(b).The maximum value of torque is 1400 N m and occurs at C and B

S o l u t i o n s t o C h a p t e r 4 P r o b l e m s

S.4.1 Initially the support reactions are calculated Only vertical reactions are present

so that, referring to Fig S.4.1(a), and taking moments about E

RB× 18 − 30 × 24 + 60 × 6 = 0i.e

RB= 20 kN (upwards)

Now resolving vertically,

RE+ RB− 30 − 60 = 0which gives

FE + FK cos θ = 0

i.e

FE + 75 × 0.6 = 0

FE = −45 kN

To check the forces in the members JK and DE take a section cutting JK, KD and DE

as shown in Fig S.4.1(b)

F IGURE S.4.1(b)

and (c)

K KJ

Also sin θ = 4/8 = 0.5 so that θ = 30◦, the remaining angles follow By inspection (or

by resolving vertically) BJ = −15 kN Further, by inspection, FB = 0 All members areassumed to be in tension

CP = −8.7 kN

Resolving parallel to CE,

CE − CA − 10 sin 30◦= 0i.e

PE = −PC = 8.7 kN

Resolving horizontally,

PF + PE cos 60◦− PC cos 60◦− PA = 0i.e

FE = −FHResolving horizontally,

GH = −8.7 kN

Resolving parallel to GE,

GJ − GE − 10 cos 60◦= 0i.e

Joint H:

Resolving perpendicularly to HJ,

HE cos 30◦+ HG cos 30◦= 0

HE = −HG = 8.7 kN

Resolving parallel to HJ,

HJ − HF + HG cos 60◦− HE cos 60◦= 0i.e

JG × 0.5 + 26.0 × 0.866 − 15 = 0

JG = −15.0 kN

Substituting for JG in Eq (i) gives

GE = −20.0 kN

S.4.3 From Fig P.4.3, by inspection, there will be a horizontal reaction of 4 kN acting

to the right at A Taking moments about B,

RA,V× 12 − 40 × 10 − 40 × 8 = 0i.e

RA,V= 60 kNResolving vertically,

RB+ 60 − 40 − 40 = 0i.e

RB= 20 kNFirst take a section cutting the members EG, EH and FH as shown in Fig S.4.3(a)

Note that tan θ = 1.5/2 so that θ = 36.9◦

F IGURE S.4.3

EG EH FH

Resolving vertically,

EH sin 36.9◦+ 20 = 0i.e

FH + EH cos 36.9◦+ EG = 0i.e

S.4.4 There will be horizontal and vertical reactions at A and a vertical reaction at B.

Then, resolving horizontally,

RA,H− 2 × 6 cos 30◦= 0

RA,H= 10.4 kN

Taking moments about B,

RA,V× 12 − 5 × 36 × 6 − 6 × 3 = 0i.e

RA,V= 91.5 kN

From symmetry at K, the forces in the members KE and KG are equal Then, ing they are tensile and resolving vertically,

assum-2 × KE cos 30◦+ 36 = 0i.e

S.4.5 The equation of the parabola which passes through the upper chord points

is y = kx2 When x = 9 m, y = 7 m so that 7 = k × 81 which gives k = 0.086 At A, when x = 3 m, y = 0.086 × 9 = 0.77 m so that BA = 7 − 0.77 = 6.23 m and when x = 6 m,